gpt pro 5.4 attempt on Math-731

Erdős Problem #251

1) ROUND-2 OBJECTIVE

Path (C): identify and fix a fatal error in Round-1, then replace it with the strongest correct rationality consequence I can extract.

Round-1 tried to deduce from "$S\in \mathbf{Q}$" that the tail quantities \[ Y_N:=\sum_{m=1}^{\infty}\frac{p_{N+m}}{2^m} \] are integers for infinitely many $N$. That implication is false. The most promising Round-2 direction is therefore not to keep pushing the old integrality argument, but to repair it and obtain a valid structural consequence of rationality.

2) Round-1 FOUNDATION USED

I use only the following Round-1 facts.

• The series \[ S:=\sum_{n=1}^{\infty}\frac{p_n}{2^n} \] converges absolutely.
• The exact tail recurrence holds: \[ Y_{N+1}=2Y_N-p_{N+1}\qquad (N\ge 0). \]
• The numerical computation from Round-1 is not used in any proof here.

I do not reuse the Round-1 claim that "$S\in\mathbf{Q}$ with odd denominator implies $Y_N\in\mathbf{Z}$ infinitely often", because that claim is incorrect.

3) NEW INSIGHT / TOOL (ROUND-2)

There are three genuinely new pieces in this round.

• First, I isolate the exact Round-1 error and prove the corrected replacement: rationality of $S$ implies only that the fractional parts of $Y_N$ are eventually periodic.
• Second, I prove an equivalent integrality statement that is correct: there exist integers $q\ge 1$ and $N_0$ such that \[ (2^q-1)Y_N\in \mathbf{Z}\qquad (N\ge N_0). \]
• Third, I rewrite $Y_N$ exactly in terms of the prime gaps $g_n:=p_{n+1}-p_n$: \[ Y_N=p_{N+1}+\sum_{r=1}^{\infty}\frac{g_{N+r}}{2^r}. \] This yields a finite-window criterion: if $S$ were rational, then short binary-weighted blocks of future prime gaps would have to lie very close to a fixed lattice for all large $N$.

4) ATTACK PLAN (ROUND-2)

The exact Round-1 gap is now clear.

• The old plan incorrectly tried to force actual integrality of $Y_N$.
• The corrected plan is to show that rationality implies eventual periodicity of the fractional parts of $Y_N$, and hence the weaker but still rigid condition $(2^q-1)Y_N\in\mathbf{Z}$ eventually.
• After rewriting $Y_N$ in terms of prime gaps, the problem becomes: can one rule out the possibility that a fixed odd multiplier $(2^q-1)$ sends these weighted gap tails to integers for all large $N$?

This overcomes the Round-1 obstruction by replacing a false lemma with a true one, and it converts the problem from "binary carries in the primes themselves" to "binary-weighted prime-gap tails".

5) WORK (ROUND-2)

Claim 1 (the Round-1 integrality deduction was false).

The statement \[ \text{``if }x=\frac{a}{b}\in\mathbf{Q}\text{ with }b\text{ odd, then }2^N x\in\mathbf{Z}\text{ for infinitely many }N\text{''} \] is false.

Proof.

Take $x=1/3$. Then $b=3$ is odd, but $2^N/3\notin \mathbf{Z}$ for every $N\ge 0$ because $3\nmid 2^N$. $\square$

Thus the specific Round-1 implication \[ S\in\mathbf{Q}\Longrightarrow Y_N\in\mathbf{Z}\text{ infinitely often} \] does not follow from rationality alone.

Claim 2 (correct rationality consequence: eventual periodicity modulo $1$).

If $S\in\mathbf{Q}$, then the fractional parts of $Y_N$ are eventually periodic.

Proof.

Write $S=a/b$ in lowest terms, and factor \[ b=2^u c \] with $u\ge 0$ and $c$ odd. Since \[ Y_N=2^N S-\sum_{j=1}^{N} p_j 2^{N-j}, \] we have \[ Y_N-\lfloor Y_N\rfloor = 2^N S-\lfloor 2^N S\rfloor, \] so the fractional part of $Y_N$ equals the fractional part of $2^N S$.

For $N\ge u$, \[ 2^N S = \frac{a2^{N-u}}{c}. \] Because $c$ is odd, the residues $2^m \bmod c$ are periodic in $m$; if $q$ is the multiplicative order of $2$ modulo $c$, then \[ 2^{m+q}\equiv 2^m \pmod c \] for every $m\ge 0$. Hence the fractional parts of $a2^m/c$ are periodic in $m$ with period $q$. Therefore the fractional parts of $Y_N$ are eventually periodic in $N$. $\square$

Claim 3 (correct eventual integrality statement).

Assume there exist integers $q\ge 1$ and $N_0\ge 0$ such that the fractional parts of $Y_N$ and $Y_{N+q}$ agree for every $N\ge N_0$. Then \[ (2^q-1)Y_N\in \mathbf{Z}\qquad (N\ge N_0). \] In particular, if $S\in\mathbf{Q}$ then such $q,N_0$ exist.

Proof.

From the exact recurrence, \[ 2Y_N=p_{N+1}+Y_{N+1}, \] and iterating this identity $q$ times gives \[ 2^q Y_N-Y_{N+q} = \sum_{m=1}^{q} 2^{q-m} p_{N+m}. \] The right-hand side is an integer.

If $Y_{N+q}$ and $Y_N$ have the same fractional part, then $Y_{N+q}-Y_N\in\mathbf{Z}$. Subtracting this integer from the previous integer identity yields \[ (2^q-1)Y_N = (2^qY_N-Y_{N+q}) + (Y_{N+q}-Y_N) \in \mathbf{Z}. \] The final sentence follows from Claim 2. $\square$

So the correct replacement for the false Round-1 statement is:

If $S\in\mathbf{Q}$, then there exist integers $q\ge 1$ and $N_0$ such that \[ (2^q-1)Y_N\in\mathbf{Z}\qquad (N\ge N_0). \] Equivalently, the fractional parts of $Y_N$ eventually lie in the finite set \[ \left\{\frac{m}{2^q-1}:0\le m<2^q-1\right\}. \]

Claim 4 (exact prime-gap expansion).

Let \[ g_n:=p_{n+1}-p_n\qquad (n\ge 1). \] Then for every $N\ge 0$, \[ Y_N = p_{N+1}+\sum_{r=1}^{\infty}\frac{g_{N+r}}{2^r}. \]

Proof.

For $m\ge 1$ we have \[ p_{N+m}=p_{N+1}+\sum_{r=1}^{m-1} g_{N+r}. \] Substituting into the definition of $Y_N$ and interchanging summation gives \[ Y_N = \sum_{m=1}^{\infty}\frac{p_{N+1}}{2^m} + \sum_{m=1}^{\infty}\sum_{r=1}^{m-1}\frac{g_{N+r}}{2^m} = p_{N+1}\sum_{m=1}^{\infty}\frac1{2^m} + \sum_{r=1}^{\infty} g_{N+r}\sum_{m=r+1}^{\infty}\frac1{2^m}. \] Now \[ \sum_{m=1}^{\infty}2^{-m}=1, \qquad \sum_{m=r+1}^{\infty}2^{-m}=2^{-r}, \] so the identity follows. $\square$

Because every prime gap after $g_1$ is even, the first gap term $g_{N+1}/2$ is an integer as soon as $N\ge 1$. Thus for large $N$ the nontrivial fractional information in $Y_N$ is carried by the deeper tail \[ \sum_{r=2}^{\infty}\frac{g_{N+r}}{2^r}. \]

Corollary 5 (rationality implies a weighted-gap lattice condition).

If $S\in\mathbf{Q}$, then there exist integers $q\ge 1$ and $N_0$ such that for every $N\ge N_0$, \[ (2^q-1)\sum_{r=1}^{\infty}\frac{g_{N+r}}{2^r}\in \mathbf{Z}. \] Equivalently, for every $M\ge 1$, \[ \min_{m\in\mathbf{Z}} \left| (2^q-1)\sum_{r=1}^{M}\frac{g_{N+r}}{2^r}-m \right| \le (2^q-1)\sum_{r>M}\frac{g_{N+r}}{2^r}. \]

Proof.

By Claim 3, $(2^q-1)Y_N\in\mathbf{Z}$ for all large $N$. By Claim 4 and the fact that $p_{N+1}$ is an integer, \[ (2^q-1)\sum_{r=1}^{\infty}\frac{g_{N+r}}{2^r} = (2^q-1)(Y_N-p_{N+1}) \in\mathbf{Z}. \] Subtracting the truncated sum gives the displayed distance estimate. $\square$

Claim 6 (finite-window approximation).

Assume $S\in\mathbf{Q}$ and let $q$ be as in Corollary 5. Then \[ \min_{m\in\mathbf{Z}} \left| (2^q-1)\sum_{r=1}^{M}\frac{g_{N+r}}{2^r}-m \right| \ll_q \frac{(N+M)\log(N+M)}{2^M} \] for all large $N$ and all $M\ge 1$. In particular, if \[ M=2\left\lceil \log_2(N+2)\right\rceil, \] then \[ \min_{m\in\mathbf{Z}} \left| (2^q-1)\sum_{r=1}^{M}\frac{g_{N+r}}{2^r}-m \right| \ll_q \frac{\log N}{N}. \]

Proof.

By Corollary 5 it suffices to bound \[ \sum_{r>M}\frac{g_{N+r}}{2^r}. \] Using $g_{N+r}\le p_{N+r+1}$ and the standard bound $p_n\ll n\log n$, we get \[ \sum_{r>M}\frac{g_{N+r}}{2^r} \ll \sum_{r>M}\frac{(N+r)\log(N+r)}{2^r}. \] Write $a:=N+M+1$ and substitute $r=M+s$: \[ \sum_{r>M}\frac{(N+r)\log(N+r)}{2^r} \ll \frac1{2^M}\sum_{s\ge 1}\frac{(a+s)\log(a+s)}{2^s}. \] Split the inner sum into $1\le s\le a$ and $s>a$.

If $1\le s\le a$, then $a+s\le 2a$, hence \[ \sum_{1\le s\le a}\frac{(a+s)\log(a+s)}{2^s} \le 2a\log(2a)\sum_{s\ge 1}\frac1{2^s} \ll a\log a. \]

If $s>a$, then $a+s\le 2s$, so \[ \sum_{s>a}\frac{(a+s)\log(a+s)}{2^s} \ll \sum_{s>a}\frac{s\log(2s)}{2^s} \ll 1, \] because the full series $\sum_{s\ge 1} s\log(2s)\,2^{-s}$ converges.

Therefore \[ \sum_{r>M}\frac{g_{N+r}}{2^r} \ll \frac{a\log a}{2^M} \ll \frac{(N+M)\log(N+M)}{2^M}. \] The special choice $M=2\lceil\log_2(N+2)\rceil$ gives $2^M\gg N^2$, so the bound becomes $\ll (\log N)/N$. $\square$

6) ADVERSARIAL VERIFICATION

• The Round-1 error is real: the counterexample $1/3$ completely breaks the old implication.
• Claim 2 uses only the standard eventual periodicity of $2^N a/b$ modulo $1$ after removing the $2$-part of the denominator.
• Claim 3 is exact: the identity \[ 2^qY_N-Y_{N+q}=\sum_{m=1}^{q}2^{q-m}p_{N+m} \] is obtained by iterating the tail recurrence and has an integer right-hand side.
• Claim 4 was checked term-by-term by expanding $p_{N+m}$ in prime gaps and interchanging absolutely convergent series.
• Claim 6 uses only the classical bound $p_n\ll n\log n$; no unproved information on prime gaps enters.

No contradiction with rationality has been obtained; what has been obtained is a corrected and strictly sharper obstruction than in Round-1.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The strict Round-2 advance is this:

• I identified a fatal flaw in the Round-1 reduction and proved that the old integrality consequence was invalid.
• I replaced it by the correct statement that rationality of $S$ forces eventual periodicity of the fractional parts of $Y_N$, equivalently \[ (2^q-1)Y_N\in\mathbf{Z} \] for all large $N$ and some fixed $q$.
• I then rewrote this as a precise weighted prime-gap condition and derived a finite-window approximation: for $M\asymp \log N$, a short binary-weighted block of prime gaps would have to lie within $O((\log N)/N)$ of a fixed lattice for every sufficiently large $N$.

The remaining missing step is now sharply formulated: prove that no fixed odd multiplier $(2^q-1)$ can make these weighted prime-gap blocks shadow the integer lattice for all large $N$.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 35%

9) REFERENCES

Only references actually used or checked.

• T. F. Bloom, Erdős Problem #251, https://www.erdosproblems.com/251, accessed 2026-03-07.
• P. Erdős, Sur certaines séries à valeur irrationnelle, Enseign. Math. $2$ 4 $1958$, 93--100.
• P. Erdős, On the irrationality of certain series: problems and results, in New advances in transcendence theory (Durham, 1986), 1988, 102--109.

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