Erdős problem #251

The prime-gap dyadic series

Is the dyadic series of consecutive primes irrational? Equivalently, is the corresponding consecutive-prime-gap dyadic series irrational?

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Read the short paper PDFOpen the PDF, 14 pages Map of results

The problem

The question

List the primes as $p_{1} = 2, p_{2} = 3, p_{3} = 5, \dots$ . Divide them by $2, 4, 8, \dots$ and add: $P = \sum_{n \geq 1} p_{n} / 2^{n}$ . Erdős asks whether this exact infinite total can be a fraction. The denominators grow fast enough for the sum to converge; its rationality is open.

An example

The first five terms give $2 / 2 + 3 / 4 + 5 / 8 + 7 / 16 + 11 / 32 = 101 / 32$ . Later terms increase the sum to about $3.6746439$ . The gap from 3 to 5 is $5 - 3 = 2$ ; that same extra 2 appears in every prime from 5 onwards.

Is $\sum \frac{p_{n}}{2^{n}}$ irrational? (Here $p_{n}$ is the $n$ th prime.)

Erdős problem #251, as erdosproblems.com/251 states it

The setup in detail

Write the primes in increasing order as $p_{1} = 2, p_{2} = 3, p_{3} = 5, \dots$ . The series is $P = \sum_{n \geq 1} p_{n} 2^{- n}$ : each next denominator is twice the previous one. The question is whether the limit of these running totals is rational. Every finite running total is a fraction, but the limit of rational numbers can be irrational.

The long record proves the elementary bound $p_{n} \leq 1250 n^{4}$ . This is enough for convergence: the ratio of consecutive bounds is

\frac{1250 (n + 1)^{4} / 2^{n + 1}}{1250 n^{4} / 2^{n}} = \frac{1}{2} (1 + 1 / n)^{4} < \frac{3}{4} (n \geq 10) .

The remaining terms are therefore bounded by a convergent geometric series. Finite changes to the primes alter the total by a rational number, so they cannot change its rationality. The difficulty lies in controlling all later terms, rather than computing a longer initial segment.

The example in detail

Putting the first five terms over a common denominator gives

\frac{2}{2} + \frac{3}{4} + \frac{5}{8} + \frac{7}{16} + \frac{11}{32} = \frac{32 + 24 + 20 + 14 + 11}{32} = \frac{101}{32} = 3.15625 .

The next terms are $13 / 64, 17 / 128, 19 / 256, \dots$ ; the full total is about $3.6746439$ . This gives its size, without deciding rationality.

Define the gaps by $g_{n} = p_{n + 1} - p_{n}$ . Their first values are $1, 2, 2, 4, 2, \dots$ , and each prime is 2 plus its preceding gaps: $11 = 2 + 1 + 2 + 2 + 4$ . In the drawing this is the height of the column for 11, split into the floor and four horizontal bands.

The highlighted gap from 3 to 5 contributes 2 to every prime from 5 onwards. Its contribution to the sum is therefore

\frac{2}{8} + \frac{2}{16} + \frac{2}{32} + \dots = 2 (\frac{1}{8} + \frac{1}{16} + \dots) = \frac{2}{4} .

The successive widths halve, and their running total approaches one quarter. This explains the red band’s width and its area, rather than introducing a new quantity alongside the prime sum.

The crux

Column $n$ has height $p_{n}$ and width $2^{- n}$ , so its area is one term. Read the same staircase horizontally: a floor of height 2, then one band per prime gap. The red band has height 2 and width $1 / 8 + 1 / 16 + \dots = 1 / 4$ , hence area $2 / 4$ .

Writing $g_{n} = p_{n + 1} - p_{n}$ , this gives $P = 2 + \sum_{n \geq 1} g_{n} / 2^{n}$ . Either both sums are rational or neither is. The theorem below constructs altered gaps with rational sums while retaining many gap properties. Their running totals need not be primes: the original question remains open.

One area, two sums. The column over the first half of the ruler has height 2, the next quarter has height 3, and so on. The sum of their areas is the prime series. Read horizontally instead: the floor contributes $2 (1 / 2 + 1 / 4 + \dots) = 2$ . The band for gap $g_{n}$ has width $2^{- n - 1} + 2^{- n - 2} + \dots = 2^{- n}$ . Consequently

P = 2 + \sum_{n \geq 1} g_{n} 2^{- n} .

All areas are nonnegative, so this regrouping preserves the total. Adding the rational number 2 does not change rationality. It is therefore enough to study the gap sum, but it is still the same open question.

What the theorem tests. Could broad properties of the gaps alone force irrationality? Add six units to two neighbouring gaps. The four splits give different weighted changes:

\frac{0}{8} + \frac{6}{16} = \frac{6}{16}, \frac{2}{8} + \frac{4}{16} = \frac{8}{16}, \frac{4}{8} + \frac{2}{16} = \frac{10}{16}, \frac{6}{8} + \frac{0}{16} = \frac{12}{16} .

Every later running total rises by six whichever split is chosen. The choice therefore changes the weighted sum without changing the later remainders modulo 2, 3 or 6. The next diagram repeats this operation at increasingly separated indices and with larger divisibility requirements.

The proved statement. Start with any convergent series $A = \sum_{n \geq 0} a_{n} 2^{- n - 1}$ of nonnegative integers, any initial cutoff $K$ , and any bound $f (n) \to + \infty$ . One fixed set of allowed indices after the cutoff supports corrections attaining every target in a nondegenerate interval above A. The corrections are nonnegative integers and satisfy $e_{n} \leq f (n)$ eventually. The allowed indices occupy a proportion tending to zero uniformly over long intervals. For each fixed positive integer q, beyond a cutoff depending on q, every correction and every cumulative correction is divisible by q. The allowed set, target interval and cutoffs are chosen before the target.

The prime boundary. Applied to prime gaps, with eventual bound $\log (n + 3)$ , this gives altered gaps with rational sums, the same eventual residues for each fixed modulus, and the same limiting distributions of fixed-length consecutive gap blocks. Their running totals still grow like $n \log n$ . These totals need not be prime: for instance, adding six replaces 19 by 25. The construction shows that the retained properties are insufficient; it proves no rationality or irrationality assertion for the actual prime series.

We construct comparison gaps with rational sums. Split six added units as $(2, 4)$ between neighbouring entries. The earlier entry has twice the weight: their contribution is $2 \cdot 2 + 4 = 8$ small units. The staircase area shows this sum. Move two units earlier. The split $(4, 2)$ contributes $2 \cdot 4 + 2 = 10$ . The area increases, but every later running total rises by six, whichever split is chosen. At indices 192 and 193, one small unit is $2^{- 194}$ . Splitting 30 added units with 14 earlier gives $30 + 14 = 44$ small units. Each bar includes all later choices. Axis zero places every movable unit later. Move two units earlier again: $30 + 16 = 46$ . The red cell adds one step of area, $w_{0} = 2^{- 193}$ . The corresponding bar shifts right by exactly that step. We use sixteen splits, putting $0, 2, \dots, 30$ units earlier. Each bar is wider than the step between neighbours, so consecutive bars overlap. Their union, shown underneath, covers one whole interval. The red dot is one target. Choose window 8, enlarge it, then choose a smaller window containing the same target. The next 32 windows also overlap, by amounts too small to see at this scale. Repeat indefinitely. The remaining uncertainty tends to zero, so the sum equals the target. Every target in the interval is attainable, including rational values. The construction keeps changes sparse and preserves eventual residues modulo each fixed integer. Altered prime gaps give comparison sequences; their running totals need not be primes.

Proposition 1.1 starts with a nonnegative integer sequence $a_{n}$ whose series $A = \sum_{n \geq 0} a_{n} 2^{- n - 1}$ converges, a starting index $K$ , and any real bound $f (n) \to + \infty$ . It constructs a fixed set $S \subseteq [K, \infty)$ of upper Banach density zero and a nondegenerate interval $I$ above $A$ . Every target $r \in I$ is attained by nonnegative integer corrections $e_{n}$ supported on $S$ , with $e_{n} \leq f (n)$ eventually. For each positive integer $q$ , both $e_{n}$ and $\sum_{i < n} e_{i}$ are divisible by $q$ after a cutoff independent of $r$ . Upper Banach density zero means the proportion of allowed correction indices tends to zero uniformly over long intervals.

The staircase height $E_{m} = \sum_{i < m} e_{i}$ is the cumulative correction before $m$ . Slot $m$ has width $2^{- m - 1}$ , and $\sum_{m} E_{m} 2^{- m - 1} = \sum_{n} e_{n} 2^{- n - 1}$ , since $\sum_{m > n} 2^{- m - 1} = 2^{- n - 1}$ . For six units shared between $n$ and $n + 1$ , the scaled contribution is $2 e_{n} + e_{n + 1} = 6 + e_{n}$ . This explains both the staircase area and the increase from 8 to 10.

The drawing uses $f (n) = n$ and initial index 188. Its first pair is $(2 d, 30 - 2 d)$ , for $d = 0, \dots, 15$ , at indices 192 and 193. $B$ denotes the original sum $A$ plus every fixed congruence correction and every variable pair placed entirely at its later index. A choice $d$ adds $d w_{0}$ above $B$ , where $w_{0} = 2^{- 193}$ . Later choices add between 0 and $F_{1}$ , where $F_{j} = \sum_{i \geq j} D_{i} w_{i}$ . The construction ensures $F_{j + 1} \geq w_{j}$ : adjacent choice intervals overlap, so recursively choosing an interval containing the target leaves a remainder tending to zero. In this example $F_{1}$ is slightly larger than $3 w_{0}$ ; the next step is $w_{1} = 3 w_{0} / 32$ . The marked target selects digits $8, 6, 6, 12, \dots$ . This marked target illustrates the choices; it is not asserted to be rational. To obtain a rational sum, choose a rational target in the covered interval. Later overlaps are positive but below drawing precision.

Fixed pair totals make every later cumulative correction independent of the target choices. A separate correction just before each pair clears its residue modulo a nested factorial modulus. These moduli eventually contain every $q$ as a divisor, while increasingly separated stages make $S$ sparse. The printed proof uses triples; the Lean proof reaches Proposition 1.1 by changing one coordinate per stage. Corrected prime gaps need not be gaps between primes, so this theorem does not settle the irrationality of the original consecutive-prime series in Erdős #251.

Results and open questions

#251. A public Lean declaration constructs a synthetic sequence of positive even digits bounded by $4 \log (n + 1) + 24$ . Its dyadic sum is exactly $6$ , every complete scaled tail and every tail shift is integral, the values $2$ and $4$ recur arbitrarily late in every residue class, and its increasing odd cumulative positions satisfy $P_{n} / (n \log n) \to 1$ . The reasoning paper states the combined theorem explicitly. This rules out a much richer package of plausible soft inputs than unboundedness or nonperiodicity alone, but these digits are not the actual prime gaps. The declaration is source-backed and does not yet have a curated claim row.

Independently, a sparse perturbation of the prime gaps can have a rational dyadic sum while retaining the prime growth scale, every fixed eventual congruence, and asymptotically the same short-block statistics. The later positions are not asserted to be prime. This is complementary to Land’s conditional result, not a refutation. Lean checks the sparse rational-target construction and the polylogarithmic schedule with growing-block transfer in SparsePaperR11.lean. The printed construction additionally gives congruence cutoffs uniform in the target; the linked Lean statement quantifies those cutoffs after the target.

The prime-growth corollary uses cited analytic results. This obstruction shows that those coarse statistics alone do not force irrationality. Lean also checks a precise conditional route: for one fixed shift, cofinally many adjacent tail shifts strictly between $- 1$ and $1$ , with unequal actual prime gaps, would prevent eventual integrality of that shift. Those prime-specific mismatches are not proved. The exact prime-gap summation-by-parts equivalence still proves neither the prime-gap series nor the original series irrational. Read the short paper, long record, and selected formal checks.

Read the public result guide

Short paper, 14 pages

Sparse Congruence-Preserving Perturbations of Dyadic Series

PDFOpen the PDF, 14 pages Open as its own page Current source

In this paper

Introduction

Let $p_{0} = 2, p_{1} = 3, \dots$ be the primes and put $g_{n} = p_{n + 1} - p_{n}$ . Erdős asked whether

Π = \sum_{n \geq 0} \frac{p_{n}}{2^{n + 1}}

is irrational [1][2] [3]. Summation by parts gives $Π = 2 + \sum_{n \geq 0} g_{n} 2^{- (n + 1)}$ , with convergence justified in Section 4. We study which properties of the gaps survive when the latter sum is changed to a prescribed value.

The support and every modulus cutoff are fixed before the target is chosen. The interval contains rational and irrational sums with the same eventual coefficient and partial-sum residues. A block changes only when it meets the support, whereas every correction enters the dyadic sum. With polylogarithmic corrections, empirical block laws agree asymptotically at lengths $o (\log \log X)$ ; for prime gaps, cumulative growth remains $n \log n$ . The comparison positions need not be consecutive primes.

We use a standard interval-covering argument for series with finite choices. Fridy’s generalised-base lemma [5] treats prescribed digit bounds and nonincreasing weights. Crmarić and Kovač [17] give the finite-choice form used here, and Kovač and Tao [6] use analogous intervals of reciprocal choices. We must arrange these choices on a sparse set while preserving cumulative congruences. To do so, we redistribute a fixed correction between two adjacent indices. The simplest useful calculation is

\begin{array}{crrrr} (e_{n}, e_{n + 1}) & (0, 6) & (2, 4) & (4, 2) & (6, 0) \\ e_{n} + e_{n + 1} & 6 & 6 & 6 & 6 \\ 2^{n + 2} (e_{n} 2^{- n - 1} + e_{n + 1} 2^{- n - 2}) & 6 & 8 & 10 & 12. \end{array}

Moving $2$ from the later index to the earlier one changes the weighted sum while leaving the ordinary sum fixed. If the cumulative correction before the pair is odd, an additional $1$ at $n - 1$ clears its residue modulo $2$ . Since the four pairs have the same sum, the choice of pair leaves every later cumulative residue unchanged. We repeat this operation with nested moduli and increasingly separated pairs. The larger separations make the weighted contributions smaller, so we enlarge the range of each pair until the later contributions fill the gaps between consecutive present choices.

Throughout, $N = {0, 1, \dots}$ , intervals of indices contain integers, and $\log$ is natural unless a base is displayed. A set $S \subseteq N$ has upper Banach density zero if

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

For integers $X, m \geq 1$ , let $μ_{a, X, m}$ denote the distribution of $(a_{n}, \dots, a_{n + m - 1})$ when $n$ is uniform on $[X, 2 X)$ . The entries are not rescaled: we sample starting indices, not distinct block values. We use $d_{T V} (μ, ν) = sup_{B} | μ (B) - ν (B) |$ , with the supremum over all sets of blocks.

Proposition 1.1 (sparse changes preserving congruences). Let $a : N \to N$ satisfy $A = \sum_{n \geq 0} a_{n} 2^{- (n + 1)} < \infty$ , let $K \in N$ , and let $f : N \to R$ tend to $+ \infty$ . There exist a set $S \subseteq [K, \infty)$ of upper Banach density zero and a nondegenerate interval $I \subset (A, \infty)$ such that, for every $r \in I$ , there is an integer correction $e : N \to N$ satisfying

supp e \subseteq S, e_{n} \leq f (n) eventually, \sum_{n \geq 0} (a_{n} + e_{n}) 2^{- (n + 1)} = r .

For each integer $q \geq 1$ there is a cutoff $N_{q}$ , independent of $r$ , such that $q ∣ e_{n}$ and $q ∣ \sum_{i < n} e_{i}$ for all $n \geq N_{q}$ . For every $ε > 0$ , the choice $f (n) = (\log (n + 3))^{ε}$ can be made with

| S \cap [X, 2 X) | = O_{ε} (\frac{X}{\log \log X}), d_{T V} (μ_{a, X, m}, μ_{a + e, X, m}) \leq \frac{m | S \cap [X, 2 X + m) |}{X} .

In particular the distance tends to zero for every integer-valued $m = m (X) \geq 1$ with $m (X) = o (\log \log X)$ , uniformly over target values.

The cutoff may depend on $q$ , but $S$ and all cutoffs are fixed before $r$ . Only $e$ varies with the target, and its support may occupy just part of $S$ . Infinitely many changes are essential: finitely many integer dyadic corrections add a rational number. The allowance may grow as slowly as $\log \log (n + 3)$ , but it cannot be bounded. A modulus exceeding the bound would force $e$ eventually to vanish; its constant cumulative sum would then be divisible by every positive integer. Nonnegativity forces $e = 0$ . Polynomially growing coefficients satisfy the convergence hypothesis; $a_{n} = 2^{n}$ does not.

Section 2 proves the construction and Section 3 transfers the stated prime-gap properties. For the actual primes, integral tail shifts characterise rationality. Two consecutive small differences with unequal gaps give a sufficient condition for irrationality. Finite enclosures certify individual witnesses without supplying arbitrarily late ones. The companion’s literature discussion contains the further comparisons, including finite subsums and other dyadic coefficient sequences.

Proof of Proposition 1.1

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

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} .

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

\sum_{n} e_{n} 2^{- n - 1} = β + \sum_{j} d_{j} w_{j} .

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

\begin{matrix} (3) & 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 (3), and their union is $[0, F_{j}]$ because $F_{j} = D_{j} w_{j} + F_{j + 1}$ . Figure 1 shows the two calculations.

This figure is drawn in the PDF.

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 last selection is the standard finite-choice covering argument of [17]. The construction above supplies its summability and overlap hypotheses while imposing the congruences and requiring the support to lie in $S$ .

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{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}

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. ◻

No limiting block law is assumed. A block-only test bounded in absolute value by $B_{0}$ changes its mean by at most $2 B_{0} d_{T V}$ . For such a test also depending on the starting index, the common-index coupling gives $2 B_{0} m | S \cap [X, 2 X + m) | / X$ instead. Vanishing absolute error can still erase a rare event; it supplies neither relative-frequency control nor a comparison of complete tails. The scale $o (\log \log X)$ is sufficient, not proved optimal. The companion’s Section 2 also distinguishes upper Banach density from ordinary density.

Application to prime gaps

The elementary bound $p_{n} \leq 1250 (n + 1)^{4}$ suffices for convergence of the prime and gap series. We use the central-binomial argument of Erdős [19]; Appendix A of the companion derives the prime-power estimate and the bound. We may therefore apply Proposition 1.1 to the gaps.

Corollary 3.1 (a rational sum with the stated prime-gap statistics). Let $p_{0} = 2, p_{1} = 3, \dots$ be the primes and $g_{n} = p_{n + 1} - p_{n}$ . Given $K \in N$ and $0 < ε \leq 1$ , there is $b : N \to N$ with rational dyadic sum such that $b_{n} = g_{n}$ for $n < K$ , $b_{n} \geq g_{n}$ , and $b_{n} - g_{n} \leq (\log (n + 3))^{ε}$ eventually. For every fixed positive modulus, both the coefficients and the cumulative positions eventually retain their corresponding residues. The empirical distributions of unnormalised blocks have total variation distance tending to zero for lengths $o (\log \log X)$ , and for every fixed nonzero $F \in Z [x_{0}, \dots, x_{k}]$ ,

| {n < N : F (b_{n}, \dots, b_{n + k}) = 0} | = o (N) .

Moreover, with $P_{n} = 2 + \sum_{i < n} b_{i}$ ,

0 \leq P_{n} - p_{n} = O_{ε} (\frac{n (\log (n + 3))^{ε}}{\log \log n}), P_{n} \sim n \log n .

These positions are not asserted to be prime.

Proof. Apply Proposition 1.1 to $a = g$ , choose a rational $r \in I$ , and fix the resulting sequence $b = g + e$ . The correction, congruence and growing-block assertions follow at once. It remains to transfer polynomial nonconcentration and cumulative growth. For fixed $k$ and nonzero $F \in Z [x_{0}, \dots, x_{k}]$ , a new zero requires a block meeting $S$ , so

\begin{aligned} | {n < N : F (b_{n}, \dots, b_{n + k}) = 0} | \\ \leq | {n < N : F (g_{n}, \dots, g_{n + k}) = 0} | + \sum_{i = 0}^{k} | S \cap [i, N + i) | . \end{aligned}

The first count is $o (N)$ by Schlage-Puchta’s Lemma 4 [18]; the finitely many support counts are $o (N)$ because $S$ has density zero. We count the event $F = 0$ , not averages of $F$ , so no bound on its values is needed. Since $b$ was chosen before $F, k$ , the same sequence satisfies every fixed-polynomial condition.

To estimate $P_{n} - p_{n} = \sum_{i < n} e_{i}$ , first observe that $| S \cap [0, n) | = O_{ε} (n / \log \log n)$ . Indeed, the indices below $\sqrt{n}$ contribute at most $\sqrt{n}$ points, and on $[\sqrt{n}, n)$ we sum the dyadic support bounds, whose lengths add to $O (n)$ and whose $\log \log X$ are comparable to $\log \log n$ . Multiplying by the pointwise correction bound gives the stated error. For $0 < ε \leq 1$ this is $o (n \log n)$ , so the prime number theorem [12] gives $P_{n} \sim n \log n$ . Finally, summing nonnegative terms in the opposite order, we obtain

\sum_{n \geq 0} \frac{P_{n}}{2^{n + 1}} = 2 + \sum_{i \geq 0} b_{i} \sum_{n > i} 2^{- n - 1} = 2 + \sum_{i \geq 0} \frac{b_{i}}{2^{i + 1}} = 2 + r \in Q .

◻

For $ε > 1$ we may use exponent $1$ , which gives smaller corrections and the error $O (n \log (n + 3) / \log \log n)$ . Thus the conclusions extend to every $ε > 0$ . For each fixed $y \geq 2$ , eventual congruence modulo $y!$ also gives $gcd (P_{n}, y!) = 1$ once $p_{n} > y$ . The cutoff depends on $y$ ; primality would require excluding every prime up to $\sqrt{P_{n}}$ at that same index. Even prime-valued positions could omit intervening primes.

Land’s draft [8] proves conditional irrationality under a uniform Hardy–Littlewood hypothesis of the kind formulated by Kuperberg [7]. Ringer’s draft [16] obtains conditional normality. Their quantitative prime-pattern hypotheses are stronger than the absolute block comparison above and are not assumed here. The companion’s Section 3 compares the hypotheses and sampling conventions.

The prime series and its actual tails

We now study complete tails of the actual prime gaps. Put $G = \sum_{n \geq 0} g_{n} 2^{- (n + 1)}$ . The polynomial bound from Section 3 proves convergence and makes the endpoint in finite summation by parts vanish.

Theorem 4.1 (prime-to-gap identity). The actual prime and gap series satisfy $Π = 2 + G$ . Their complete tails

T_{N} = \sum_{j \geq 1} g_{N + j} 2^{- j}

satisfy $T_{0} = 2 G - 1$ and $T_{N + 1} = 2 T_{N} - g_{N + 1}$ . Thus $Π$ , $G$ and $T_{0}$ have the same rationality status.

Proof. Finite summation by parts gives

\sum_{i = 0}^{n} \frac{p_{i}}{2^{i + 1}} = 2 + \sum_{i = 0}^{n - 1} \frac{g_{i}}{2^{i + 1}} - \frac{p_{n}}{2^{n + 1}} .

Since $p_{n} = O ((n + 1)^{4})$ , the endpoint tends to zero and we obtain $Π = 2 + G$ . We obtain the tail recurrence by splitting off the first term of $T_{N}$ ; removing $g_{0} = 1$ from $G$ gives $T_{0} = 2 G - 1$ . ◻

Tao noted this reduction in the problem’s forum discussion [15].

Corollary 4.2 (exact irrationality reformulation). The prime-value dyadic series is irrational if and only if the prime-gap dyadic series is. Both series converge by the polynomial bound proved above; neither side is proved irrational.

Proof. Adding the rational number $2$ preserves rationality and irrationality. ◻

The index $N$ is the last omitted coefficient, so $T_{N}$ starts with $g_{N + 1} / 2$ . Rescaling distinguishes these tails from the remainders of $G$ . The recurrence also admits an extra term $C 2^{N}$ ; the vanishing boundary condition below excludes it.

Lemma 4.3 (the boundary condition for the tail recurrence). Let $\sum_{j \geq 1} | a_{j} | 2^{- j} < \infty$ and $U_{N + 1} = 2 U_{N} - a_{N + 1}$ . Then $U_{N} = \sum_{j \geq 1} a_{N + j} 2^{- j}$ for every $N$ if and only if $2^{- N} U_{N} \to 0$ .

Proof. Iteration gives $2^{- N} U_{N} = U_{0} - \sum_{j = 1}^{N} a_{j} 2^{- j}$ . The limit is zero exactly when $U_{0}$ is the full series. Subtracting its first $N$ terms and multiplying by $2^{N}$ then gives the tail formula. Conversely the formula implies $2^{- N} U_{N} = \sum_{k > N} a_{k} 2^{- k} \to 0$ by absolute convergence. ◻

Integral shifts

For $U_{N + 1} = 2 U_{N} - a_{N + 1}$ with $a_{n} \in Z$ , write $D_{h} (N) = U_{N + h} - U_{N}$ . Modulo integers, each step doubles, so an integral shift is a return to the same fractional part. For rational $U_{0}$ , the power of two in its reduced denominator determines when returns of positive length can begin; the odd part determines their lengths.

Theorem 4.4 (exact rationality classification). For a real integer-coefficient recurrence, the following are equivalent: $U_{0} \in Q$ ; $D_{h} (N) \in Z$ for some $h \geq 1, N \geq 0$ ; and, for some fixed $h \geq 1$ , $D_{h} (N) \in Z$ at every sufficiently large $N$ . More precisely, if $U_{0} = u / (2^{s} d)$ is in lowest terms, with $d$ odd, then

den (U_{N}) = 2^{max (s - N, 0)} d, D_{h} (N) \in Z ⟺ N \geq s and d ∣ 2^{h} - 1 (h \geq 1) .

Consequently $U_{0}$ is irrational exactly when every positive shift is nonintegral at every index, equivalently when for each fixed $h \geq 1$ there are arbitrarily late nonintegral shifts.

Proof. Iteration gives

U_{N} - 2^{N} U_{0} \in Z, D_{h} (N) - 2^{N} (2^{h} - 1) U_{0} \in Z .

An integral shift with $h \geq 1$ makes the nonzero integer multiple $2^{N} (2^{h} - 1) U_{0}$ integral, and therefore forces $U_{0} \in Q$ . Conversely, write $U_{0} = u / (2^{s} d)$ in lowest terms. Multiplication by $2^{N}$ cancels exactly $min (N, s)$ powers of two from the denominator. The remaining factor $2^{h} - 1$ is odd, so the second relation is integral exactly when $N \geq s$ and $d ∣ 2^{h} - 1$ . For $d > 1$ , Euler’s congruence provides such a shift $h = φ (d)$ , and for $d = 1$ we take $h = 1$ . This shift works for every $N \geq s$ . Negating the pointwise and eventual criteria gives the two assertions about irrationality. ◻

The classification applies to any integer-coefficient recurrence; Lemma 4.3 is needed only to identify complete tails. The periodicity is of fractional parts, not necessarily tail values. For the positive even coefficients $2, 4, 2, 4, \dots$ starting at $n = 1$ , the tails alternate between $8 / 3$ and $10 / 3$ . Length- $1$ shifts are nonintegral, but length- $2$ shifts vanish: one shift length cannot suffice.

Small tail differences and finite tests

Fix $h \geq 1$ and put $D_{N} = T_{N + h} - T_{N}$ and $δ_{N} = g_{N + h + 1} - g_{N + 1}$ . Then $D_{N + 1} = 2 D_{N} - δ_{N}$ . Both gap indices are positive, so neither involves the exceptional odd gap $g_{0} = 1$ ; hence $δ_{N}$ is even. Two integral differences of absolute value less than one must both vanish, forcing $δ_{N} = 0$ . Evenness yields the sharper signed test below.

Proposition 5.1 (two consecutive differences of absolute value less than one). Let $D, D^{'} \in R$ , $δ \in 2 Z$ and $D^{'} = 2 D - δ$ . The conditions $| D | < 1$ , $| D^{'} | < 1$ and $δ \neq 0$ hold exactly when, for some $s \in {- 1, 1}$ ,

δ = 2 s, \frac{1}{2} < s D < 1.

In that case $s D^{'} \in (- 1, 0)$ and both $D$ and $D^{'}$ are nonintegral.

Proof. The bounds on $D, D^{'}$ imply $- 3 < δ < 3$ , so a nonzero even $δ$ equals $2 s$ with $s \in {- 1, 1}$ . Substituting in $D^{'} = 2 D - 2 s$ gives $1 / 2 < s D < 1$ . Conversely this interval gives $s D^{'} = 2 s D - 2 \in (- 1, 0)$ . ◻

The endpoints cannot be included: when $δ = 2$ , $D = 1 / 2$ gives $D^{'} = - 1$ , and $D = 1$ already violates $| D | < 1$ . The zero triple $D = D^{'} = δ = 0$ shows why the mismatch is also required.

Problem 5.2 (two small tail differences at arbitrarily large indices). For every integer $h \geq 1$ and every cutoff $N_{0}$ , exhibit $N \geq N_{0}$ with

\begin{matrix} (4) & | T_{N + h} - T_{N} | < 1, | T_{N + h + 1} - T_{N + 1} | < 1, g_{N + h + 1} \neq g_{N + 1} . \end{matrix}

Theorem 4.4 requires nonintegrality, not smallness; (4) is only a sufficient condition. All three requirements must hold at the same arbitrarily late indices. Separate infinite sets need not intersect.

By Proposition 5.1, such an index must have $δ_{N} = \pm 2$ . Schlage-Puchta’s lemma [18], applied to $x_{h} - x_{0} \pm 2$ , shows that these eligible indices already have density zero. Density zero is compatible with arbitrarily late witnesses; it supplies no lower bound for their occurrence. The absolute block comparison in Proposition 1.1 likewise gives no relative-frequency estimate for this rare event.

Finite separation from the integers

An enclosure certifies nonintegrality only when it misses every integer. Choose $M (n) \geq g_{n}$ with $\sum_{n} M (n) 2^{- n} < \infty$ , and write

S_{h, N, L} = \sum_{j = 1}^{L} (g_{N + h + j} - g_{N + j}) 2^{- j}, R_{h, N, L} (M) = \sum_{j > L} (M (N + h + j) + M (N + j)) 2^{- j} .

Splitting the complete tails after $L$ terms gives

D_{N} = S_{h, N, L} + 2^{- L} D_{N + L}, | D_{N} - S_{h, N, L} | \leq R_{h, N, L} (M) .

The uncomputed term is a rescaled later complete tail difference with the same shift $h$ . An arbitrary real majorant gives an enclosure, but need not give a computable radius.

For the actual prime gaps, one computable choice is $M (n) = 1250 (n + 2)^{4}$ . For $P (x) = x^{4} + 8 x^{3} + 36 x^{2} + 104 x + 150$ , the identity $2 P (x) = (x + 1)^{4} + P (x + 1)$ telescopes to $\sum_{j = 1}^{J} (x + j)^{4} 2^{- j} = P (x) - 2^{- J} P (x + J)$ . The endpoint vanishes as $J \to \infty$ , giving

R_{h, N, L} (M) = 1250 2^{- L} (P (N + h + L + 2) + P (N + L + 2)) .

Proposition 5.3 (finite separation criterion). If for every $h \geq 1$ and every cutoff $N_{0}$ there are $N \geq N_{0}, L \geq 1$ with

\begin{matrix} (5) & dist (S_{h, N, L}, Z) > R_{h, N, L} (M), \end{matrix}

then $Π$ is irrational.

Proof. The enclosure gives $dist (D_{N}, Z) \geq dist (S_{h, N, L}, Z) - R_{h, N, L} (M) > 0$ . Thus each fixed positive shift is nonintegral at arbitrarily large indices, and Theorem 4.4 applies. ◻

Remark (what arbitrary truncation does not gain). Whenever $| D - S_{L} | \leq R_{L} \to 0$ , nonintegrality of $D$ is equivalent to $dist (S_{L}, Z) > R_{L}$ for some $L$ . Indeed, for $δ = dist (D, Z) > 0$ , take $R_{L} < δ / 2$ and use the $1$ -Lipschitz property of distance. Unrestricted depth therefore gives an exact reformulation, not an additional analytic estimate. At a prescribed logarithmic depth one must also control the margin from the integers as $N$ varies; a shrinking remainder alone does not do so. The companion, Appendix D.2, gives an explicit example distinguishing refinement at a fixed value from enclosures of changing values.

A nonintegral value need not lie in either signed interval of Proposition 5.1. To certify (4), suppose $δ_{N} = 2 s$ , with $s \in {- 1, 1}$ , and enclose $D_{N}$ using integers $A, Q, B$ with $Q > 0$ , $B \geq 0$ and $| Q D_{N} - A | \leq B$ . Then $s D_{N} \in [(s A - B) / Q, (s A + B) / Q]$ . Placing its lower endpoint above $1 / 2$ and its upper endpoint below $1$ gives, respectively,

2 s A - Q > 2 B, Q - s A > B .

We retain the signed numerator: reduction modulo $Q$ preserves distance to $Q Z$ but loses the enclosure’s location. The gap difference fixes $s$ , so it cannot be chosen to fit the enclosure. At $h = 1, N = 2, L = 40$ , one has $δ_{2} = g_{4} - g_{3} = - 2$ and

Q = 2^{40}, A = - 662838684750, B = 11764181250, s = - 1.

Here $A / Q = S_{1, 2, 40}$ and $B / Q = R_{1, 2, 40} (M)$ , and the two strict margins are $202637379224$ and $424908761776$ . The companion’s integer certificates give the prime-index calculation and remainder derivation. This finite pair leaves the quantifiers in Problem 5.2 unresolved.

Further consequences and questions

We can place the interval in Proposition 1.1 inside $(A, A + η)$ for any $η > 0$ . Choose $K^{'} \geq K$ such that $(K^{'} + 1) 2^{- K^{'}} < η$ , and start the construction after $K^{'}$ . The bound $e_{n} \leq n$ from the proof gives

0 < \sum_{n} e_{n} 2^{- (n + 1)} \leq (K^{'} + 1) 2^{- K^{'}} < η .

Thus rational and irrational targets remain available arbitrarily close to $A$ , with the prescribed prefix unchanged. The estimate controls the dyadic change, not the ordinary total.

Nonperiodic coefficients

Rationality forces eventual periodicity of the binary digits obtained after carrying, not of the original integer coefficients. For example,

\sum_{i = 0}^{n - 1} \frac{i - 1}{2^{i + 1}} = - \frac{n}{2^{n}} ⟶ 0.

Its coefficients are unbounded and its first coefficient is $- 1$ . We obtain a positive even example by taking $c_{n} = 2 (n^{2} + 4 n + 2)$ and $U_{n} = 2 (n + 4)^{2}$ . Direct substitution gives $U_{n + 1} = 2 U_{n} - c_{n + 1}$ , while the polynomial growth gives $2^{- n} U_{n} \to 0$ . Lemma 4.3 therefore identifies $U$ as the complete tail and yields

\sum_{j \geq 1} c_{j} 2^{- j} = 32, \sum_{n \geq 0} c_{n} 2^{- (n + 1)} = 18.

The coefficients increase strictly and $c_{n + 1} - c_{n} = 4 n + 10$ is never $\pm 2$ . The companion’s Section 8 gives further examples, including a bounded nonperiodic sequence in Appendix D. None of these examples constructs consecutive primes.

Proposition 6.1 (prime gaps do not become periodic). For every positive $h$ , the actual consecutive-prime-gap sequence is not eventually periodic with period $h$ .

Proof. If the gaps were eventually periodic, they would be bounded. For each $m \geq 2$ , however, the $m - 1$ integers $m! + 2, \dots, m! + m$ are composite. The consecutive primes on either side therefore have a gap of at least $m$ , contradicting boundedness. ◻

The results on bounded prime gaps [13], clusters of each fixed size [11][20], and extreme large gaps [4] do not give the simultaneous inequalities in (4). This remains so with Polymath’s bound $\underset{n}{lim inf} g_{n} \leq 246$ [20]. For fixed $h$ and $ε > 0$ , taking $L = ⌈ (4 + ε) \log_{2} (N + 2) ⌉$ in the polynomial remainder gives an error $O_{h, ε} (N^{- ε})$ . The finite sum must still clear the relevant boundary by more than this error to supply a witness for Problem 5.2. The perturbation theorem rules out no argument using additional primality information or quantitative rare-event estimates.

Verification and reproducibility

The linked Lean declarations use Lean 4 [9] and mathlib [10]. The formal proof of Proposition 1.1 changes one coordinate at each stage, whereas the proof here uses triples. Both choose the set containing the supports, the interval and the congruence cutoffs before the value of the sum. For Corollary 3.1, the formal statement covers every $ε > 0$ and assumes Schlage-Puchta’s lemma only for nonconcentration and the prime number theorem only for the asymptotic $P_{n} \sim n \log n$ . The source links record formal support for the linked statements, rather than verification of the different printed proofs. The companion’s Appendix G retains the complete source index at this note’s original pin, including the finite telescoping identities and their nonperiodicity consequences. Its Section 3 retains the broader subsum and automatic-sequence comparisons.

Acknowledgements

The author received no external funding and declares no competing interests. The numbering follows Bloom’s catalogue [14].

References

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Paul Erdős, On the irrationality of certain series: problems and results. New Advances in Transcendence Theory, Cambridge University Press, 1988, pp. 102–109, doi:10.1017/CBO9780511897184.009.
Kevin Ford, Ben Green, Sergei Konyagin, James Maynard and Terence Tao, Long gaps between primes. Journal of the American Mathematical Society 31 (2018), 65–105, doi:10.1090/jams/876; arXiv:1412.5029v3.
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Vjekoslav Kovač and Terence Tao, On several irrationality problems for Ahmes series. Acta Mathematica Hungarica 175 (2025), 572–608, doi:10.1007/s10474-025-01528-0; arXiv:2406.17593v4. Statement and section numbers refer to arXiv version 4.
Vivian Kuperberg, Sums of singular series with large sets and the tail of the distribution of primes. The Quarterly Journal of Mathematics 74 (2023), no. 4, 1457–1479, doi:10.1093/qmath/haad030; arXiv:2210.09775v2. Statement numbers refer to arXiv version 2, 15 June 2023.
Johan Land, A conditional proof of the irrationality of $\sum_{n \geq 1} p_{n} 2^{- n}$ under a uniform Hardy–Littlewood prime-tuples conjecture. Research draft, 5 September 2026. Formalisation material accompanies the draft.
Leonardo de Moura and Sebastian Ullrich, The Lean 4 theorem prover and programming language. Automated Deduction – CADE 28, Lecture Notes in Computer Science, Springer, 2021, pp. 625–635, doi:10.1007/978-3-030-79876-5_37.
The mathlib Community, The Lean mathematical library. Proceedings of the 9th ACM SIGPLAN International Conference on Certified Programs and Proofs, ACM, 2020, pp. 367–381, doi:10.1145/3372885.3373824. Describes a Lean 3-era snapshot; the repository lock identifies the library used by the present Lean 4 sources.
James Maynard, Small gaps between primes. Annals of Mathematics 181 (2015), 383–413, doi:10.4007/annals.2015.181.1.7.
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Thomas F. Bloom, Erdős Problem #251. 2026. Accessed 6 September 2026.
Erdős Problems contributors, Erdős Problem #251 discussion thread. 2026. Accessed 6 September 2026: Tao comment of 7 October 2025 and Land comments of 6 September 2026.
Stefan Ringer, Local gap statistics, telescoping, and normality: a local-pattern approach to Erdős problem 251. Preprint, 11 September 2026. Mutable main-branch TeX consulted 16 September 2026; the cited conditional statements were rechecked 18 September 2026. Formalisation material accompanies the draft.
Tonći Crmarić and Vjekoslav Kovač, On the irrationality of certain super-polynomially decaying series. Colloquium Mathematicum 179 (2025), 55–68, doi:10.4064/cm9628-5-2025; arXiv:2504.18712v1. Lemma 4 is cited using arXiv version 1.
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Long paper, 47 pages

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

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GitHub lean/ErdosProblems/Erdos251/ActualPrimePaperR11.lean
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GitHub lean/ErdosProblems/Erdos251/FreePairReduction.lean
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GitHub lean/ErdosProblems/Erdos251/NonconcentrationConsequencesR11.lean
GitHub lean/ErdosProblems/Erdos251/NonconcentrationCoreR11.lean
GitHub lean/ErdosProblems/Erdos251/OrderLatticeDiagonal.lean
GitHub lean/ErdosProblems/Erdos251/PaperBoundedCarryR7.lean
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GitHub lean/ErdosProblems/Erdos251/PaperCompleteR20/LocalTargetInterval.lean
GitHub lean/ErdosProblems/Erdos251/PaperCompleteR20/RealPropagation.lean
GitHub lean/ErdosProblems/Erdos251/PaperCompleteR20/SignedWindow.lean
GitHub lean/ErdosProblems/Erdos251/PaperCompleteR20/SparseConstructionAudit.lean
GitHub lean/ErdosProblems/Erdos251/PaperCompleteR20/SparseNonconcentration.lean
GitHub lean/ErdosProblems/Erdos251/PaperCompleteR20/TrueTail.lean
GitHub lean/ErdosProblems/Erdos251/PaperCompleteR21/JointPrimeGapCountermodel.lean
GitHub lean/ErdosProblems/Erdos251/PaperCompleteR21/PerturbedPrimePositions.lean
GitHub lean/ErdosProblems/Erdos251/PaperCompleteR21/TwoWindowSparsity.lean
GitHub lean/ErdosProblems/Erdos251/PaperCoreR7.lean
GitHub lean/ErdosProblems/Erdos251/PaperCorrespondenceAudit.lean
GitHub lean/ErdosProblems/Erdos251/PaperFiniteCertificatesR7.lean
GitHub lean/ErdosProblems/Erdos251/PaperNonconcentrationR7.lean
GitHub lean/ErdosProblems/Erdos251/PaperSparseCouplingR7.lean
GitHub lean/ErdosProblems/Erdos251/PaperTailBoundsR7.lean
GitHub lean/ErdosProblems/Erdos251/PerturbationGrowthR11.lean
GitHub lean/ErdosProblems/Erdos251/PolynomialGapSeriesValue.lean
GitHub lean/ErdosProblems/Erdos251/PrimeGapDyadicTail.lean
GitHub lean/ErdosProblems/Erdos251/PrimeSourceR11.lean
GitHub lean/ErdosProblems/Erdos251/RealPrimeGapTail.lean
GitHub lean/ErdosProblems/Erdos251/ResidueFeedbackCore.lean
GitHub lean/ErdosProblems/Erdos251/ReturnedSparseWindow.lean
GitHub lean/ErdosProblems/Erdos251/ShiftedGapCountingR9.lean
GitHub lean/ErdosProblems/Erdos251/SparseAmbientR9.lean
GitHub lean/ErdosProblems/Erdos251/SparseIntervalR8.lean
GitHub lean/ErdosProblems/Erdos251/SparsePaperR11.lean
GitHub lean/ErdosProblems/Erdos251/SparsePolylogR11.lean
GitHub lean/ErdosProblems/Erdos251/SparseRationalisationCore.lean
GitHub lean/ErdosProblems/Erdos251/SparseScheduleDensityR8.lean
GitHub lean/ErdosProblems/Erdos251/SparseScheduleR8.lean

Statements with Lean proofs

Lean is software that verifies a formal proof against its theorem statement and assumptions. These proofs establish only the statements listed here.

The finite summation-by-parts identity relating the prime dyadic partial sums to the prime-gap dyadic partial sums, with an exact endpoint term and no convergence premise.
The factor-of-two normalisation between the two displayed indexings, at every finite horizon.
The unconditional prime-to-gap identity, using an elementary polynomial bound on primes to prove summability; irrationality of the prime series is equivalent to irrationality of the prime-gap series.
For an abstract integer-digit dyadic tail recurrence: the exact block identity, and that a shift of fixed length is integral exactly when an explicit scaled state is.
That one integral shift propagates to every later index.
That repeated doubling removes the power-of-two denominator, so every rational-valued dyadic recurrence has one fixed positive shift integral at all sufficiently late indices.
For a real-valued dyadic recurrence, cofinal non-integrality of every fixed positive shift implies irrationality of the initial state.
That the actual consecutive-prime-gap sequence is unbounded, by the factorial prime-free-interval construction, and therefore is not eventually periodic with any positive period.
For any integer-digit dyadic tail recurrence, two adjacent fixed-length shifts lying strictly between -1 and 1 cannot both be integral when the corresponding digits differ.
For the actual prime gaps, a cofinal supply of adjacent small-shift pairs with a mismatching gap excludes eventual integrality of that fixed shift.
The generic distance-to-integers core of the finite-truncation criterion: an approximation farther than its error bound from every integer certifies non-integrality.

What these proofs do not establish

The target irrationality.
A cofinal nonintegrality theorem for the concrete prime-gap tails; defining the series and proving its recurrence do not supply this.
A theorem supplying cofinally many adjacent actual-prime-gap tail shifts in the open unit interval with the required gap mismatch.
Any theorem forcing periodicity or automaticity of the prime gaps from rationality; the free-carry identity disconfirms that route.

The formal record: 5 main results, 2 open obligations, 1 selected result module, Comparator review

Main results

Statements and their limits are copied from the public research record. Recorded status: registered_declarations

formalisation of existing result family: coefficient_only_no_go

The consecutive prime gaps are unbounded.

Unbounded coefficients alone do not force irrationality.

ErdosProblems.Erdos251.exists_primeGap0_gt

formalisation of exact finite identity and nonperiodicity contrast; novelty and significance unassessed family: coefficient_only_no_go

The coefficient-only wrapper records the exact finite carry partial-sum identity together with non-eventual-periodicity of the synthetic coefficient stream and the actual prime-gap stream, exposing the limits of coefficient-only reasoning.

The wrapper proves the finite identity and the two non-eventual-periodicity statements only. It does not assert the infinite rational-sum limit of the synthetic countermodel, identify the synthetic coefficient stream with actual prime gaps, establish a prime-gap tail bridge, prove #251 irrationality, or make novelty, priority, significance, or external-review claims.

ErdosProblems.Erdos251.carryPartialSum_natCast_eq

formalisation of existing result; novelty unassessed family: prime_gap_reformulation

Lean checks that the prime-value dyadic series is irrational exactly when the prime-gap dyadic series is, discharging summability by an elementary polynomial prime bound rather than the prime-number theorem. Neither side is proved irrational. The already-known identity is Tao's (7 October 2025) and is recorded separately as cited only.

The equivalence does not prove irrationality of either series.

ErdosProblems.Erdos251.irrational_tsum_primeDyadicTerm_iff_primeGap

abstract Lean-checked classification and normal form; novelty unassessed family: dyadic_tail_integrality_classification

For any integer-digit real dyadic tail recurrence T(N + 1) = 2*T(N) - g(N + 1), the initial value T(0) is irrational exactly when every positive-length tail shift is nonintegral.

This exact normal form covers arbitrary integer-digit dyadic tail recurrences with the declared real state type; it does not supply a cofinal nonintegral shift for the consecutive-prime-gap recurrence. The prime-specific producer remains open, so no prime-gap irrationality, novelty, significance, or external-review claim is made and Erdős #251 remains open.

ErdosProblems.Erdos251.irrational_initial_iff_all_positive_tailShifts_nonintegral

conditional actual-prime-gap endpoint reduction; novelty and significance unassessed family: small_mismatch_criterion

For the actual prime-gap dyadic tail, cofinally many adjacent pairs of strictly small h-shifts with unequal corresponding prime gaps force failure of eventual RatIntegral h-shifts.

The representative assumes T : ℕ → ℚ, the DyadicTailRecurrence for the actual prime-gap digits, and for every N₀ an $N \geq$ N₀ whose adjacent h-shifts both lie strictly in (-1, 1) while the corresponding prime gaps differ. Under that cofinal small-mismatch supply it excludes eventual RatIntegral h-shifts; it does not prove the supply or actual smallness. rationalPrimeGapTail_has_positive_shift_not_eventually_small records the contrary rational-state obstruction for at least one positive shift. This is a conditional reduction only: no #251 irrationality, infinite rational-sum limit, universal producer, novelty, priority, significance, or external-review claim is made; #251 remains open. The actual-prime-gap specialization is distinct from coefficient_only_no_go and does not duplicate prime_gap_reformulation.

ErdosProblems.Erdos251.primeGapTailShift_not_eventuallyIntegral_of_cofinal_small_mismatch

Assumptions still needing proofs

Prove any one of these and the corresponding reduction above becomes unconditional.

prime_gap_cofinal_shift_escape For every fixed $h \geq 1$ and every N0, find N ≥ N0 for which $\sum_{j \geq 1} (g_{N + h + j} - g_{N + j}) / 2^{j}$ is not an integer.
cofinal_adjacent_small_mismatch For every fixed $h \geq 1$ and every N0, find N ≥ N0 such that both adjacent full tail shifts have absolute value less than 1 and $g_{N + h + 1} \neq g_{N + 1}$ . The checked local consumer then excludes eventual integrality of the h-shift.

Selected Lean proof files

The public result record selects 1 module. Browse all 49 source files in this programme’s source map, including shared result sources and local imported dependencies.

Module	Lines	Declarations	Theorems
`ErdosProblems.Erdos251.PrimeGapDyadicTail`	1665	116	92

Comparator review

Comparator runs selected Lean proofs against separately supplied theorem statements using recorded software versions. A successful run establishes that those proofs satisfy those statements; it does not prove the unanswered questions above.

Replayed by the Comparator

Statement and type parity replayed against the pinned toolchain.

Exact finite summation by parts, the unconditional infinite prime-gap identity, and irrationality equivalence.

The equivalence does not prove irrationality of either series.

prime_gap_reformulation: targeted
For any integer-digit dyadic tail recurrence T(N+1) = 2*T(N) - g(N+1), the rational h-shift is integral exactly when the reduced denominator of T(N) divides $2^{h} - 1$ , equivalently when $2^{h}$ is congruent to 1 modulo that denominator. For the real recurrence, the load-bearing normal form says that irrationality of T(0) is equivalent to every positive tail shift being nonintegral; the eventual-integrality equivalence and denominator/order laws are supporting mechanism evidence.

These exact statements cover arbitrary integer-digit dyadic tail recurrences with the declared rational and real state types; they classify integrality and give an irrationality normal form but do not supply a cofinal nonintegral shift for the consecutive-prime-gap recurrence. The generic hypotheses therefore do not close the prime-specific producer gap: no prime-gap irrationality, novelty, significance, or external review is claimed, and Erdős #251 remains open. This is one abstract classification family, distinct from prime_gap_reformulation, not a duplicate frontier claim.

dyadic_tail_integrality_classification: targeted
For the actual prime-gap dyadic tail, cofinally many adjacent pairs of strictly small h-shifts with unequal corresponding prime gaps force failure of eventual RatIntegral h-shifts. The adjacent-pair zero obstruction is the load-bearing mechanism; exact non-eventual periodicity of consecutive prime gaps supplies the actual-prime-gap contrast.

The representative assumes T : ℕ → ℚ, the DyadicTailRecurrence for the actual prime-gap digits, and for every N₀ an $N \geq$ N₀ whose adjacent h-shifts both lie strictly in (-1, 1) while the corresponding prime gaps differ. Under that cofinal small-mismatch supply it excludes eventual RatIntegral h-shifts; it does not prove the supply or actual smallness. rationalPrimeGapTail_has_positive_shift_not_eventually_small records the contrary rational-state obstruction for at least one positive shift. This is a conditional reduction only: no #251 irrationality, infinite rational-sum limit, universal producer, novelty, priority, significance, or external-review claim is made; #251 remains open. The actual-prime-gap specialization is distinct from coefficient_only_no_go and does not duplicate prime_gap_reformulation.

small_mismatch_criterion: targeted
The exact finite carry identity and the two non-eventual-periodicity statements expose the coefficient-only barrier: irregularity and growth of a coefficient stream do not by themselves force irrationality, while the synthetic carry stream remains distinct from the actual prime-gap stream.

CoefficientOnlyNoGo checks the exact finite partial-sum identity and the two non-eventual-periodicity statements. It does not assert the infinite limit of the synthetic countermodel, identify its coefficient stream with actual prime gaps, establish a prime-gap tail bridge, prove #251 irrationality, or make novelty, priority, significance, or external-review claims.

coefficient_only_no_go: targeted

Represented indirectly

Covered through a neighbouring selected interface, not replayed one-to-one.

Odd rational denominators give totient-length integral shifts, and integrality propagates through the recurrence.

This describes rational states and supplies no contradiction for actual prime gaps.

totient_shift_propagation: represented_by_integral_shift_family

Outside the Comparator selection

Stated in the corpus; not part of the replayed interface set.

Block identities and exact denominator criteria classify rationality through integral positive tail shifts.

The concrete prime-gap producer remains missing.

integral_shift_classification: not_selected_abstract_family_represented_in_manifest

The result statements, their limits and the counts are copied from the public Lean repository’s published records identified by b0ebeb9add157584. These records are copied into the page when the website is built. Introductory explanations and labels are written for this site. The diagram at the top is the exception: a program draws it from numbers the papers print and tests the numerical relationships shown. Its caption and explanation are written for this site.