GPT 6 Astra Ultra attempt on Math-731
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Contents
Problem 251: prime-like coefficients can still telescope to a rational
1) OBJECTIVE AND SOURCE REVIEW
Attempt dated 2026-09-23. The target is irrationality of $S=\sum_{n\ge1}p_n/2^n$. I read251.tex, 251_v2.tex, 251_v3.tex, and 251_v4.tex fully. They concern corrected tail periodicity, continued-fraction denominator exclusions, nested intervals, and a larger Stern--Brocot computation. I do not inherit their large numerical certificates without rerunning them. I test whether the elementary features of the prime coefficients can contradict rationality. 2) WORK
Define $f(x)=\tfrac12(x+1000)\log(x+1000)$, $b_n=\lfloor f(n)\rfloor$, and, for $n\ge1$, \[ c_n=2(2b_{n-1}-b_n)+1. \] These are positive odd integers, strictly increasing, and $c_n\sim n\log n$. Here are direct verifications. Let $g(x)=2f(x-1)-f(x)$. For $x\ge1$, \[ g'(x)=2f'(x-1)-f'(x) \ge\tfrac12(\log1000+1)-\frac1{2000}>3. \] This uses $f''(t)=1/(2(t+1000))$. Since $2b_{n-1}-b_n-g(n)$ lies in $(-2,1)$, consecutive values of $2b_{n-1}-b_n$ differ by more than zero. Positivity follows already at $n=1$ from $g(1)>2$ and then monotonicity. Finally $g(n)=f(n)-2(f(n)-f(n-1))\sim\tfrac12 n\log n$, proving the asymptotic.Nevertheless the series is an integer. For every $N$, \[ \sum_{n=1}^N\frac{2b_{n-1}-b_n}{2^n}=b_0-\frac{b_N}{2^N}. \] Letting $N\to\infty$ yields \[ \sum_{n\ge1}\frac{c_n}{2^n}=2b_0+1\in\mathbb Z. \] Thus increasing odd integer coefficients with precisely the prime first-order scale $n\log n$ can satisfy all the tail lattice relations and still give a rational sum. A successful proof must use additional arithmetic of the primes.
3) ADVERSARIAL VERIFICATION
This sequence is not asserted to consist of primes. The floor errors have been included in proving strict monotonicity. The source251_v2.tex's unqualified open-interval continued-fraction rule has an endpoint exception: $x=[0;2]=1/2$, $y=[0;1,2]=2/3$ produce the proposed candidate $[0;2]=x$, outside the open interval. Endpoint-aware algorithms can repair this, so the example alone does not refute the source's reported numerical values. It does show why an exact implementation and certificate are necessary.