The joint Dickman law for consecutive integers

Proof. We prove convergence by computing all mixed factorial moments of the bin counts. We first record the elementary prime estimates used in this calculation. Put \vartheta(z)=\sum_{p\le z}\log p. For z\ge2 we have

Indeed, the primes in (m,2m] divide \binom{2m}{m}, giving \vartheta(2m)-\vartheta(m)\ll m. Dyadic summation proves the first estimate, and separating the primes at \sqrt z proves the second. Moreover,

because the terms with \nu\ge2 are O(\sqrt z\log z). Expanding \log n over prime powers now gives

The floor errors use the preceding prime-power bound, and the terms with \nu\ge2 use \sum_p\sum_{\nu\ge2}(\log p)/p^\nu<\infty. Integral comparison on the left proves the third estimate in (2.7). Partial summation yields

as well as \sum_{p\le z}1/p=\log\log z+O(1).

The number of integers n\leq\beta x divisible by p^2 for some prime p>x^{1/J} is at most

This remains negligible relative to the size of any one fixed progression in the lemma. We may therefore replace multiplicity counts by counts of distinct primes.

For the mixed factorial moments of these distinct-prime counts, write (b)_r=b(b-1)\cdots(b-r+1) and (b)_0=1. Fix nonnegative integers r_1,\ldots,r_{J-1}, and let r=r_1+\cdots+r_{J-1}. The case r=0 is immediate. For r>0, a term in the factorial-moment expansion selects, in order, r_k distinct primes from bin k, for each k. Denote their product by d; only d\leq\beta x can contribute. The number of such selections is O_{J,\beta,r}(x/\log x). To see this, fix every selected prime except one, and write d_1 for their product. If the last prime has any admissible choice, its upper bound \beta x/d_1 exceeds x^{1/J}. The prime-counting upper bound therefore gives at most

choices. Summing this bound over the other primes costs a bounded product of reciprocal prime sums over the bins. We may drop distinctness and product restrictions when taking this upper bound.

For sufficiently large x, every selected prime is coprime to q. The Chinese remainder theorem then gives

The sum of the errors over all selections is o(x). After normalization, the factorial moment is consequently the reciprocal sum over the selections with d\leq\beta x, up to o(1).

In the variable v=\log p/\log x, (2.8) says that the reciprocal prime measure on bin k converges to dv/v on (k/J,(k+1)/J]. The product restriction is

The limiting product measure is absolutely continuous, so the boundary \sum v=1 has measure zero. Repeated selections do not change the limit: their reciprocal contribution is bounded by a constant times \sum_{p>x^{1/J}}p^{-2}=o(1). Returning from distinct-prime counts to multiplicity counts also costs o(1) in the normalized moment, because the exceptional set has size o(x) and the counts are bounded.

List the selected bin indices as k_1,\ldots,k_r, with r_k occurrences of k. We have proved the explicit limit

For r=0, both sides are one. The right-hand side depends only on J and the multi-index (r_k). All count vectors lie in a fixed finite set, and mixed factorial polynomials span the functions on that set. The asserted distributional convergence follows.

Taking the expectation of the function (b_k)\mapsto\prod_k\zeta_k^{b_k} gives (2.4); using the vector (\xi_k) gives the mean \mu_g. Finally, if every prime factor of u is at most x^{1/J}, then f_x(u)=g_x(u)=1. Complete multiplicativity of f_x and g_x proves (2.6).□

Proof. Put y=x^{1/J} and \sigma=1+1/\log x. For large x we have y<X. Since the summands defining the distance are nonnegative, we can restrict to p\leq y and delete E and the primes dividing the modulus of \chi. Deleting those finitely many primes costs only a constant in the lower bounds below.

We record a prime-sum comparison, uniform in any factors of absolute value at most one. Replacing the weight 1/p on p\leq y by p^{-\sigma} on all primes has absolute error O_J(1). Below y, the error is at most

For the tail, partial summation and the prime-counting bound in (2.7) give

The prime-power terms in the logarithm of either a zeta or a Dirichlet L Euler product are also uniformly O(1) for \sigma>1.

For |t|\leq1 it follows that

Here we used only an upper bound for the logarithm: a nonprincipal Dirichlet L-function has no pole at 1 and is bounded on the compact region under consideration. Since (2.8) and the accompanying partial-summation estimate give \sum_{p\leq y}1/p=\log\log x+O_J(1), this proves a lower bound \log\log x-O_{J,\chi,E}(1) for this frequency range.

For |t|>1, let m be the order of \chi on the units. The elementary inequality |1-z^m|\leq m|1-z|, for |z|=1, gives

The Vinogradov–Korobov bound, with any fixed logarithmic exponent strictly between 2/3 and 1, gives

The bound on the line 1 follows from [8, (1.2)]. Its extension to the right can be seen by applying the Phragmén–Lindelöf principle in 1\leq\sigma\leq2 to

The pole at 1 is removed, the logarithmic power has an analytic branch in this strip, and both vertical boundaries are bounded. For |\operatorname{Im}s|\geq1, the removed rational factor is bounded away from zero, which gives (2.17). For \sigma\geq2 the same estimate follows from absolute convergence. Applying the prime-sum comparison just proved, we obtain uniformly for 1<|t|\leq CX,

Combining the two frequency ranges proves

which implies (2.16).□

Proof of Lemma 2.3. Teräväinen [26, Section 4, proof of Theorem 1.11] uses real multiplicative generating functions for large-prime counts and recovers event coefficients from them. Here the finite count range gives a pointwise tensor interpolation of the centered function of all bin counts by completely multiplicative functions. We prove this interpolation with coefficients independent of B,x. Choose J+1 distinct numbers in (0,1]. The Vandermonde matrix whose entries are their powers of orders 0,\ldots,J is invertible. Taking tensor products over the J-1 count coordinates shows that there are finitely many complex constants c_\ell and vectors \boldsymbol z_\ell=(z_{\ell,1},\ldots,z_{\ell,J-1})\in(0,1]^{J-1} such that

The coefficients and vectors depend only on the fixed labels and J. Define

Each h_{\ell,x} is real, nonnegative, 1-bounded, and completely multiplicative. On every interval m\leq C_Bx with C_B fixed for fixed B, the identity F_x(m)=\sum_\ell c_\ell h_{\ell,x}(m) holds for all sufficiently large x.

We next prove the needed estimate for each fixed residue class of the integer argument. For a class r'\pmod D, put d=\gcd(r',D) and q=D/d, and write m=dk. Then m\equiv r'\pmod D is equivalent to k\equiv r'/d\pmod q, a unit class modulo q. For all sufficiently large B, no prime factor of d belongs to P_B, so G_{B,v}(dk)=G_{B,v}(k). For fixed B and sufficiently large x, (2.6) also gives F_x(dk)=F_x(k), and the same equality holds for every interpolant h_{\ell,x}.

By character orthogonality, the restricted k-sum is a fixed linear combination of sums of

All these functions are 1-bounded and multiplicative. The sum length after division by d is H_B=L_B/d. Normalization by L_B is d^{-1} times normalization by H_B. Changing integer endpoints introduces an error O_D(L_B^{-1}).

Consider first the principal character \chi_0\pmod q. The functions in (2.19) are then real, so Theorem 2.4 applies to each interpolant. Although their individual long means need not vanish, their centered linear combination does. More precisely, on any dyadic interval [X,2X] with X=c_Bx and c_B>0 fixed for fixed B,

To justify the last limit, split the sum into residue classes modulo the fixed common period of G_{B,v} and \chi_0, and apply Lemma 2.1 to every class. The number and the sizes of these residue classes may depend on B; they are fixed in this x-limit. The mean square of the centered short average thus has inner upper limit at most a fixed multiple of r(H_B), where the multiple depends on the interpolation coefficients and D, but not on B.

For a nonprincipal \chi, the function in (2.19) equals \chi(p) at all primes p\leq x^{1/J} outside the finite set P_B. Lemma 2.6 therefore applies for each fixed B, with X=c_Bx and the frequency range |t|\leq X. For all sufficiently large B we have H_B\geq10, so Theorem 2.5 applies once x is sufficiently large. Taking the inner upper limit in (2.15) leaves at most a constant times (\log\log H_B/\log H_B)^2. This tends to zero as B\to\infty. There are only finitely many interpolants and characters, so the same holds after their linear combination.

Finally, for a given origin n, the condition i\equiv r\pmod D in (2.12) is the condition m=n+i\equiv n+r\pmod D. There are only D possible argument residues r', and it suffices to sum the bounds just proved for these possibilities. After m=dk, the short-interval origin is (n+a_B)/d. For fixed B and all sufficiently large x,

Choose a dyadic cover of the fixed scaled interval [s_0T/(2d),2s_1T/d]. It gives intervals [c_{B,j}x,2c_{B,j}x] with every c_{B,j}>0 fixed in the inner x-limit, and with a number of intervals bounded in terms of s_0,s_1,D independently of B. Their scales are comparable to Tx/d. Mapping the origins to their integer parts has multiplicity at most d+1, and replacing an origin by its integer part costs O(H_B^{-1}). The integer-origin version of the preceding estimates consequently applies. Normalizing by Tx instead of the length of each dyadic range changes only fixed factors. Since H_B=L_B/d\to\infty for every d\mid D, this proves (2.12).

All assertions used an unrestricted upper limit as x\to\infty; restricting that upper limit to a subsequence preserves them.□

Proof. The first statement is the classical fixed-order Selberg–Delange expansion; see [11, Theorem 1] and [14, Theorem 13.2]. The normalization and the moving roughness cutoff required here are recorded explicitly below. In \Re s>1,

Define the local powers by their power-series logarithms. The logarithm of each local factor of G_z is O(p^{-2\Re s}). Consequently G_z is analytic, nonzero, and bounded on \Re s\ge0.9, with all derivatives bounded on fixed smaller compact sets. Near s=1, put u=s-1. The analytic factor

has a convergent Taylor series and has value G_z(1) at u=0. Truncated Perron inversion, followed by a contour around the cut to the left of 1, integrates its jth Taylor term against Y e^{u\log Y}u^{j-z}. Hankel’s reciprocal-gamma formula gives the factor (\log Y)^{z-j-1}/\Gamma(z-j). Taylor’s remainder, integrated on that contour, is O_H(Y(\log Y)^{z-H-2}). For completeness, the contour may have right edge 1+1/\log Y, height \exp(c\sqrt{\log Y}), left edge 1-c'/\sqrt{\log Y}, and small circular part of radius 1/\log Y. The classical zero-free region for \zeta permits fixed sufficiently small c,c'>0; the other contour pieces and the Perron truncation error are O_H(Y\exp(-c''\sqrt{\log Y})(\log Y)^{O_H(1)}) and are absorbed in the displayed remainder. The local expansion just described also identifies c_0(z)=G_z(1)/\Gamma(z).

We spell out the character uniformity. The twisted series is

The same cancellation of the linear local term bounds G_{z,\chi} uniformly in \chi on \Re s\ge0.9. Write L_Y=\log Y and \Theta=\exp(c\sqrt{L_Y}), and use the rectangle

The classical Dirichlet zero-free region excludes zeros in this rectangle except possibly a real zero of a real primitive character inducing \chi. Siegel’s bound states that for every fixed \varepsilon>0 such a zero obeys 1-\beta\gg_\varepsilon d^{-\varepsilon}; see [14, Theorems 12.3 and 12.10] for the zero-free region and exceptional-zero bound. Choose \varepsilon<1/(2C). Since d\le L_Y^C, this distance is eventually larger than c'/\sqrt{L_Y}, uniformly in the allowed moduli. The Euler factors removed for imprimitive characters have no zeros in \Re s>0. Thus an analytic logarithm of L(s,\chi) exists throughout the rectangle, agreeing with its Euler-product logarithm on its intersection with \Re s>1.

There is also a uniform polynomial bound in L_Y on this rectangle. For s=\sigma+it truncate the Dirichlet series at V=\lceil d(2+|t|)\rceil. The periodic character sums have modulus at most d, so partial summation bounds the tail by O(d(1+|s|)V^{-\sigma}). The initial segment is at most O((1+\log V)V^{\max(1-\sigma,0)}). Since \log V\ll_C\sqrt{L_Y} and (1-\sigma)\log V=O_C(1) in the rectangle, these estimates bound L(s,\chi) by a fixed power of L_Y. Because z is real, the modulus of the chosen analytic power is |L(s,\chi)^z|=|L(s,\chi)|^z.

The coefficients have modulus at most one. At an endpoint Y\in\mathbb Z+\tfrac12, truncated Perron therefore has error O(Y(1+\log Y)^2\exp(-c\sqrt{L_Y})); this follows also by summing its usual \min(1,(\Theta|\log(Y/n)|)^{-1}) error. Shift the Perron contour to the left edge. There is no pole or branch cut in the twisted case. The new vertical integral is bounded by Y\exp(-c'\sqrt{L_Y}) times a fixed power of L_Y, and the horizontal integrals have the additional factor \Theta^{-1}. This proves an exponential square-root-log saving and hence the claimed saving of any fixed logarithmic power. For arbitrary Y\ge3, let Y' be the least element of \mathbb Z+\tfrac12 with Y'\ge Y. Then 0\le Y'-Y<1 and d\le(\log Y)^C\le(\log Y')^C. Replacing Y by Y' changes the sum by at most one term and any smooth main term by an amount absorbed in the stated errors. This gives both estimates for arbitrary real Y.□

Proof. Define the multiplicative function h_z by h_z(p^e)=(-z)^e for p\le P_0 and e\ge1, and by h_z(p^e)=0 for p>P_0 and e\ge1. Its Euler factors give the exact identity g_{z,P_0}=g_z*h_z. If \epsilon_0=1/\log P_0, then

Indeed p^{\epsilon_0}\le e in the product, its logarithm is bounded by a constant times \sum_{p\le P_0}p^{-1+\epsilon_0} plus a bounded sum of square terms, and Mertens’ estimate bounds that first sum by O(\log\log P_0). Increasing the absolute constant C_0 if necessary gives the displayed assertion for both values of z. Since (\log v)^k\le k!\epsilon_0^{-k}v^{\epsilon_0}, it also gives

Put L_Y=\log Y. In the convolution sum the terms v>\sqrt Y contribute, after division by Y, at most

because the inner untwisted or twisted sum is at most Y/v in absolute value. For v\le\sqrt Y, apply Lemma 3.1 at Y/v. For the untwisted sum expand each factor by Taylor’s formula:

This remainder is uniform for 0\le\log v\le L_Y/2. After summing with h_z(v)/v, the total Taylor and Selberg–Delange remainders are at most

The moment sums in the polynomial coefficients may be extended to all v. To see that the resulting error is negligible, combine (\log v)^k\ll_k\epsilon_0^{-k}v^{\epsilon_0/2} with the preceding Rankin bound: the tail of each such moment is O_k((\log P_0)^{C_0+k}\exp(-L_Y/(4\log P_0))). Thus explicitly

The moment bounds prove the claimed coefficient bounds. Since \log P_0=1000\log B and L_Y\ge B^{0.89}, choosing the fixed integer H sufficiently large makes all these errors O_D(YB^{-D}). The zeroth moment is \prod_{p\le P_0}(1+z/p)^{-1}. Combining it with the Euler product for c_0(z) gives

The second product tends to one, and Mertens’ product formula proves its stated asymptotic and positivity.

For a nonprincipal character, use the same convolution with h_z(v)\chi(v) and the second estimate of Lemma 3.1. On v\le\sqrt Y we have \log(Y/v)\ge L_Y/2 and, for all large B, d\le B^{15}\le(\log(Y/v))^{18}. Take an arbitrarily large fixed saving exponent in that lemma. The factor \sum_v|h_z(v)|/v\ll(\log P_0)^{C_0} is absorbed by increasing that exponent. The terms v>\sqrt Y have already been bounded independently of \chi. This proves the required uniform character estimate.□

Proof. Since P_0>B^{15}, all rough integers are units for every modulus in the statement. Character orthogonality and Lemma 3.2 give, for each unit class,

There is no loss of a factor \varphi(d) here: it is cancelled by the normalization in character orthogonality. Partial summation, multiplication by (\log P_0)R^{1/2} when z=1/2, and summation over the unit residues against e(ur/q) now prove (3.2). The sum of the latter phases is the Ramanujan sum \mu_{\rm Mob}(q) because (u,q)=1. For explicit error accounting, summation over residues costs at most q, the normalizing factor is at most B eventually, and differentiating w_3(t)e(\xi t) costs at most a constant times (1+|\xi|)(\|w_3\|_\infty+\|w'_3\|_1+\|w_3\|_1) on the fixed support. Thus the initial B^{-200} saving exceeds all these losses by much more than the claimed B^{-50}. On that support \log(Xt) is in the range of the preceding lemma once B is sufficiently large.

For a squarefree product n of the allowed primes, independence gives the exact mass

On n\le\exp(3.3B) the last product is 1+O(B/P_0), uniformly: its logarithm is O(\omega(n)/P_0)=O(B/P_0). In this range every rough prime factor is below the upper cutoff \exp(4B). Hence the mass without that last product is exactly Q_{z,B}g_{z,P_0}(n)/n. Partial summation of the unit-class formula between any two endpoints e^{Ba},e^{Bb}, with B^{-1/10}\le a\le b\le3.2, gives

Indeed each endpoint error is O(B^{-200}), and integrating the error O(tB^{-200}) against dt/t^2 over a log interval of length O(B) costs O(B^{-199}); also Q_{z,B}\le1. The O(B/P_0) relative mass correction contributes at most O(B/P_0) in total, since the uncorrected masses are bounded above by the true probabilities. Endpoint atoms have exponentially small mass because n\ge\exp(B^{0.9}). This proves (3.3) for arbitrary endpoint conventions and arbitrarily short or partial intervals. No relative error for a short interval is asserted or needed.

If \delta<1/R, the event s\le\delta forces the product to be empty, so its probability is Q_{z,B}\ll R^{-z}. If 1/R\le\delta\le1, the event forces all primes with \log p>\delta B to be absent. Mertens’ product estimate therefore bounds its probability by

These estimates prove (3.4), including \delta=0. Finally, on a squarefree rough n in the range of (3.5),

uniformly, and also for n=1. This proves (3.5); outside the squarefree rough support both masses vanish.□

Proof. At a squarefree modulus d, the one-dimensional forbidden intersection has at most k^{\omega(d)} residue classes, each counted with error O(1). Its remainder is therefore O(k^{\omega(d)}). In the rectangle there are at most k^{\omega(d)}d residue pairs: at each prime there are at most kp pairs, and the Chinese remainder theorem multiplies these bounds. Each residue pair has count

Thus its remainder is O(k^{\omega(d)}(N_1+N_2+d)). For fixed k, \sum_{d\le D}k^{\omega(d)}\ll_k D(1+\log D)^{O_k(1)}; one may obtain this by bounding k^{\omega(d)} by a fixed divisor function and applying the elementary hyperbola bound to its sum. Use the preceding upper sieve with D=N^{1/2} in an interval, or D=N_*^{1/2} in a rectangle, and choose c_k<1/(2s_k). The interval remainder is N^{1/2+o(1)}. The rectangle remainder is at most

for sufficiently large N_*. Enlarge the constants to cover smaller lengths.

The preceding sieve application temporarily omitted the primes p\le P_k. For each unweighted choice of forbidden conditions, restore the factors of those primes whose restrictions the application retains. If one such prime forbids the whole residue space, the fully sifted count is zero. Otherwise its surviving fraction 1-\theta_p is at least 1/p in an interval and at least 1/p^2 in a rectangle. The product of the reciprocals of these fractions over the bounded initial set is therefore bounded in terms of k. Enlarging C_k restores all their factors in the displayed main term, without changing the remainder. This argument is uniform when conditions coincide and when Z includes only part of the initial set. Primes that an application chooses to omit remain absent from both its restrictions and its product.

For the weighted assertion, at every prime independently choose whether to forbid each specified condition, using probability one minus its local weight. Make these choices independently between conditions too, even if some coincide. For a fixed integer point, its probability of avoiding all the randomly chosen conditions is exactly the product of its local weights. Apply the unweighted bound for each choice and take expectations. The constants and remainders are uniform because every choice has at most k residue conditions or proper lines. Independence between primes makes the expectation of the local-density product the product of the expected local densities. The latter is precisely the residue average stated in the lemma. The initial-prime factors were restored before this averaging, so no lower bound for an averaged local factor is needed. This proves the weighted version, including coincident conditions. Squarefreeness conditions and all factors in [0,1] above Z may be dropped when applying an upper bound.□

Proof. Choose a sufficiently small fixed c>0 and sieve up to Z=\exp(cB), as permitted by Lemma 3.4 throughout the fixed log-size range. For a single form, local weight zero at p\le P_0 and local weight t\in\{1/2,1/4\} above P_0 give residue averages 1-1/p and 1-(1-t)/p, respectively. Their product is

Dropping squarefreeness only increases the sum. Put H_B=(\log P_0)R^{1/2}. Multiplication by H_B for the first moment gives one. Multiplication by H_B^2 for the second moment gives

The sieve remainders remain negligible after these polynomial factors. The omitted primes above Z up to any of the fixed log-size endpoints have bounded reciprocal sum; in particular using this fixed small c does not alter any power of R in the bound.□

Proof. Use the same sieve limit Z=\exp(cB) with c sufficiently small. For (3.7), at p\nmid jb the roots of c and b+jc are distinct. At a prime at most P_0 the local density of avoiding them is 1-2/p; at a larger prime the residue average of the two weights 1/2 is 1-1/p. Apart from a factor 1+O(p^{-2}), these are the squares of the single-form factors in the preceding proof. Bounded small primes may be omitted. If p\mid j, then p<P_0 and b is a unit at p. Only the root of c is forbidden. Its loss compared with two ordinary roots is at most 1+O(1/p) outside the omitted bounded set. The product of these losses is covered by \Sigma(j) after fixing C_2 sufficiently large. If p\mid b, then p>P_0 and we may drop both local conditions; their reciprocal sum is

Thus these exceptional primes have bounded total cost. Lemma 3.4 now gives density at most C\Sigma(j)(\log P_0)^{-2}R^{-1} in an interval of length comparable to X. Its normalization H_B^2 proves (3.7).

For (3.8), use the rectangle in (b,c) of area comparable to TX^2. At p\nmid j, the equations b=0, c=0, and b+jc=0 are three distinct lines. Their pairwise and triple intersections have density p^{-2}. Inclusion–exclusion, also with the local reducing weights, therefore gives the product of the three single-form averages up to 1+O(p^{-2}). At p\mid j the lines for a and b coincide; since p<P_0, the loss from three ordinary rough exclusions to two is at most 1+O(1/p). Again their product is covered by \Sigma(j). The rectangular sieve gives density at most C\Sigma(j)(\log P_0)^{-3}R^{-3/2}, which is cancelled by H_B^3. All sieve remainders are negligible because X is exponential in B, whereas T, the normalizations, and the possible singular losses are polynomial in B. Extending a support to a rectangle or interval for the sieve is harmless: the local congruence restrictions continue to hold on the original positive support, and the extension is used only for an upper bound.□

Proof. We give direct upper bounds relative to the scales in the statement; no division by the actual mass of a weighted sum is used. Sieve with the fixed Z=\exp(cB) used in Lemmas 3.5 and 3.6. For a coefficient whose count is being tested, replace its local weight 1/2 on a subset \mathcal Q\subset\mathcal P by (1/2)e^s or (1/2)e^{-s}, where s>0 is fixed and small enough that (1/2)e^s<1. The random-root sieve still applies. At ordinary primes the logarithm of its product, relative to the unmodified single-form factor, changes by

The O(1) is uniform for small fixed s. In the three-form rectangle, line intersections alter it by a convergent sum of O(p^{-2}); the factors at primes dividing j are still bounded by \Sigma(j), independently of such s. The one-form case and independent two-variable case have the same conclusion with their respective scale bounds. All modified local weights lie in [0,1], so dropping those above Z preserves the direction of the bound, even for the positive tilt.

For a prefix g<1, take \mathcal Q=\{p\in\mathcal P:\log p\le B^g\}; for g=1 take all of \mathcal P. Mertens’ estimate gives

For example, for g<1 the left side is g\log B-\log\log P_0+O(1) once B is large, and its difference from g\ell is O(\log\log B)=o(\ell). There are only fixed finitely many grid points. The exponential Markov factor for a count exceeding (g/2+\tau)\ell is e^{-s(g/2+\tau)\ell}; the tilted sieve therefore bounds its weighted contribution by the appropriate base scale times

For a count below (g/2-\tau)\ell the corresponding bound is

If the latter threshold is negative the event is empty. Otherwise choose s sufficiently small in terms of the fixed \tau; since both exponential remainders in brackets are O(s^2) and g\le1, both brackets are at most -s\tau/2. Thus every prefix violation saves a fixed positive power of B. A union bound over the grid and the at most three coefficients contributes o(1) times the scale bound.

For a tail endpoint Y=Y_i<B, use the negative tilt on \mathcal Q=\{p\in\mathcal P:\log p>Y\}. Uniformly in these endpoints,

When Y\le cB this is Mertens’ estimate with upper endpoint e^{cB}. When cB<Y<B the sum is zero and \log(B/Y)=O_c(1), which gives the same formula. Exponential Markov at the threshold 0.4\log(B/Y)-C_* gives, relative to the appropriate base scale,

Now choose a small absolute s>0, independently of the grid. The bracket is -0.1s+O(s^2) and is negative. With a_s=\frac12(1-e^{-s})-0.4s>0, the sum of these bounds over the dyadic endpoints Y<B is

For Y\ge B the required lower threshold is nonpositive, so there is no failure. Taking c_3=s proves the asserted tail loss.

The sieve remainders remain negligible in each application: their savings are powers of intervals exponential in B, whereas normalization and Markov factors are only fixed powers of B, and there are O(\log B) tail endpoints. For C_*\ge0, the Markov factors involving C_* are at most their values at zero. Thus the residual function \epsilon_B can be chosen independently of C_*. The prefix factors depend only on the fixed grid and \tau. This proves (3.11) and its one- and two-weight variants.

Finally let X_p be the independent indicators in the probability statement, with parameters \lambda_p=1/(2p)+O(p^{-2}). For either sign and any subset \mathcal Q,

Its logarithm is (e^{\pm s}-1)\sum_{p\in\mathcal Q}\lambda_p+O(\sum_p\lambda_p^2). The latter error is bounded uniformly, while the parameter sums are g\ell/2+o(\ell) for a prefix and \tfrac12\log(B/Y)+O(1) for a tail with Y<B. Omitting E_B changes any parameter sum by at most O_{C_E}(B/P_0)=o(1), uniformly for fixed C_E. Apply exactly the two Markov calculations above, now to this product moment-generating function and with no sieve remainder. This proves all the probability assertions.□

Proof. For Haar distributed w, let

At each prime these are disjoint hits, each of probability 1/p. Their joint law differs by O(1/p^2) in total variation from two independent Bernoulli(1/p) indicators. Independence across primes therefore couples them to the independent sets S_1,S_2 above, with total failure probability

The probability that any p\in\mathcal P has p^2\mid w or p^2\mid w+1 is also O(1/P_0).

On the complement of these exceptional events, coefficient divisors are subset products of the corresponding site sets, and the prime set of each quotient is exactly the complementary subset. By (3.1), a divisor and its complement then have untruncated weight B2^{-|S_\nu|}. The truncated weight inserts exactly the regularity indicators for those two subsets. Summing the two fair splits shows that the value of (4.1) is precisely (4.5) on this successful coupling.

We can discard the exceptional events in both moments. Any nonzero representation in (4.1) forces each entire site to have at most (1+2\tau)\ell distinct primes, by the total regularity cutoffs for the coefficient and quotient. This remains true in the presence of squares: their two prime sets still cover the site’s distinct primes, although they need not be disjoint. Hence there are at most 2^{2(1+2\tau)\ell} coefficient choices across the two sites. Combining this with (3.9) gives, for example, the uniform bound D_B\ll B^4. Together with \widetilde D_B\ll B, the coupling and site-square errors change its first two moments by at most O(B^8/P_0)=o(1). This proves the comparison.□

Proof. If the intersection is empty there is nothing to prove. Otherwise enclose it in an interval [v,v+H] with v\ge Y/2, and set X=e^v. The probability of a squarefree product n of the permitted primes is

Set w_Y(n)=0 for other integers. Since Y/2\le U\le Y and U\ge L, Mertens’ estimate gives

Let c_{\mathrm{s}}>0 be an admissible exponent for the one-dimensional upper sieve of Section 3, with a remainder O(N^{.8}) on intervals of length N. Choose a fixed c>0 sufficiently small that c\le1/4 and 2c<c_{\mathrm{s}}, and put Z=e^{cY}. Then Z\le e^U and, since X\ge e^{Y/2}, Z\le X^{c_{\mathrm{s}}}. Dropping squarefreeness, the restriction on prime factors greater than Z, and the reducing weights above Z majorizes w_Y(n) by

Apply the random-weight form of that sieve on an interval containing [X,e^H X] and of length comparable to X, with constants depending only on H. It yields

where

If Z\ge P_0, the relation 1-a_p=3/4+O(1/p) and Mertens’ estimate imply

Consequently q_Y\mathcal D\ll_c1/Y. If Z<P_0, ordinary rough exclusion instead gives \mathcal D\ll1/\log Z=1/(cY), and the same conclusion follows from q_Y\le1. Finally,

Using 1/n\le1/X on the summation interval proves

as required.□

Proof of Lemma 4.1. By Lemma 4.2, it suffices to prove the asserted mean and second-moment bounds for \widetilde D_B.

The first moment. Without the regularity indicator in (4.5), the products a,c are independent and each has law \nu_{1/2}. Let f_B=f_{1/2,B} and f(s)=c_{1/2}s^{-1/2} on the compact positive log ranges under consideration. For fixed s=\log c/B on the support of \rho_0, (3.3) and partial summation give, uniformly there,

To justify the use of (3.3), the relevant interval in \log a/B has length O(1/B); its absolute distribution-function error is O(B^{-80}), and partial summation against the smooth window, whose total variation is bounded, leaves an error o(1/B) before the displayed multiplication by B. Since \log T/B\to0 and f_B converges uniformly with derivatives on these compact intervals, another application of (3.3) to c shows that the untruncated first moment tends to

The same calculation, or its upper-bound version for fixed intervals containing the supports, shows that the size-and-ratio event \mathcal H_1 has probability O(1/B) under the independent coefficient-product law.

We next control the regularity losses in this normalization. Cover the coefficient support by O(B) dyadic-size boxes c\asymp X, a\asymp TX. All their logarithmic scales lie in a fixed compact subinterval of the ranges for the arithmetic estimates. From (3.5), the joint product probability at (a,c) is bounded by

Within one such box ac\asymp TX^2. The independent-coefficient version of (3.11) bounds the untruncated weighted sum where either coefficient is nonregular by

After the harmonic denominators, summing the O(B) boxes, and multiplying by the B in (4.5), the coefficient-regularity loss is therefore o(1)+O(e^{-c_3C_*}).

Conditional on a selected coefficient subset at a site, the remaining indicators at primes not selected for that coefficient are independent, with probabilities

The selected coefficient primes have reciprocal sum O(B/P_0), uniformly on the coefficient support, because the logarithm of their product is O(B). Thus the independent-indicator regularity estimate of Section 3 applies uniformly after these primes are omitted. The conditional chance of a remaining subset being nonregular is o(1)+O(e^{-c_3C_*}). Multiplying by the bounded smooth weights and the coefficient-event probability O(1/B) shows that the remaining-subset loss in (4.5) is again o(1)+O(e^{-c_3C_*}) after its outside factor B.

It follows that

where the error constant in the exponential term does not depend on C_*. This proves the asserted positive lower bound after choosing C_0 large, for instance with a fixed d_0<d_*/2 and then allowing the negligible coupling error. The upper bound \int D_B\ll1 follows by omitting all regularity indicators in the independent model. Both bounds can use constants independent of C_*. For the lower bound one may also note that increasing C_* only relaxes the tail cutoffs.

The second moment. We use the two-split notation introduced before Lemma 4.3. By (4.6), the required bound is the global probability estimate stated there.

Compare the two assignments of each site prime. If there is a change, its largest logarithm lies in one of the disjoint intervals (Y/2,Y], where

Here 4B\le2^{i_{\max}}\log P_0<8B for large B. Write \mathcal A_Y for the event that this is the largest changed interval. Every change in it is either a move from remaining to coefficient or a move in the reverse direction. The joint law and \mathcal E_1\cap\mathcal E_2 are invariant under exchanging the splits. Hence

where \mathcal U_{Y,\nu} requires at least one prime at site \nu with logarithm in (Y/2,Y] to move from the first remaining set to the second coefficient set. This orientation inequality has been taken in the original symmetric two-split law.

For the upper bound on its right side, couple the four first-split sets \mathcal C_\nu^{(1)},\mathcal R_\nu^{(1)} to four mutually independent prime processes, each with inclusion probabilities 1/(2p). At a prime the two classes at one site were mutually exclusive, so the cost of this coupling is O(1/p^2), and the total cost is O(1/P_0). Assign independent fair second-split coins to each occurrence in these processes. On the successful, collision-free coupling this constructs exactly the original second split. On the exceptional event there may be two occurrences of a prime, for which we still assign independent coins and define coefficient products with multiplicity. This provides a convenient coupled model for upper bounds. Even if the coupling error is counted separately in every one of the O(\log B) intervals, its contribution after multiplication by B^2 is O(B^2\log B/P_0)=o(1). No uniform coupling assertion conditioned on a rare coefficient value is needed.

Work now in that independent model. Its first coefficient products have independent \nu_{1/2} laws, so the first-moment calculation gives \Prob(\mathcal H_1)=O(1/B). Fix first coefficient sets satisfying \mathcal H_1 and the first coefficient regularity conditions. For each term on the oriented right side, we will retain the needed first-remainder restrictions while bounding its second ratio condition. The sum of these conditional bounds over the changed scales, together with the no-change contribution, must be O_{C_*}(1/B) uniformly in the fixed coefficient sets. The orientation factor and coupling error remain outside this conditional calculation.

For a prime set S let N_S(>Y)=\#\{p\in S:\log p>Y\}, and put

The first coefficient tail cutoffs give N_{\mathcal C_\nu^{(1)}}(>Y)\ge q(Y). Expose the first remaining sets above Y, and retain the two necessary tail restrictions N_{\mathcal R_\nu^{(1)}}(>Y)\ge q(Y). Conditional on all these first high-prime sets, the chance that none of their second assignments changes is exactly

On the retained restrictions this is at most

If q(Y)<0 this upper bound exceeds one, which remains a valid bound. Averaging over the first remaining high-prime sets does not increase it.

All first remaining restrictions below Y may now be omitted. For each site, a prime below this threshold is newly added to the second coefficient exactly when it lies in the independent first remaining process and its second coin selects the coefficient. These new-addition indicators are independent and have probabilities 1/(4p). The new additions at the two sites are independent of one another, of the first coefficient sets and their second coins, and of all the exposed variables above Y.

Fix a recipient site \nu. Condition on the retained first-coefficient primes at both sites and on the additions at the other site. The second ratio condition in \mathcal H_2 then restricts the logarithm of the addition product at site \nu to an interval of length at most

For example, for recipient site 1 the second coefficient products have the form a^{(2)}=a_{\mathrm{ret}}b_{\mathrm{new}} and c^{(2)}=c_{\mathrm{ret}}c_{\mathrm{new}}, and solving a^{(2)}/(Tc^{(2)})\in I_\psi gives precisely such an interval for \log b_{\mathrm{new}}. The same calculation with the inequality inverted applies at site 2. On \mathcal U_{Y,\nu} there is a new prime with logarithm greater than Y/2, so this addition product also has logarithm at least Y/2. There are no new additions above Y on \mathcal A_Y.

We can drop the second coefficient size restriction and all second regularity restrictions for this upper bound. Lemma 4.3 then bounds the conditional chance of the remaining addition-product conditions by O(1/Y), uniformly in every value on which we have conditioned. To make the conditioning order explicit, write \mathcal C for the fixed first coefficient sets, G_Y for the retained first remaining tail restrictions, and H_Y for the event of no change above Y. Let \mathcal F include \mathcal C and reveal all second coins of the first coefficient sets, all first remaining data and second coins above Y, and the additions at the other site. The recipient addition product Q_\nu below Y is independent of this information, and the ratio condition specifies an \mathcal F-measurable interval J(\mathcal F). For the oriented event under consideration, the tower property gives

In the last step the high second coins are averaged after the uniform small-ball bound is applied; no conditional bound on \Prob(H_Y\mid\mathcal F) is asserted. Thus, conditional on any good first coefficient sets, the contribution for this orientation at scale Y is at most

The constants here may depend on the fixed C_*, but not on the first coefficient sets, B, or Y.

If there is no change at all, use the tail cutoff at Y=\log P_0. All four first subsets then have at least .4\log R-C_* primes. The same calculation bounds the conditional probability, retaining the first remaining tail restrictions, by

The dyadic scales satisfy

because \alpha_0>1 and the largest Y is less than 8B. Moreover,

Requiring first coefficient regularity only decreases the O(1/B) mass of \mathcal H_1 established above. Combining the preceding bounds, including the orientation factor and the coupling errors, gives

Multiplication by B^2 proves \E\widetilde D_B^2\ll_{C_*}1. Lemma 4.2 transfers these bounds to the Haar divisor weight and completes the proof of (4.4).□

Proof. The function H is bounded and supported on a fixed compact interval in t, so it is square-integrable. Conditional expectations with respect to t and a finite residue coordinate w\bmod q approximate H in L^2 as these residue coordinates increase. Equivalently, for every \varepsilon>0 there is a function H_q(t,w) depending only on t and w\bmod q such that

This follows, for example, by first approximating in the dense space of finite sums of products of L^2 functions of t and locally constant functions of w, and then taking conditional expectation.

For all sufficiently large B, no prime dividing q belongs to \mathcal P. The finite-residue function D_B is therefore independent of w\bmod q under Haar measure. Consequently

By Cauchy–Schwarz and (4.4), the error on replacing H_q by H is O_{C_*}(\varepsilon), uniformly in large B; the length of the fixed t-support is absorbed in the constant. Since \varepsilon is arbitrary, this proves (4.7).□

Proof. Choose 0<\alpha_\phi<\beta_\phi<\infty with \operatorname{supp}\phi\subset[\alpha_\phi,\beta_\phi], and write u_- =\inf I_\psi, u_+=\sup I_\psi. If the inner sum in (4.9) is nonzero, then

Let \mathcal I_{c,x} denote this interval. Use the nonnegative weights x^{-1}\rho_0(\log c/B)A(c)K(m) on this interval. Cauchy–Schwarz gives

where |g_x(m)|=1 supplies the unit factor. No multiplicativity of the centered function F_x is needed.

For fixed B, the weight K_0 is periodic and has mean

Progression counting, K\le K_0, and the length of \mathcal I_{c,x} therefore imply

For the last bound, partition the compact log support of \rho_0 into O(B) dyadic-size boxes and apply (3.6) in each box. Thus we may fix C_4>0, independently of C_*, such that

for all sufficiently large B. Combining this with (4.8) proves (4.10), with any fixed

This makes the uniformity of the positive lower constant explicit.

For the diagonal, set

where a ranges over the coefficient support and l over positive integers for which the truncated weights are nonzero. The total-count cutoff and (3.9) give

using the strict inequality 2(h_0+\tau\log2)<.32. Since all smooth weights here are nonnegative,

Bounding one of the two factors A(a)K(l_a) by Q_B and |F_x(am+1)|^2 by 4 yields

where

The same divisor reindexing as before gives

Here no label is present, so finite progression counting directly gives

by the first-moment upper bound in Lemma 4.1. It follows that

which completes the proof.□

Proof. Replace all truncated weights by their untruncated majorants. For fixed pairwise coprime a,b,c, the two endpoint congruences have Haar probability 1/(ab). At a prime p\in\mathcal P not dividing abc, the three residue weights have distinct zero classes 0,-j,b/c for u. Here p>|j|, and b/c\not\equiv-j\pmod p follows from a=b+jc. Their local mean is 1-3/(2p). At primes dividing abc discard the local reducing factors; this changes the Euler-product bound by at most \exp(O(\sum_{p\mid abc}1/p))=O(1), since that sum is O(B/P_0). Independence over primes and Mertens’ formula give

On a dyadic box c\asymp X, the denominator is ab\asymp T^2X^2. The three-form estimate (3.8) bounds its contribution before the factor 1/B by O(\Sigma(j)/T). There are O(B) boxes. This proves the assertion using only untruncated upper bounds.□

Proof. For an admissible fair split b at the first endpoint, A(b)K(S_i\setminus b)=B2^{-|S_i|}. At the other endpoint, the analogous identity and then averaging over its site set give the split-product law \nu_{1/2}. By (3.5), its mass at a is at most CA_0(a)/(Ba). Dropping all remaining restrictions for an upper bound yields

using (3.7). This is a restricted expectation of mass at most one, with no division by its probability. If the fixed type has no admissible split, its kernel is zero. The same estimate with signed lag reversed handles either endpoint. Summing the bounded averages of \Sigma proves the degree assertions.□

Proof. In the absence of site squares, quotient prime sets at the endpoints equal the complementary sets used in (5.8). For a candidate at i,k=i+j, the remaining argument in (5.2) is

We show that, at a negligible total error, the set of primes dividing this argument can be replaced for each unordered candidate by an independent 1/p set, independent of the site sets and of the other candidates’ extra sets. The following argument first bounds two obstructions determined by the endpoint sets: coincident rational roots and a root forced to hit a third occupied site. It then fixes a surviving actual site configuration and couples the remaining prime tests, without conditioning on absence of site squares. The site-square event is paid for in the matrix weights, and the final step controls the independent extra-weight fluctuations in cut norm.

Coincident rational roots. Since b/c is reduced, equal r for two candidates forces the same denominator c. Given r,c, the coefficient at any position s' is fixed, namely b+(s'-i)c. On the same pair this determines the candidate uniquely. Any other unordered pair with this root uses at least a third position with a prescribed positive coefficient at least Tc. Conditional on the first two independent site sets, that coefficient is available at the third site with probability at most 1/(Tc): it has probability zero unless it is a squarefree product of \mathcal P primes, and otherwise the probability is its reciprocal. The polynomial candidate and position bounds, and c\ge e^B, make the union of these events negligible. The site coupling transfers the conclusion to the actual model.

Forced hits at a third endpoint. Also discard the event that, for some candidate and s'\ne i,k, an occupied prime p at site s' satisfies p\nmid abc and r\equiv-s'\pmod p. In the independent-site model, condition on the candidate’s two endpoint sets. Such a prime divides b+(s'-i)c, a nonzero integer of logarithmic size O(B). It is nonzero because c>1 and (b,c)=1. For any nonzero integer of logarithmic size O(B), the sum of reciprocals of its prime divisors above P_0 is O(B/P_0). Thus the conditional union probability is O(B/P_0) per candidate and third position. A polynomial union bound and the site coupling again suffice.

Coefficient primes and occupied primes. The two structural exclusions above depend only on the endpoint sets, and their actual-law probabilities have been bounded through the site coupling. Fix an actual site configuration avoiding them; its hits at each prime are disjoint. We still do not condition on absence of site squares. At an unoccupied prime, u\bmod p is uniform outside the M forbidden classes -1,\ldots,-M. At an occupied prime, its residue is fixed and the higher p-adic digits remain uniform. These conditional laws are independent over primes.

If p\mid ab, then the coefficients are squarefree and pairwise coprime, and the test p\mid m in (5.11) has conditional probability 1/p, determined by the next digit. If p\mid c, it is impossible. Set these exceptional tests to zero temporarily, in both the actual model and the independent comparison model. The total error per candidate is O(B/P_0). This use of uniform higher digits is made before removing site-square events from the weights; we do not condition those digits on absence of squares.

At p\nmid abc, the root test is u\equiv r\pmod p. If the prime is occupied at an endpoint, such coincidence would force p\mid b or p\mid a. At any other occupied site it is excluded by the preceding discarded event. The actual root test is therefore zero at the remaining occupied primes. The cost of setting the independent test to zero there is bounded in expectation by the deterministic candidate bound times

This domination does not require candidates to be independent of occupied primes.

Unoccupied primes. The roots are now distinct rational numbers. A collision between two of their residues modulo p, after excluding coefficient primes, requires p to divide their nonzero cross-multiplied difference. That integer has logarithmic size O(B). A collision with a forbidden site class has the same bound, or concerns a coefficient prime already handled. For each candidate pair or candidate–site pair, these defective primes have reciprocal sum O(B/P_0). At all such primes, discard all involved tests. Conditional on the sites, each defined test has probability at most 1/(p-M)\le2/p, so another polynomial loss is harmless.

At a remaining unoccupied prime the root classes are distinct and avoid all forbidden classes. Their joint law is a disjoint-choice law with individual hit probabilities 1/(p-M). If there are N_{\rm cand}=O(B^4) tests, its total variation distance from independent Bernoulli(1/p) tests is

For example, compare first to independent tests of probability 1/(p-M), at cost O(N_{\rm cand}^2/p^2) from multiple hits, and then change each probability to 1/p, at cost O(N_{\rm cand}M/p^2). Summing over primes constructs the claimed conditional coupling. The preceding union bounds give total failure probability O(B^{C}/P_0)+O(B^{C}e^{-B}) for a fixed crude exponent C. On these failures, use the deterministic polynomial matrix-mass bounds. After including that loss and enlarging the crude exponent, their contribution to the expected cut norm is at most O(B^{50}/P_0)+O(B^{50}e^{-B})=o(1). Equation (5.7) is handled in the resulting weights, as just explained.

Fluctuations of the extra weights. The preceding coupling has separated the extra roots from the endpoint types. It remains to show that their independent fluctuations are small even after maximizing the testing signs. The comparison model has independent endpoint types and an additional independent set S_{\rm root} for each unordered candidate. In its weight, replace k_B in (5.8) by K(S_{\rm root}). Conditional on the endpoints, the candidate weights are independent, and their means give precisely \mathcal L. By (3.9), each full candidate weight, including its six truncated factors and the factor 1/B, is at most

for large B, with harmless enlargement of constants. For fixed sign vectors in (5.5), an unordered candidate has sign coefficient of absolute value at most two. Positivity bounds the conditional variance sum by a constant times B^{-0.07} times \sum_{i,k}\mathcal L_{ik}.

Bernstein’s inequality, union over at most 2^{2M} sign choices, and integration of the tail show that the conditional expected cut norm of the centered fluctuation is at most

Indeed, before division by M, a threshold

has tail O(e^{-y}) for y\ge0, with C absorbing the sign enumeration. Lemma 5.2 and Jensen’s inequality make the expectation of (5.12) tend to zero. The triangle inequality now proves (5.10); the deterministic polynomial mass bounds control the coupling failures for both kernels. All estimates used only uniform bounds for the smooth factors on \mathcal S.□

Proof. Every alternative candidate has a unique decomposition

Here all the products are squarefree, and y is coprime to b while z is coprime to a. The coefficient c' is determined once the four products are chosen. We may discard any of the alternative candidate conditions for upper bounds; however, we retain the coefficient regularity conditions that are used below. Since a',b'\le2T\exp(2B), we have \log y,\log z\le3B for all sufficiently large B.

Set \Delta=1/L, g_0=3\Delta, and U=\max(\log y,\log z). We count the alternatives with U\le B^{g_0} directly from prefix regularity. When U>B^{g_0}, fixing the omissions and the new product at one endpoint places the other new product in one residue class modulo |j|, through e_1z-d_1y=jc'. The resulting progression density is the gain that will pay for the remaining subset choices.

For a grid point g<1, define

and write \omega_g(v)=|\mathcal P(v)\cap\mathcal Q_g|, with the same notation for a prime set. Regularity gives

for each of the first and alternative coefficients, and for each regular first remainder. Mertens’ estimate, on this fixed grid, gives

The o(\ell) is uniform over the finitely many grid points.

Omissions above an addition prefix. Suppose that all primes of z and y lie in \mathcal Q_g. Let D_a=\mathcal P(a)\setminus\mathcal P(e_1) be the primes omitted from a. Regularity of the total counts of a,a' gives

Regularity at g gives the analogous inequality

Subtracting shows that at most 4\tau\ell primes above the prefix can be omitted. The same conclusion holds for the omissions from b. When g=1 there are no primes above the prefix.

Write

with the endpoint values defined by continuity. For completeness, if m\le8\tau\ell, the number of such subsets is at most 2^m\le2^{8\tau\ell}. For 8\tau\ell<m\le(1/2+\tau)\ell, the binomial bound

applies: H is increasing up to 1/2, and mH(k/m) is increasing in m for fixed 0<k<m. Thus the choices at both coefficients together cost B^{\varepsilon_{\mathrm{high}}(\tau)+o(1)}, where \varepsilon_{\mathrm{high}}(\tau)\to0 as \tau\to0. The estimate is uniform also when the relevant set is empty.

Small additions. If

all additions lie in \mathcal Q_{g_0}. On the two first-remainder regularity events, the four relevant prefix sets—the prime sets of a,b and the two first remainders—have at most (2g_0+4\tau)\ell primes altogether. Allowing arbitrary subset choices there, and then the higher omissions just bounded, gives the pointwise estimate

This also bounds its expectation with the first-remainder regularity indicators. By taking L large and then \tau small, its exponent apart from o(1) can be made strictly smaller than .008.

Large-addition classes and combinatorial choices. For U>B^{g_0} use the classes

where g ranges over grid points. All addition primes again belong to \mathcal Q_g. It suffices to treat the portion of a class in which \log z>B^{g-\Delta}. The portion with \log y>B^{g-\Delta} is estimated by the same argument with the endpoints exchanged; the numeric sieve below permits either sign for its second slope. If both inequalities hold, counting the alternative twice only increases the upper bound.

Define the exact normalized counts and their clipped values by

On the first-remainder regularity events, 0\le r_z^*,r_y^*\le1/2+\tau/g. There are O(\ell^2) possible pairs of integer counts. We estimate one such pair at a time.

The number of omitted primes in the a prefix is r_z^*g\ell+O(\tau\ell), among g\ell/2+O(\tau\ell) available primes. Applying \binom mq\le\exp(mH(q/m)), and summing over the O(\ell) permissible values of q, gives, with the higher-omission factor included,

The function \varepsilon_L(\tau) can be chosen to tend to zero for fixed L. To justify this uniformly in the counts, divide m,q by \ell and use uniform continuity of H on [0,1]; the perturbations are O(\tau/g), and g\ge g_0>0 is fixed. The clipping from r_z^* to r_z changes these arguments by the same amount. Also \ell/\log B\to1, and all polynomial factors in \ell contribute B^{o(1)}. We use \varepsilon_L(\tau) below for a possibly enlarged function with this same limiting property.

The choices of d_1 have the corresponding bound with r_y. Conditional on a regular first remainder \mathcal T_b, the number of its subsets y of the prescribed cardinality is at most

This bound is uniform in that remainder set. We may condition on it and sum over these y while the remainder at the other endpoint still has its independent tilted law.

Availability of a fixed numeric addition. The entropy bounds count the possible omitted prime sets. To control large additions, we also need the probability that each numerically specified product is present in the other remainder set. The next bound keeps the first remainder’s regularity requirement in this probability. For a fixed squarefree z in this class, with \mathcal P(z)\subset\mathcal Q_g\setminus\mathcal P(a), its inclusion probability in \mathcal T_a is

The last bound is uniform because \log z\le3B and all its primes exceed P_0. Conditional on those inclusions, the remaining available indicators in \mathcal Q_g are independent with parameter sum

The excluded primes have reciprocal sum O(B/P_0), so this estimate is uniform in the fixed coefficient and numeric addition. First prefix regularity requires the number of these other inclusions to be at most (1/2-r_z^*)g\ell+\tau\ell.

If X is a sum of independent Bernoulli variables with parameter sum \Lambda, then for t\ge0,

For 0\le k\le\Lambda, minimizing gives exponent -\Lambda+k-k\log(k/\Lambda), with its continuous value at k=0. For k\ge\Lambda use the bound 1, and for k<0 the event is empty. Uniform continuity of the resulting rate, including the endpoints, and the fixed-grid relation g\ge g_0 therefore imply

Here u\log u=0 at u=0. Retaining only prefix regularity on the left gives an upper bound for retaining all of first-remainder regularity, which is what the multiplicity expectation requires.

A harmonic sieve for the numeric addition. We now sum the availability bound over numeric additions satisfying the alternative coefficient relation. The factor 1/z in that bound is the reason for the harmonic normalization below. Fix e_1,d_1,y from the preceding choices. The alternative size conditions imply

with absolute comparison constants; we may enlarge the interval to [Z'/2,2Z'].

For these fixed data and the fixed cardinality class, let \mathcal Z consist of the positive integers z with the following properties:

The congruence makes c'(z) an integer. Every valid alternative in this designated class yields an element of \mathcal Z. The inclusion \mathcal P(z)\subset\mathcal T_a remains the event estimated in (6.5); the two alternative remainder regularity conditions are discarded. Equation (6.5) applies to every z\in\mathcal Z, because each is a possible product in the tilted remainder at the a endpoint. If \mathcal Z is empty there is nothing to prove; otherwise its interval and log conditions give Z'\ge\tfrac12\exp(B^{g-\Delta}).

All of e_1,d_1,y are units modulo |j|, since their prime factors exceed P_0>T. Thus the displayed congruence is one unit residue class. We claim

where r_{\min}\in(0,1/2) is any fixed number. The excess exponent 3\Delta+r_{\min}+\tau\log2+o(1) can be made small by the later choices of grid, clamp, and tolerance.

To prove the deterministic estimate, put \lambda=\max(r_{\min},r_z), so r_{\min}\le\lambda\le1/2. On \mathcal Z the fixed cardinality gives

The membership of (e_1z,d_1y,c'(z)) in \mathcal C_j also gives the upper prefix cutoff for c'(z), and hence

The prefix factor is introduced to supply a second reducing root in the sieve below. At the leading g-scale its saving -g/2 combines with the cutoff cost g\log2/2, leaving -gh_0 in the exponent. The calculation below verifies the root conditions and retains the grid and tolerance losses. Insert these two factors before enlarging the summation domain, and keep only the reducing prime factors up to

Deleting the other reducing factors increases each summand. Since z\ge Z'/2 on \mathcal Z, we obtain the deterministic majorant

The finite congruence product is defined for every integer value of c'(z), including zero. The hard prefix cutoff, the fixed-cardinality condition, and all other candidate restrictions have been removed only in this deterministic sum. Equation (6.5) is applied only to z\in\mathcal Z; the displayed enlargement bounds the weighted sum that results.

For explicit congruence accounting put J_1=|j| and \sigma=j/J_1, and choose a unit z_0 modulo J_1 satisfying the progression condition. Write

The parameter t runs through an interval of length N\asymp Z'/J_1. Its logarithm is at least B^{g-\Delta}-O(\log B). Consequently

so the interval upper sieve of Section 3, in fixed dimension two, applies uniformly.

The displayed product already omits every prime at most P_0, including every prime of j. Among the remaining primes also drop those dividing e_1d_1y. Their reciprocal sum is O(B/P_0), uniformly, since \log(e_1d_1y)=O(B). At each retained prime the two slopes are invertible, and the two roots are distinct. Indeed their determinant is

which is nonzero modulo such a prime. The local weights of the roots are \lambda and 1/2, so the average local weight is exactly

This also explains why no coefficient-height factor enters the sieve: only the number and distinctness of residue roots are used. If one keeps primes dividing j instead, the first form is a fixed unit and the second has invertible slope; dropping them avoids any need for a separate local factor.

The random-weight interval sieve bounds the sum of the retained reducing weights by a constant times

The first term is N B^{(g-2\Delta)(\lambda-3/2)+o(1)}, by Mertens’ estimate. Indeed the reciprocal sum in that product is (g-2\Delta)\log B-\log\log P_0+O(1) =(g-2\Delta)\log B+o(\log B), and the sum of the squared reciprocals is negligible. Discarding the small-prime conditions costs at most powers of \log P_0 relative to their possible sieve factors, and those powers are B^{o(1)}; alternatively, the displayed large-prime product already proves the upper bound directly.

Using z\asymp Z' and restoring the threshold factors, the harmonic normalization of the main sieve bound is N/Z'\asymp1/J_1. The remainder contributes at most

which is negligible uniformly in the class, since g\ge4\Delta and \log N\ge B^{g-\Delta}-O(\log B). The total power of B in the main term is

If r_z^*\le1/2, then r_z=r_z^*, and this power equals

We used -3/2+(\log2)/2=-1-h_0. The clamp contribution in parentheses vanishes when r_z\ge r_{\min} and lies between 0 and r_{\min} when 0\le r_z<r_{\min}, including the value r_{\min} at r_z=0. The grid contribution is at most 3\Delta. If r_z^*>1/2, our choice is r_z=\lambda=1/2; the Rankin factor is then exactly 1 and the same upper bound follows directly with the clipped r_z. These observations prove (6.6). If y is the designated large numeric variable, the identical proof uses the slope -d_1 for c' instead, and its determinant has the same nonvanishing property outside the already discarded coefficient primes.

Combining the bounds and choosing parameters. Condition on the first remainder at the b endpoint. If it is not regular its contribution vanishes. If it is regular, enumerate the choices of d_1,y by the two entropy bounds with r_y, and enumerate e_1 by (6.4). For each such choice, sum the availability bound (6.5) using (6.6). Independence of the other first remainder justifies this conditioning, and every bound is uniform in the conditioned regular remainder. Thus we may average it out without any additional factor.

A direct expansion of H gives the exact identity

The three combinatorial factors have entropy exponents gH(2r_z)/2, gH(2r_y)/2, and gH(2r_y)/2. Adding the probability and sieve exponents therefore bounds the contribution of one class, one orientation, and one count pair, with both first-remainder regularity indicators, by

The tolerance error here includes the preceding \tau\log2 and the finitely many entropy and rate perturbations. The last inequality uses H\le\log2 and 0<g\le1. If the bracket in the first exponent is negative, multiplying it by g still gives a nonpositive number; otherwise its maximum is at most 2\log2-1-h_0=\kappa_1.

Here is an explicit consistent order of choices. First choose L large enough that 6\Delta\log2<.004 and 3\Delta<.01. Next choose a fixed r_{\min}<.01. Finally make \tau small enough that the small-addition exponent is strictly below .008, that

and that all the strict weight inequalities in (3.10) hold. These choices are possible because \varepsilon_L(\tau) and \varepsilon_{\mathrm{high}}(\tau) tend to zero for the fixed grid. They impose no condition on C_*, since only the prefix and total regularity conditions were needed for this estimate.

There are finitely many grid classes and orientations and O(\ell^2)=B^{o(1)} count pairs. Since |j|\ge\eta T\asymp_\eta B^{.32} and

the sum of all the large-addition bounds is o(1). The small-addition bound, with a strict exponent below .008, is O(B^{.009}) after the o(1) in its exponent is absorbed. This proves the lemma, including the first candidate itself when valid.□

Proof of Proposition 6.1. For an allowed edge i,k=i+j, apply Lemma 6.2 to (6.3) and use the uniform bound on \Xi. This gives

The remaining coefficient sum is bounded using (3.8). Cover the support of c by O(B) dyadic boxes c\asymp X, so a,b\asymp TX in each box. On such a box, dropping coefficient cutoffs for an upper bound gives

Summing the boxes yields O(\Sigma(j)/T). The same proof works for negative j by symmetry. Consequently

The choice in (3.10) gives \kappa_1-5\tau\log2=1-5h_\tau>.229. Thus the exponent on the right is strictly below -.220 before the o(1) term. Finally

so summing over a row proves (6.1), with room to absorb the o(1) in the exponent.

For the last assertion set D_i=\sum_k\mathcal L_{ik}. Given S_i, the summands with different k\ne i are independent, because the other site types are independent and the roots have already been averaged out. Therefore

The conditional mean is bounded for every S_i by (5.9). The variance identity consequently gives

Notice that only an integrated conditional variance bound has been used; a pointwise conditional second-moment estimate is unnecessary. Jensen’s inequality across the rows now gives

For the kernel restricted to I as in the statement, nonnegativity can only reduce the conditional means, row-square sums, and total mass. Keeping the same normalization M therefore gives the stated bounds.□

Proof. At a fixed prime p, the categories common to both splits, exclusive to the first, and exclusive to the second have probabilities 1/(4p) each, and are mutually exclusive. The joint law of three independent Bernoulli indicators with these marginal probabilities differs from this categorical law in total variation by O(p^{-2}). For example, the probability of two or more independent successes is O(p^{-2}), and each remaining probability differs from its categorical value by O(p^{-2}). Couple these laws independently over the primes. A union bound gives total failure probability

On successful coupling the products agree. Collisions that would put a squared prime in one of the independent products are included in the failure event.

If two joint probability laws differ in total variation by \varepsilon, each cell-pair probability differs by at most a constant times \varepsilon. Their step kernel difference is thus bounded pointwise by O(\varepsilon\mu_d^{-2}). The total measure of the output space is |I|, so Schur’s inequality bounds the operator difference by the same expression times |I|. Now

Taking \varepsilon=O(P_0^{-1}) proves the assertion.□

Proof. It suffices to consider i=2. Put

where the last inequality is (3.4). The row integral at A=J\times\{r\} of the restricted kernel is at most

Here E_2 is independent of C,E_1. The marginal law of CE_1 differs from \nu_{1/2} by O(P_0^{-1}) in total variation: the independent common and exclusive indicators can be coupled to their mutually exclusive counterparts prime by prime exactly as in the preceding lemma. Thus (3.3) and \mu_d^{-1}\ll B^{21/10} bound the last display by

For the column integral at A'=J'\times\{r'\}, first omit the condition that CE_1 has logarithm in I. Condition on E_2=e with \ell_B(e)\le\theta. The remaining condition on C is

The shifted interval lies in [2/5,3], has length \delta_B, and the prescribed residue is a unit. Applying (3.3) with z=1/4, uniformly in e, gives conditional probability at most

Integration over the event \ell_B(E_2)\le\theta and division by \mu_d therefore give a column bound C\sigma_\theta+O(B^{-70}). Schur’s inequality proves the claimed operator bound. The i=1 case follows on transposing the kernel. The kernel of the union is nonnegative and is entrywise at most the sum of the two restricted kernels, so the row and column estimates also prove its bound.□

Proof. Integrating (7.1) with d=1 over I_l sums precisely the fine cells contained in I_l. The result is \E[g(S)\Prob_{\rm split}(x_b\in I_l\mid S)]. Divide by |I_l|. The bounds on V_l follow from its definition; the final uniformity follows because the preceding results are operator bounds rather than fixed-input limits.□

Proof. Applying the subset identity above at both endpoints in (5.8) turns their subset sums into fair-split expectations, still with regularity indicators attached to the selected and unselected sets, and changes the coefficient k_B/B into k_BB. The remaining coefficient is c=(a-b)/j, and its weight is A_0(c) times its regularity indicator.

All weights other than g,h are nonnegative. We may therefore bound the cost of removing an indicator by putting \lvert g\rvert,\lvert h\rvert\le1 and summing the corresponding positive mass. On the support under consideration, c\asymp X, a,b\asymp TX, and 0.9B\le\log X\le2.2B. There are O(B) dyadic choices of X. Formula (3.5) bounds the split-product probabilities by

Consequently the untruncated mass in one such box is at most

by (3.8), since ab\asymp T^2X^2. If any one of the three coefficients is nonregular, (3.11) gives the same bound with the additional factor r_B+e^{-c_3C_*}. Summing over the boxes proves the required estimate for coefficient failures.

It remains to treat the two unselected endpoint sets. At a given prime p, the probabilities of the three possibilities “selected”, “unselected”, and “absent” are respectively 1/(2p),1/(2p),1-1/p. Conditional on the selected product being a, the unselected indicators at primes not dividing a are therefore independent with probabilities

At primes dividing a they are zero. The same statement holds for b, independently at the other site. Each selected coefficient on our support has log size O(B), and all its prime factors exceed P_0. In particular, the reciprocal sum of its prime factors is O(B/P_0). The uniform cutoff probability estimate established after (3.11) thus bounds either unselected-set failure by r_B+O(e^{-c_3C_*}), uniformly in the chosen coefficients. Multiplying by the preceding positive mass bound and summing the boxes proves (8.1). This argument uses (3.8) and (3.11) for 1\le j\le T; it has made no use of j\ge\eta T or of the block length M. This proves the asserted independence of its constant.

Finally, if one of (a,b),(a,c),(b,c) exceeds one, a prime dividing both a and b must occur. Indeed a-b=jc, and every prime factor of a coefficient is greater than P_0>j. The split products at the two sites are independent, so the probability of a common selected prime is at most

For a fixed pair of selected products there is at most one value of c, and the remaining positive integrand is at most C B\sup_c A_0(c). Since \sup A_0\le(\log P_0)R^{1/2}=B^{1/2+o(1)}, the resulting error is O(B^{3/2+o(1)}/P_0). After removal of these restrictions, the fair splits and independence of the two sites give (8.2) exactly.□

Proof. Every estimate above was uniform in a pair of tests with absolute value at most one. Apart from (8.1), the pre-projection arc errors are uniform o(1/T). The histogram and residue projection errors have the explicit common upper bound

The series tail is bounded by (8.14), and the archimedean and coarse replacements have total error

All these bounds hold separately for every allowed edge. They thus remain valid if the endpoint tests differ from edge to edge, and in particular for the collection of tests in (8.19).

For a fixed positive lag there are at most M ordered pairs with that lag; including the negative lag at most doubles this number. Division by M therefore reduces their total error to at most twice the sum of the edge error over \eta T\le j\le T. The first-moment bound for \Sigma gives

Consequently the sum of (8.1) is O(r_B+e^{-c_3C_*}), with a constant independent of \eta,C_6,C_*. A uniform o(1/T) error sums to o(1). For the other errors use the bounded mean of j/\varphi(j) and

Since \log(2Q)(B\delta_B+B^{-c_7})\to0, the denominator sum contributes o(1). Absolute convergence makes the series tail tend to zero as well. Finally \mathfrak S(j)\le j/\varphi(j) controls the archimedean and coarse errors, giving the stated C_\eta\epsilon_1+o(1). There is no factor of C_6, because the count of pairs for each lag has already been divided by M. This proves (8.19).

For the deterministic row estimates, boundedness of k_B, w, and the finitely many features gives the pointwise bound

on allowed pairs. Each row has at most two entries for each positive lag. The first and second powers of j/\varphi(j) have bounded means. For the second power this was proved before (8.11), and the first follows by Cauchy–Schwarz. Summing the displayed pointwise bound and its square over 1\le j\le T proves (8.20), uniformly in every collection of site types and in the block length.□

Proof. Put D_i=\sum_k\mathcal H_{ik} and d_B=K B^{0.07}, where K is any fixed sufficiently large constant. Conditional on S_i, the summands of D_i are independent, since each off-diagonal summand depends on one other independent site. Consequently

For sufficiently large B, d_B\ge2C_0. Conditional Chebyshev followed by averaging the row type gives

Thus the event

has complement of probability O(Mq_B/d_B^2)=O(B^{-0.03}). Markov’s inequality gives

Take

This implies (9.2). Finally, Jensen’s inequality across rows gives

In particular, for any events \mathcal B_B with probabilities tending to zero uniformly in s, \E[A_B\ind_{\mathcal B_B}]=o(1) by Cauchy–Schwarz.□

Proof. Fix the realized matrix. Choose column signs \delta_k attaining the cut norm and choose the best-response row signs, so the unnormalized maximum is the positive quantity \sum_i|r_i|, where r_i=\sum_k\mathcal E_{ik}\delta_k. For this deterministic matrix only, sample a uniform subset J' of m_2 columns and form the unbiased estimate

The variance formula for sampling without replacement yields

For real numbers r and \widehat r, |r|-r\sgn(\widehat r)\le2|r-\widehat r|. Applying this inequality and then Cauchy–Schwarz across rows, the expected loss in the unnormalized bilinear sum is at most

Here m_2\asymp M B^{-0.18} for large B. Hence at least one subset has this loss bound, with \sgn(\widehat r_i)=\widetilde g_i. Its signs on J' are the restrictions of the chosen optimizing column signs.

Deleting every ordered pair with an endpoint in J' changes any such bilinear sum in absolute value by at most 2m_2d_B, using symmetry and the degree event. After deletion, maximizing the column signs gives exactly \sum_{k\in R'}|\sum_{i\in R'}\mathcal E_{ik}\widetilde g_i|. Since m_2d_B/M=O(B^{-0.11}), the claim follows.□

Proof. Only the types in the deterministic set J' have been conditioned on. In particular, no event describing which signs an optimizer would choose has been imposed. The types in R' remain independent with their usual laws, and |g_i|\le1.

For k\in R' and a possible value v of S_k, put

Choose h_k(v)=\sgn(a_k(v)), and set both families of tests equal to zero on removed indices. Equation (8.19), which is uniform over all bounded single-site tests, gives

This application is valid for every fixed value of \mathbf t, however the resulting functions g_i depend on that value.

Conditional on S_k, the centered terms

are independent. The diagonal contributes zero. Conditional variance followed by Cauchy–Schwarz therefore gives

Symmetry supplies the squared-column bound from the squared-row bound. Averaging this estimate over k costs at most q_B^{1/2}=O(B^{-0.105}). This proves the asserted mean bound, including its uniformity in the conditioned choice.□

Proof of Proposition 9.1. The skeleton lemma reduces the adaptive cut norm to a finite family of optimizations, each with conditional mean at most E_0+o(1). The remaining step is a concentration estimate strong enough to hold for the entire family. The common degree event will be excluded only once. We prove concentration for each of the deterministic subset/sign assignments in Lemma 9.4, and then take a union over those assignments. Fix such an assignment and condition on \mathbf S_{J'}=\mathbf t. Abbreviate its random quantity by Z. Let \mathcal G_{R'} be the set of configurations of the remaining site types satisfying

The observation after Lemma 9.2 shows that

uniformly in the conditioning and assignment.

We next verify a Lipschitz bound between any two configurations \mathbf u,\mathbf v\in\mathcal G_{R'}, rather than only between good configurations differing in one coordinate. Let A=\{i\in R':u_i\ne v_i\}. The row functions g_i have already been fixed by \mathbf t and \sigma. Thus a term \mathcal E_{ik}(u_i,u_k)g_i(u_i) can change only if i or k belongs to A. The reverse triangle inequality for each column sum gives

The last inequality uses both endpoint degree bounds and symmetry. It also includes the change of g_i when i\in A. No connecting path of good configurations is required.

Choose a fixed L'>E_0+2+\epsilon_2. On \mathcal G_{R'} the function f=\min(Z,L') is nonnegative and Lipschitz for the site Hamming distance d_H, with constant c_B=4d_B/M. For sufficiently large B the good set is nonempty. We use the McShane extension [19] in its infimum form [3, Section 3, Equation (3.2)], followed by clamping:

The site spaces are finite for every B, so the formula is measurable. The infimum defines a c_B-Lipschitz function by the triangle inequality, and its restriction to the good set is f: the Lipschitz inequality for f gives the lower bound, and taking \mathbf v=\mathbf u gives the upper bound. Clamping preserves the Lipschitz bound. Hence 0\le\widetilde Z\le L' everywhere and \widetilde Z=\min(Z,L') on \mathcal G_{R'}. By Lemma 9.4,

again uniformly in every fixed choice.

Changing any one of the independent remaining site types changes \widetilde Z by at most c_B. McDiarmid’s bounded-differences inequality [18] therefore gives, for every a>0,

Taking a=\epsilon_2/4 and using the uniform mean bound shows that, for large B,

On \mathcal G_{R'}, the event Z>E_0+\epsilon_2/2 is equivalent to this event for \widetilde Z, since the threshold is strictly below L'.

There are at most

deterministic subset/sign assignments. For each assignment, integrate the last conditional tail bound over its subset types. We may then union-bound the extension tails, because

The original full-sample event \Omega_{\rm deg} implies \mathcal G_{R'} for every subset J' simultaneously, by nonnegativity of the envelope. Consequently

Only the exponential extension tails have been union-bounded; the probability of \Omega_{\rm deg}^c is counted once.

With probability 1-o(1), the squared-mass event also holds, and Lemma 9.3 now bounds the fully adaptive cut norm by E_0+\epsilon_2/2+O(B^{-0.01})+O(B^{-0.11}). On the exceptional event use \|\mathcal E\|_\square\le A_B. Lemma 9.2 makes its expected contribution o(1). This proves (9.1). All probabilities and errors used here were uniform for each fixed s in the compact range, which proves the stated uniformity without constructing a simultaneous event over s.□

Proof. Write x_b=(\log b)/B for the normalized logarithm of the fair split product b, and put q_l^0(y)=\ind_{I_l}(y)/|I_l| for y\ge0. The endpoints of the coarse cells lie in [1/2,3], where (3.3) bounds the marginal split-product densities uniformly for large B. Choose small endpoint neighborhoods, still in a fixed compact subinterval of (0,3.2), and replace q_l^0 by a continuous compactly supported function q_l. We may arrange 0\le q_l\le1/|I_l|, with equality to q_l^0 outside these neighborhoods. By (3.3), their \nu_{1/2} mass is at most a constant times their total length plus O(B^{-80}). Consequently the neighborhoods can be fixed so that

is as small as desired, simultaneously for the finitely many cells.

The function t\mapsto q_l(-\log t) on (0,1], assigned value zero at t=0, is continuous on [0,1]. Indeed q_l is zero for all sufficiently large arguments. Uniform polynomial approximation on [0,1] therefore gives a real polynomial p_l for which p_l(e^{-y}) approximates q_l(y) uniformly for every y\ge0. Both this approximation and the preceding endpoint smoothing can be chosen so that

for all sufficiently large B. Taking the uniform polynomial error at most one also ensures \sup_{0\le t\le1}|p_l(t)|\le1/|I_l|+1.

Conditional on a site set S, fair splitting gives

The second identity follows by making the fair split choices separately for each prime. Conditional expectation is a contraction in L^2, which proves the asserted error bound and the uniform bound on the approximants. The last identity applies to the distinct primes dividing n+i, irrespective of their multiplicities, and hence gives exactly G_{B,v}(n+i) at the integer sites.□

Proof. Write

The omitted nondividing-prime factors form a product of numbers in (0,1]. Since P_1\ge2, its difference from one is at most

uniformly in j. All the retained nondividing-prime factors are in [0,1], including the possible zero factor at p=2. Consequently

The positive divisor expansion gives

Its mean up to T is bounded by

The unrestricted positive series equals \prod_p(1+1/(p(p-1)))<\infty, so this tail tends to zero. The same expansion bounds the mean of a(j) by that product, uniformly in T. The required conclusion follows.□

Proof. Let \gamma>0 be arbitrary. First choose the approximants in Lemma 10.1 with sufficiently small \zeta that the feature replacement has iterated upper-limit error at most \gamma/3. Their degrees and coefficients are now fixed, and the approximants have the stated bounds depending only on the fixed coarse partition. Next choose a fixed P_1 by Lemma 10.2. Replacing \mathfrak S(|k-i|) by \mathfrak S_{P_1}(k-i) in the energy costs at most \gamma/3 in upper limit. To check the normalization, the absolute error is bounded by a fixed constant times

The label and origin bounds have been absorbed in that fixed constant.

Choose a fixed 0<\delta<\eta/4, to be made small after P_1 has been fixed. Partition \{1,\ldots,M\} into a fixed number R_0 of consecutive blocks J_a, 1\le a\le R_0, of lengths L_a differing by at most one. Choosing R_0 sufficiently large in terms of \delta,C_6, for all sufficiently large B we have

with a positive fixed lower constant. In particular all these lengths tend to infinity.

Call an ordered block pair fully allowed when every (i,k)\in J_a\times J_b satisfies \eta T\le|k-i|\le T. Discard any block pair containing both an allowed and a nonallowed pair of positions. The values of |k-i| on a block pair vary by at most 2\delta T. Thus each discarded allowed pair has lag within 2\delta T of \eta T or T. For each row there are only O(\delta T+1) such positions. The factors now present, including the periodic lag factor, are bounded for the fixed P_1 and coarse partition. With the normalization 1/(MT), this discarding costs O(\delta)+O(1/T), with a constant allowed to depend on these already fixed parameters.

Choose a representative position i_a\in J_a for every block, and for every fully allowed block pair put \lambda_{ab}=|i_b-i_a|/T\in[\eta,1]. Smoothness of w on the fixed compact s-range and on \eta\le\lambda\le1 gives

Replacing the weight by this representative value therefore has another O(\delta) cost. We fix \delta sufficiently small that the two block errors together have upper limit at most \gamma/3. Neither the representatives nor their scaled lags have to converge as B grows; only this uniform bound is used.

It remains to show that the fully factored expression tends to zero. For 0\le r<D and 1\le l\le m_1, define

By the feature identity, this is a fixed finite linear combination of

The lengths L_a tend to infinity and the starting indices are fixed for each B. The modulus D, monomials v, and polynomial coefficients are all fixed before B grows. Lemma 2.3 applies with origins n\in[s_0Tx,s_1Tx], and yields

for every one of the finitely many indices.

Set c_{r,r'}=\mathfrak S_{P_1}(r'-r), using the periodic extension in Lemma 10.2, including at the zero residue. On a block pair, splitting each position by its residue class gives the factored expression

The feature polynomials are real, so the conjugation here matches exactly the conjugation in the original energy. The factors L_aL_b/(MT) are uniformly bounded. The numbers of cells, blocks, and residue classes are fixed, and k_B and the representative smooth weights are uniformly bounded. Cauchy–Schwarz in the origin variable and (10.4) make every term in (10.5) tend to zero in the asserted iterated upper-limit sense. This remains true when the bounded coefficients depend on n/(Tx) or on B.

The original main energy consequently has iterated upper limit at most \gamma in absolute value. Since the approximation parameters were fixed before the limits and \gamma>0 was arbitrary, the proposition follows.□

Proof of Proposition 2.2. Suppose mixed decorrelation fails for some fixed J and two fixed phase vectors. Fix the resulting subsequence, the limiting profile W and bump \phi from Section 2, and the smooth functions \rho_0,\psi from Section 4. Choose the regularity grid and tolerances as required for the estimates of Sections 3 and 6. The amplification then supplies the positive constant c_5 in (4.10). This constant is independent of every sufficiently large fixed C_*.

We collect the upper bounds, recording the dependencies needed to choose parameters without a cycle. The diagonal contribution to I_{2,x}/(BT) tends to zero in the prescribed order. Equations (5.1)–(5.4) express the remaining contribution as the block energy, up to errors bounded in iterated upper limit by

Here C_0 can be chosen independently of C_*, \eta and C_6, by the untruncated mean bound (5.3).

The root comparison (5.10) and the sampling bound (9.1) allow replacement of this block kernel by \mathcal M. The resulting upper-limit error is bounded by

The constants C_1,C_2 do not depend on the later choices \eta,C_6 or the coarse partition. Indeed the truncation term in (8.19) has a coefficient independent of the lag range and block length, and converting a real cut-norm bound to an energy of labels of absolute value at most two costs only an absolute factor. For example, splitting each label into its real and imaginary parts gives a factor at most 16. Integrating over \mathcal I_s costs only its fixed length. The finite-residue counting passage and the uniformity in s were established in Sections 5 and 9. The remaining o(1) errors vanish as B\to\infty after the inner x-limit, with all the displayed parameters fixed.

For any such fixed choices, Proposition 10.3 makes the main energy zero in upper limit. Consequently

First choose a sufficiently large fixed C_* so that the first term is less than c_5/10. Next choose a sufficiently small fixed \eta and a sufficiently large fixed C_6 so that the next two terms are each less than c_5/10. Then choose positive fixed \epsilon_1,\epsilon_2 so that their terms are each less than c_5/10, and choose the corresponding coarse partition in (7.3). All feature, periodic and block approximations in Proposition 10.3 are made subsequently with these parameters fixed. Thus (10.6) is strictly smaller than c_5, contradicting the lower bound (4.10).

Finally, the bad subsequence was extracted from an arbitrary sequence of integer scales on which x^{-1}\sum_{1\le n<x}\overline{g_x(n)}F_x(n+1) stays a fixed positive distance from zero. Every inner limit above uses the final subsequence selected for the profile. At each fixed B, all multipliers, residue moduli and shifts are finite, so the argument never requires their invariance or counting estimates while B and x grow simultaneously. The contradiction rules out every original sequence of this kind. Thus the mixed correlation tends to zero along the full sequence of integer scales, for every fixed pair of phase vectors.□

Proof. First suppose that c=k_0/J for integers J\ge2 and 1\le k_0<J. For m\le x, set

Then \Pplus(m)\le x^c exactly when N_{c,x}(m)=0, and 0\le N_{c,x}(m)\le J. For a nonnegative integer z, write (z)_h=z(z-1)\cdots(z-h+1), with (z)_0=1. The finite inclusion–exclusion identity is

For each 0\le h\le J, the falling-factorial multinomial identity gives

Apply the initial-segment form of (2.9) to every term, with r_k=0 for k<k_0. The multinomial coefficient counts the assignments of the h ordered variables to the bins. Summing the resulting simplex integrals therefore joins those bins into (c,1] and gives

For h=0, the empty integral is one, and the left-hand limit is also one. The initial-segment limit used here holds through real x, as does the factorial-moment limit in Section 2.

In this integral put s_i=cv_i. The upper bound on each v_i is then implied by v_1+\cdots+v_h\le1/c, and changing the lower boundary from v_i>1 to v_i\ge1 does not change the integral. Averaging the inclusion–exclusion identity shows that the rational smoothness density is H(1/c), where for u\ge0 we put

The sum is locally finite because I_h(u)=0 when h>u. In particular, 1/c\le J ensures that the sum at u=1/c agrees with the finite inclusion–exclusion sum above. Put I_0(u)=1 for u\ge0.

For h\ge1 and y>z\ge1, symmetry and the identity 1=(v_1+\cdots+v_h)/(v_1+\cdots+v_h) give

To verify this, cancel the last denominator after using symmetry and set t=v_1+\cdots+v_h. The remaining variables satisfy v_1+\cdots+v_{h-1}\le t-1. The same formula holds for h=1. Thus the I_h are continuous, H=1 on [0,1], and

Successive integration on the intervals [k,k+1] uniquely determines a continuous solution from its values on [0,1]. Hence H=\rho, proving (11.2) for rational c.

For an arbitrary real c\in(0,1), choose rational c_-<c<c_+ in (0,1). The smoothness event at x^{c_-} is contained in the event at x^c, which is contained in the event at x^{c_+}. The rational limits bound the lower and upper limits of the middle normalized count by \rho(1/c_-) and \rho(1/c_+). Letting c_-,c_+ approach c and using continuity of \rho at the finite argument 1/c proves (11.2).□

Proof of Theorem 1.1. Fix a,b\in(0,1) and choose fixed exponents c_-<a<c_+ and d_-<b<d_+ in (0,1). For each fixed \varepsilon\in(0,1), all sufficiently large real X satisfy, uniformly for \varepsilon X\le n\le X,

On this interval, the fixed-scale event at the lower exponents is contained in the moving-threshold event of Theorem 1.1, which is contained in the fixed-scale event at the upper exponents. The discarded initial interval costs at most \varepsilon+O(1/X) in normalized counting. The limit through all real X in (11.7) therefore bounds the lower and upper limits of the desired count by the corresponding products at the lower and upper exponents, with that error. Let \varepsilon\downarrow0, then let the exponents approach a,b. Continuity of D gives the claimed product through all real endpoints, with ordinary, unweighted counting.□

Proof of Corollary 1.2. For n\ge2, put

The empirical probability law means uniform sampling from 2\le n\le\lfloor X\rfloor; its normalization 1/(\lfloor X\rfloor-1) differs from 1/X by a factor tending to one. Recall the continuous distribution function D from (11.5).

Although 0\le U_n\le1, the second coordinate can slightly exceed one. This overshoot vanishes and does not alter the limiting law: for every n\ge N\ge2,

Consequently the empirical laws are tight and every limiting law is supported on [0,1]^2. More explicitly, clamp V_n to \widetilde V_n=\min(V_n,1). For 0<a,b<1, the joint distribution functions of (U_n,\widetilde V_n) are exactly those of (U_n,V_n), hence converge by Theorem 1.1 to D(a)D(b). These interior rectangles determine the limiting probability law: their masses approach one as a,b\uparrow1, and finite differences give the masses of all interior grid cells. Approximating continuous functions on [0,1]^2 by such grids proves weak convergence to the product law with marginal distribution function D. Removing the clamping does not affect this convergence, by the displayed bound and the vanishing proportion of n<N.

Let U,V be independent with distribution function D. Continuity gives \Prob(U=V)=0, and exchangeability of this limiting pair gives

The boundary of the set \{(u,v):u<v\} is the diagonal, which has zero product mass. Weak convergence therefore yields

Since both logarithmic sizes have the same positive denominator, U_n<V_n is exactly \Pplus(n)<\Pplus(n+1). Applying the same continuity-set argument to v<u gives the reverse ordering limit as well. The symmetry used here belongs to the independent limiting law; no symmetry of the finite consecutive-integer pairs is assumed.□