Hankel Determinants of Geometric Moments and Rational Lambert Values

Proof of Theorem 4.1. The source forms and polynomial cancellation. Fix $n\ge 1$ and set

In this proof $N$ denotes the cyclotomic cutoff. We use the coefficient pair $A_n,B_n$ of the source identities [4], with $H_n=A_{n}F-B_n$. Let ${\mathrm{\Phi }}_{\ell }$ be the cyclotomic polynomial whose roots are the primitive $\ell$th roots of unity. Put $D_N(X)=\prod _{\ell =1}^N{\mathrm{\Phi }}_{\ell }(X)$, $M_n=266n^2+34n+1$, and

where the periodic function

has period $1$: the coefficients of the floor arguments in each difference have equal sums, $14+13=12+15$ and $2\cdot 14=13+15$. On $[0,1)$ it is zero or one, with support

These are the intervals $\mathcal{I}$ used to define $J$. The function $\omega$, these thirteen intervals and the exponents ${\nu }_{\ell }=\omega (n/\ell )$ of the source’s (22) are printed at [4].

First isolate the part that does not create decay. At $x>1$, put $q=1/x$. Identity (9) on p. 156 of the source gives

The denominator length is $\beta -a_2$, as in the gamma expression (7) and residues (8) of [4]; the unnumbered display of $R(T)$ on that page has length $\beta -a_2-1$ instead. Each finite product lies in $[P,1]$, where $P=(q;q{)}_{\mathrm{\infty }}>0$. The $t=0$ term gives the lower bound; summing $q^{a_{0}t}$ gives the upper bound:

In particular $H_n(x)>0$ and $\log H_n(x)=O_x(1)$.

The required polynomial integrality is the inclusion in Zudilin’s Lemma 7, display (23) [4]. Its comparison of cyclotomic valuations under six admissible parameter permutations gives divisibility in $\mathbb{Z}[X]$, not merely at integer values of $X$. Its parameter vector is $n(13,14,12,14,15,13)$, its maximum is $15n$, and $\beta -a_1-a_2=n>0$, which is the positivity condition $s>0$ of the lemma. The conditions (14) of [4], $a_1\le a_2$ and $a_1+a_2\le \beta \le a_0+a_2$, hold as well, so the exponent in (23) is the integer (16), which equals $M_n$. Lemma 7 therefore gives, for every $n\ge 1$,

This is the polynomial conclusion before integer specialisation in the source’s (24), p. 162. The coefficients are the cancelled source coefficients:

Indeed, coefficients of $F$ and $1$ over $\mathbb{Q}(X)$ are unique because $F$ is not a rational function. To see this, we split $F(e^h)$ at $m=1/h$ as $h\downarrow 0$. For $m\le 1/h$, the inequalities $y^{-1}-1\le (e^y-1{)}^{-1}\le y^{-1}$ for $0<y\le 1$ give $h^{-1}\sum _{m\le 1/h}{m}^{-1}+O(h^{-1})$. Bounding the remaining terms by a geometric tail gives a further $O(h^{-1})$, and therefore $F(e^h)=h^{-1}\log (1/h)+O(h^{-1})$, which has no rational-function pole order at $X=1$. This establishes uniqueness of the coefficient functions; it assumes nothing about an individual value $F(a/b)$. The cancelled remainder is positive for $x>1$, since $D_N/{\mathrm{\Omega }}_n$ is a product of cyclotomic polynomials positive there.

The degrees after cancellation. The Gaussian binomial polynomial ${[\genfrac{}{}{0pt}{}{m}{k}]}_X=\prod _{j=1}^k(1-X^{m-k+j})/(1-X^j)$ is monic of degree $k(m-k)$ for integers $0\le k\le m$. The source’s (8) and (10) on p. 156 give the coefficient whose degree we need:

Let $d_{n,k}$ be the degree of the $k$th summand. Adding the monomial exponent and the degrees of the two Gaussian factors gives

Hence $d_{n,k+1}-d_{n,k}=40n+1-k>0$ for $a_2\le k\le \beta -2=27n$. The last summand alone therefore determines the leading term, giving

Here $\phi (\ell )=\mathrm{deg}{\mathrm{\Phi }}_{\ell }$ is Euler’s totient function, and ${\nu }_1=\omega (n)=0$. In particular,

The monomial removes $M_n$ degrees and the surviving cyclotomic product restores $\mathrm{deg}(D_N/{\mathrm{\Omega }}_n)$; their difference is the net removal $K_n-W_n$.

To determine the other coefficient’s degree, fix $n$ and let $x\to \mathrm{\infty }$. The bounds for $H_n$ are uniform for $x\ge 2$, since $P\ge \prod _{j\ge 1}(1-2^{-j})>0$ and $(1-x^{-a_0}{)}^{-1}\le 2$. Hence $H_n(x)=O(1)$ and ${\mathrm{\Lambda }}_n(x)=O(x^{W_n-K_n})$. The unique top summand of $A_n$ has leading coefficient $(-1{)}^{a_1+a_2+\beta }=(-1{)}^n$. Since the Gaussian factors and $D_N/{\mathrm{\Omega }}_n$ are monic, $U_n$ has that leading coefficient too. Moreover $F(x)=x^{-1}+O(x^{-2})$ and $K_n\ge 2$, so

Here $W_n\ge K_n-M_n=(559n^2+13n)/2>0$. Thus $\mathrm{deg}{V}_n=W_n-1$, with the same unit leading coefficient as $U_n$. For every prime $p\mid b$, reduction of the cleared first coordinate gives

Thus $U_n(a/b)$ has reduced denominator exactly $b^{W_n}$, which is the least common clearing denominator for the pair. Any common integer divisor of the cleared pair is coprime to $b$. This exact cost concerns these coefficients only: common factors at other primes and combinations of rows remain separate questions.

The quadratic degree limits. We now keep the base fixed and let $n\to \mathrm{\infty }$, extracting the quadratic part of the net degree removal. These are the cyclotomic limits of [4]. We use the reciprocal intervals, summatory totient estimate and trigamma function from [3], truncating the block sum before passing to the limit. The elementary summatory estimate $\sum _{\ell \le y}\phi (\ell )=3y^2/{\pi }^2+O(y\log y)$ gives

For $[u,v)\in \mathcal{I}$, the condition $\{n/\ell \}\in [u,v)$ is the disjoint union of intervals $n/(k+v)<\ell \le n/(k+u)$ for $k\ge 0$. Apply the summatory estimate to finitely many blocks. Instead of summing its errors over all $k$, bound the omitted blocks together: for $k\ge K\ge 1$, all their indices satisfy $\ell \le n/(K+u)$. The bound $\sum _{\ell \le y}\phi (\ell )\le y^2$ gives

For $y\ge 1$, sum $\phi (\ell )\le \ell$; for $0\le y<1$ the sum is empty. The tail bound is uniform in $n$, so we may let $n$ and then $K$ tend to infinity:

Here no contributing index exceeds $14n$, since $u\ge 1/14$. Consequently

The cleared remainder. It remains to turn polynomial degrees into values at fixed $x>1$. Writing ${\mu }_{\mathrm{M}\mathrm{o}\mathrm{b}}$ for the Möbius function, we have

Each $-\log (1-x^{-d})$ occurs only for multiples $\ell$ of $d$, at most $15n/d$ times. The total absolute error is therefore at most

The series converges because $-\log (1-x^{-d})\le x^{-d}/(1-x^{-1})$, so the error is $O_x(n)$. Using $0\le 1-{\nu }_{\ell }\le 1$ and $\log H_n(x)=O_x(1)$ gives

At $x=a/b$, the coefficients $b^{W_n}{U}_n(a/b)$ and $b^{W_n}{V}_n(a/b)$ are integers, and

The quadratic coefficient is negative under the hypothesis. If $F(a/b)=r/s$ with integers $r,s$ and $s>0$, then $s{b}^{W_n}{\mathrm{\Lambda }}_n(a/b)$ would be a positive integer tending to zero. This contradiction uses positivity of the same remainder, not independence of successive forms. ◻

Proof. The exact inequalities ${31}^2<4^5$ and $4^{200}<{31}^{81}$ give $2/5<\log 4/\log 31<81/200$. The first term of each trigamma difference yields

Using $\pi >157/50$ gives the rational lower bound

Taking a common positive integer power preserves both the logarithmic ratio and coprimality, so the same criterion applies for every $r\ge 1$. ◻

Proof. We use Zudilin’s passage from linear forms to an exponent bound [4], at each fixed base $(a/b{)}^r$. Take $U_n,V_n$ and $W_n=\mathrm{deg}{U}_n$ from Theorem 4.1. A measure needs more than positive remainders tending to zero: it must bound how small they are relative to their coefficients. The logarithmic remainder asymptotic supplies a lower bound; we now bound $U_n$ above. In that construction the coefficient of $F$ before cancellation is a sum of $O(n)$ Laurent monomials times two Gaussian binomial polynomials. Each Gaussian polynomial has nonnegative coefficients summing to at most $2^{27n+2}$. Thus the sum of the absolute coefficients is $\exp (O(n))$. Its largest exponent is $K_n=(1091n^2+81n+2)/2$, so its absolute value at a fixed $x>1$ is at most $x^{K_n}\exp (O(n))$. The cyclotomic estimate in the same proof bounds the normalising multiplier by $x^{W_n-K_n}\exp (O_x(n))$. Multiplication gives

Set $\xi =F(a/b)$, $Q_n=b^{W_n}{U}_n(a/b)$ and $P_n=b^{W_n}{V}_n(a/b)$. With

we have $\log (Q_n\xi -P_n)=-\tau n^2+o(n^2)$ and $|Q_n|\le \exp (\alpha n^2+o(n^2))$.

Here $q$ denotes an approximation denominator, not the geometric base of Sections 2 and 3. The comparison uses the integer $Ap-Bq$, allowing it to vanish. For integers $A,B,p,q$ with $q>0$, if $L=A\xi -B$ and $2q|L|\le 1$, then

If $Ap-Bq=0$, this is equality; otherwise $1/q\le |L|+|A||\xi -p/q|$ gives the inequality. Thus no rational $p/q$ is excluded from the comparison. Fix $0<\eta <\tau$ and write

for all sufficiently large $n$. The upper remainder bound determines our choice: for a sufficiently large denominator $q$, take $n=\lceil \sqrt{\log (2q)/(\tau -\eta )}\rceil$. Then $2q(Q_n\xi -P_n)\le 1$. The preceding integer argument now applies to $(A,B)=(Q_n,P_n)$ for every numerator $p$. Also $Q_n\ne 0$: otherwise $Q_n\xi -P_n$ would be an integer strictly between $0$ and $1$. The lower remainder bound and the coefficient bound now give

since $n^2=\log (2q)/(\tau -\eta )+O_{\eta }(\sqrt{\log q}+1)$. The rounding error is $o(\log q)$, so it does not change the exponent. This choice covers every sufficiently large approximation denominator $q$; no monotonicity assumption on the coefficients $Q_n$ is used.

Letting $\eta \downarrow 0$, we deduce the bound $1+\alpha /\tau =(1-\theta )/({\theta }^{\ast }-\theta )$. A common power multiplies $\alpha$ and $\tau$ by $r$, so the quotient is unchanged, although the constants in the approximation inequality may depend on $r$. For $31/4$, the exact interval bounds in the companion record, Section 2.5, give $\theta <0.4036982$ and ${\theta }^{\ast }>0.40568$. The quotient increases with $\theta$ and decreases with ${\theta }^{\ast }$ in this range, so

The same section gives the rational certificates for these bounds and sharper enclosures of the constants. ◻