Prime-Gap Dyadic Series: Perturbations, Exact Criteria and Certificates

Will Cook

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Prime-Gap Dyadic Series: Perturbations, Exact Criteria and Certificates

Erdős #251 By Will Cook September 2026 Authorship, AI use and citation

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Summary. A convergent dyadic series with nonnegative integer coefficients can be made rational by sparse nonnegative corrections of upper Banach density zero, eventually bounded by any f(n)→∞, while preserving every fixed modulus and, for a logarithmic allowance, short-block empirical distributions. For prime gaps the cumulative positions remain asymptotic to n log n; no primality conclusion is asserted. Summation by parts reduces the prime series to the gap series, and the tails $T_{N} = \sum_{j \geq 1} g_{N + j} 2^{- j}$ satisfy $T_{N + 1} = 2 T_{N} - g_{N + 1}$ . That recurrence characterises rationality and yields finite tests with explicit remainder bounds. An exact finite certificate shows that a rational value of the prime series would have denominator at least $2^{39998} > 10^{12040}$ . The bound does not establish irrationality.

This paper covers the complete problem-specific reasoning surface for Erdős #251, including all registered result families and their boundaries.

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In this paper

Density and block conventions.

All intervals of indices contain integers, and $\log$ is natural unless a base is displayed. Upper Banach density zero means

lim_{H \to \infty} sup_{u \in N} \frac{| S \cap [u, u + H) |}{H} = 0.

For integers $X \geq 1$ and $m \geq 1$ , let $μ_{a, X, m}$ be the empirical probability measure obtained by choosing an integer $n \in [X, 2 X)$ uniformly and observing $(a_{n}, \dots, a_{n + m - 1})$ . The entries are not rescaled, and equal blocks are counted once for each starting index. We use $d_{T V} (μ, ν) = sup_{B} | μ (B) - ν (B) |$ , with the supremum over all sets of blocks. For the same function of the block, bounded in absolute value by $B_{0}$ , its means under the two measures differ by at most $2 B_{0} d_{T V} (μ, ν)$ . No rescaling of the coefficients is implicit.

Theorem 2.1 (sparse changes preserving congruences). Let $a_{n} \in N$ and $A = \sum_{n \geq 0} a_{n} 2^{- n - 1} < \infty$ . For every cutoff $K$ and every $f : N \to R$ tending to infinity, there are a set $S \subseteq [K, \infty)$ of upper Banach density zero and a nondegenerate interval $I \subset (A, \infty)$ with the following property. Every $r \in I$ has the form

r = \sum_{n \geq 0} (a_{n} + e_{n}) 2^{- n - 1},

where $e_{n} \in N$ is supported on $S$ , $e_{n} \leq f (n)$ eventually, and, for every integer $q \geq 1$ , there is an $N_{q}$ , chosen independently of $r$ , such that

q ∣ e_{n} and q ∣ \sum_{i < n} e_{i} (n \geq N_{q}) .

For every $ε > 0$ , the construction can instead be chosen with $e_{n} \leq (\log (n + 3))^{ε}$ eventually and $| S \cap [X, 2 X) | = O_{ε} (X / \log \log X)$ . In that case the empirical distributions of original and corrected unnormalised blocks of length $m (X) = o (\log \log X)$ , sampled at the same integer starts in $[X, 2 X)$ , have total variation distance tending to zero, uniformly over $r$ . Precisely the coupling error is at most $m | S \cap [X, 2 X + m) | / X$ ; with $‖ Φ ‖_{\infty} \leq B_{0}$ the bounded-test error is at most twice this quantity times $B_{0}$ , also for tests depending on the starting index.

The hypotheses allow any nonnegative integer sequence of polynomial growth, but not $a_{n} = 2^{n}$ , whose dyadic series diverges. The bound on the corrections can grow as slowly as $\log \log (n + 3)$ . A fixed bound is not possible here: eventual boundedness together with eventual divisibility by every fixed modulus would make the corrections eventually zero; their fixed total would then have to be divisible by every modulus and hence be zero. Nonnegative corrections could not change the sum. The cutoff may depend on the modulus, but the common set $S$ and all cutoffs are fixed before the target. The support of a correction may occupy only part of $S$ . Upper Banach density zero is stronger than ordinary density zero. For example, the union of the integer intervals $[k!, k! + k)$ , $k \geq 3$ , has ordinary density zero but upper Banach density one. If $k! \leq x < (k + 1)!$ , at most $k (k + 1) / 2$ indices have occurred, so their proportion is at most $k (k + 1) / (2 k!) \to 0$ . Nevertheless each interval is filled and their lengths are unbounded. Upper Banach density zero rules out this concentration in long intervals, however far apart those intervals lie.

Proof. Fixed totals. Use $n_{j} - 1$ to cancel a cumulative residue and $n_{j}, n_{j} + 1$ for a pair with fixed ordinary sum. Varying the pair changes its weighted contribution without altering any later residue to be cancelled. Choose an initial index $n_{- 1} \geq K$ and centres with $s_{j} = n_{j} - n_{j - 1} \geq 4$ , so the triples do not overlap. Let $M_{j}$ be positive integers with $M_{j} ∣ M_{j + 1}$ , and let $D_{j}$ bound the digit at stage $j$ . The spacing, moduli and digit bounds will be chosen below.

Let $C_{j}$ record the ordinary correction before the $j$ th triple. Starting from $C_{0} = 0$ , prescribe

c_{j} = (- C_{j}) mod M_{j} (0 \leq c_{j} < M_{j}), C_{j + 1} = C_{j} + c_{j} + M_{j} D_{j} .

For integers $0 \leq d_{j} \leq D_{j}$ , set

\begin{matrix} (1) & e_{n_{j} - 1} = c_{j}, e_{n_{j}} = M_{j} d_{j}, e_{n_{j} + 1} = M_{j} (D_{j} - d_{j}), \end{matrix}

and put $e_{n} = 0$ elsewhere. The two sums of a triple are

\begin{aligned} \sum_{r = n_{j} - 1}^{n_{j} + 1} e_{r} & = c_{j} + M_{j} D_{j}, \\ \sum_{r = n_{j} - 1}^{n_{j} + 1} e_{r} 2^{- r - 1} & = c_{j} 2^{- n_{j}} + M_{j} D_{j} 2^{- n_{j} - 2} + d_{j} M_{j} 2^{- n_{j} - 2} . \end{aligned}

The ordinary total contains no $d_{j}$ , so the recurrence fixes every $c_{j}$ in advance. In the weighted total only the last term varies: transferring $M_{j}$ from $n_{j} + 1$ to $n_{j}$ adds

w_{j} = M_{j} 2^{- n_{j} - 2} .

After $c_{j}$ has been inserted, the cumulative correction is divisible by $M_{j}$ ; the two entries of the pair preserve this divisibility. Fix $q ∣ M_{J}$ . At index $n_{J}$ both the coefficient correction and the cumulative correction are divisible by $q$ . At every later triple, $q$ divides $C_{j}$ and $M_{j}$ , so it also divides $c_{j}$ and both pair entries. The intervening corrections are zero. Hence

q ∣ e_{n}, q ∣ \sum_{i < n} e_{i} (n \geq n_{J}) .

The entry $c_{J}$ need not be divisible by $q$ , which is why it is placed at $n_{J} - 1$ , before the cutoff. The index $n_{J}$ depends on $q$ and the chosen moduli, but not on the digits or the value to be attained.

Spacing and size. The choices at stage $j$ are $0, w_{j}, \dots, D_{j} w_{j}$ . The covering argument will work if the later ranges total at least $w_{j}$ . Taking $D_{i} = 2^{s_{i}} - 1$ compensates for separation by making each later range a scaled dyadic difference:

D_{i} w_{i} = (2^{n_{i} - n_{i - 1}} - 1) M_{i} 2^{- n_{i} - 2} = M_{i} (2^{- n_{i - 1} - 2} - 2^{- n_{i} - 2}) .

Because $M_{i} \geq M_{j}$ for $i > j$ , replacing every later modulus by $M_{j}$ gives the finite estimate

\begin{matrix} (2) & \begin{aligned} \sum_{i = j + 1}^{J} D_{i} w_{i} & \geq M_{j} \sum_{i = j + 1}^{J} (2^{- n_{i - 1} - 2} - 2^{- n_{i} - 2}) \\ = w_{j} - M_{j} 2^{- n_{J} - 2} (J > j) . \end{aligned} \end{matrix}

No monotonicity of the weights is needed. Each entry of (1) is at most $M_{j} 2^{s_{j}}$ , so the size cost of separating the stages is exponential. We must fit this product below the allowance at all three indices, while $s_{j} \to \infty$ and every fixed positive integer eventually divides $M_{j}$ .

To allow a nonmonotone bound $f$ , put

h (n) = inf_{m \geq n} min (f (m), m) .

This nondecreasing lower envelope tends to infinity even when $f$ oscillates. A bound at $n_{j - 1}$ therefore controls all three later coordinates; the cap by $m$ ensures $e_{n} \leq n$ and hence summability for every digit choice. Enlarge $n_{- 1} \geq K$ until $h (n_{- 1}) \geq 32$ , and set

\begin{matrix} (3) & \begin{aligned} k_{j} & = max {k \geq 2 : k! 2^{k + 2} \leq h (n_{j - 1})}, \\ M_{j} & = k_{j}!, s_{j} = k_{j} + 2, n_{j} = n_{j - 1} + s_{j} . \end{aligned} \end{matrix}

The maximum exists: $k = 2$ is admissible because $2! 2^{4} = 32$ , and $k! 2^{k + 2} \to \infty$ . Since $n_{j} \to \infty$ and $h$ is nondecreasing and divergent, $k_{j}$ is nondecreasing and tends to infinity. Thus $M_{j} ∣ M_{j + 1}$ , every fixed $q$ eventually divides $M_{j} = k_{j}!$ , and $s_{j} = k_{j} + 2 \to \infty$ . At every coordinate $n$ of the $j$ th triple, this schedule also gives

0 \leq e_{n} \leq M_{j} 2^{s_{j}} \leq h (n_{j - 1}) \leq min (f (n), n) .

Let $S = ⋃_{j \geq 0} {n_{j} - 1, n_{j}, n_{j} + 1}$ . For $R \geq 4$ , all sufficiently late centres are at least $R$ apart. Every interval of length $H$ contains at most $3 (H / R + 2)$ points from those triples; the earlier triples add a fixed number, independent of the interval’s position. Divide by $H$ , take the supremum over positions, then let $H \to \infty$ . The limiting upper bound is $3 / R$ ; letting $R \to \infty$ proves upper Banach density zero.

Attaining the interval. The support and congruence requirements are now settled independently of the target. Only the digits remain to be chosen. Define

β = \sum_{j \geq 0} (c_{j} 2^{- n_{j}} + D_{j} w_{j}), F_{j} = \sum_{i \geq j} D_{i} w_{i} .

Even zero digits contribute $β > 0$ ; $F_{j}$ is the maximum variable contribution from stage $j$ onwards. The size bound makes both finite, with $F_{0} > 0$ and $F_{j} \to 0$ . The total correction is

\begin{matrix} (4) & \sum_{n} e_{n} 2^{- n - 1} = β + \sum_{j} d_{j} w_{j} . \end{matrix}

Taking $J \to \infty$ in (2) now gives

\begin{matrix} (5) & F_{j + 1} \geq M_{j} 2^{- n_{j} - 2} = w_{j} . \end{matrix}

These bounds do not yet prove that every point of $[0, F_{j}]$ is attained. A digit $d$ leaves a remainder in $[0, F_{j + 1}]$ precisely for current remainders in

[d w_{j}, d w_{j} + F_{j + 1}], d = 0, \dots, D_{j} .

These intervals meet or overlap by (5), and their union is $[0, F_{j}]$ because $F_{j} = D_{j} w_{j} + F_{j + 1}$ .

Given $x \in [0, F_{0}]$ , start with $ρ_{0} = x$ . Having chosen the first $j$ digits, choose $d_{j}$ so that

ρ_{j + 1} = ρ_{j} - d_{j} w_{j} \in [0, F_{j + 1}] .

The covering just proved makes this possible at every stage. Since $0 \leq ρ_{j} \leq F_{j}$ and $F_{j} \to 0$ , we obtain

x - \sum_{i = 0}^{j} d_{i} w_{i} = ρ_{j + 1} ⟶ 0.

Thus every $x \in [0, F_{0}]$ is attained, and we may take $I = (A + β, A + β + F_{0})$ . The target $r \in I$ determines only the digits, through $x = r - A - β$ ; it changes neither $S$ , $I$ nor the cutoff $N_{q} = n_{J}$ chosen using $q ∣ M_{J}$ . This is the finite-choice covering argument of [27]: the $j$ th choice set has largest gap $w_{j}$ and diameter $D_{j} w_{j}$ , its maxima are summable, and the sum of subsequent diameters is at least $w_{j}$ . Equality is allowed. Fridy’s generalised-base lemma [9] and the reciprocal-choice argument in [11] are antecedents. The finite-choice argument uses (5); the required comparison of moduli was checked in (2).

The logarithmic scale. For $f (n) = (\log (n + 3))^{ε}$ , the size cost $2^{s_{j}}$ leads to the iterated logarithm. We choose separations of order $\log \log n_{j}$ and reserve part of the allowance for growing moduli. Keep the same triples and $D_{j} = 2^{s_{j}} - 1$ , and set

\begin{matrix} (6) & \begin{aligned} s_{j} & = ⌊ \frac{ε}{2} \log_{2} \log (n_{j - 1} + 3) ⌋, \\ k_{j} & = max {k \geq 2 : k! \leq (\log (n_{j - 1} + 3))^{ε / 4}}, \end{aligned} \end{matrix}

with $M_{j} = k_{j}!$ and $n_{j} = n_{j - 1} + s_{j}$ . Start far enough out that $s_{j} \geq 4$ and all choices exist. The exponents $ε / 4$ and $ε / 2$ add to $3 ε / 4 < ε$ , so the corrections satisfy the prescribed bound:

M_{j} 2^{s_{j}} \leq (\log (n_{j - 1} + 3))^{3 ε / 4} \leq (\log (n + 3))^{ε}

at each coordinate $n$ of the triple. After enlarging the initial index, this is also at most $n$ . The moduli remain nested, and every fixed positive integer divides all sufficiently late moduli. Summability, the congruence induction and (2) therefore apply without change. Since $s_{j} ≍_{ε} \log \log n_{j}$ , the relevant centres near $[X, 2 X)$ are separated by $≫_{ε} \log \log X$ . Each contributes at most three support points, giving

| S \cap [X, 2 X) | = O_{ε} (X / \log \log X) .

Choose the same uniform starting index $n \in [X, 2 X)$ for both blocks. They agree when the block avoids $S$ . A changed coordinate $t \in S \cap [X, 2 X + m)$ affects only starts in $[t - m + 1, t]$ , at most $m$ choices. The union bound gives

\begin{aligned} d_{T V} (μ_{a, X, m}, μ_{a + e, X, m}) & \leq P (e_{n + i} \neq 0 for some 0 \leq i < m) \\ \leq \frac{m | S \cap [X, 2 X + m) |}{X} . \end{aligned}

For $m \leq X$ , two dyadic bands cover $[X, 2 X + m)$ , giving $O_{ε} (m / \log \log X)$ . A block of length $o (\log \log X)$ therefore avoids $S$ with probability tending to one. The convergence is uniform over targets because the same $S$ contains every correction support. For a test bounded in absolute value by $B_{0}$ that also depends on the starting index, the coupled values agree at unchanged blocks and differ by at most $2 B_{0}$ elsewhere. Their mean difference is consequently at most $2 B_{0} m | S \cap [X, 2 X + m) | / X$ . This uses the common starting index, not merely the marginal distributions. No limiting distribution, randomness or independence of the original coefficients is assumed. ◻

Identical marginal distributions alone would not control an index-dependent test. For example, let $a_{n} = n mod 2$ and $b_{n} = 1 - a_{n}$ . For every even $X$ , their one-coordinate empirical distributions on $[X, 2 X)$ are equal, but the test $1_{{x = n mod 2}}$ has mean $1$ for $(n, a_{n})$ and $0$ for $(n, b_{n})$ . This example is not a sparse perturbation; it isolates why the proof couples blocks at the same index.

The estimate gives a sufficient scale, not an optimal threshold. It tends to zero for fixed block lengths and for $m (X) = ⌊ \sqrt{\log \log X} ⌋$ , but gives no vanishing error for $m$ comparable to $\log \log X$ . Its error is also absolute, not relative: probabilities $1 / \log X$ and $0$ differ by $o (1)$ despite the disappearance of the event. Relative preservation would require an error small compared with the event’s probability. Finally, agreement on a finite block does not identify its complete weighted tail; later coefficients still contribute.

Corollary 2.2 (target intervals arbitrarily close to the original sum). For the sparse theorem for an arbitrary sequence, the target interval can additionally be required to lie in $(A, A + η)$ for any prescribed $η > 0$ .

Proof. In the construction for a general bound $f$ , every nonzero correction satisfies $e_{n} \leq n$ . Choose an integer $K^{'} \geq K$ sufficiently large that $\sum_{n \geq K^{'}} n 2^{- (n + 1)} = (K^{'} + 1) 2^{- K^{'}} < η$ , and start the construction beyond $K^{'}$ . This still leaves the originally prescribed prefix unchanged. The complete interval of corrections has positive lower endpoint and upper endpoint at most this sum. Here $η$ is chosen before the construction, so the common support and interval may depend on $η$ . Within that interval only the digits depend on the target. This does not make every target above $A$ attainable under one fixed size bound. ◻

About this paper

Authorship and AI use. Will Cook built and directed the research infrastructure and maintains the public release. He reviewed claims when he could. AI agents did most of the research and drafting. Cook did not independently verify every claim.

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