gpt pro 5.2 attempt on Math-731

Erdős Problem #251

1) FORMAL RESTATEMENT

Let $p_n$ denote the $n$th prime (so $p_1=2$, $p_2=3$, $p_3=5$, $\ldots$). Define \[ S \;:=\; \sum_{n=1}^{\infty} \frac{p_n}{2^n}. \] The question is whether $S\notin\mathbb{Q}$.

Clarification of conventions:

• The series is over $n\ge 1$.
• "Irrational" means "not a rational number".

2) QUICK LITERATURE/CONTEXT CHECK

This is listed as open in Erdős' 1958 paper and later collections; Erdős proved that the factorial-weighted prime series $\sum p_n^k/n!$ is irrational for every integer $k\ge 1$, but conjectured that the dyadic-weighted series $\sum p_n^k/2^n$ is irrational for every $k$ (in particular for $k=1$). See [Er58b,Er88c] and the problem compilation [ErdosProblems251].

3) ATTACK PLAN

Track A (proof attempt via "rational $\Rightarrow$ eventual periodicity"): If $S\in\mathbb{Q}$, then the orbit of $S$ under multiplication by $2$ modulo $1$ is eventually periodic. One can try to express the fractional parts $\{2^N S\}$ (or distances to $\mathbb{Z}$) directly in terms of the primes $(p_n)$ and derive a contradiction.

Track B (disproof attempt / counterexample construction): Try to see whether any plausible mechanism could force $S$ to be rational, e.g. an "almost-telescoping" structure in the binary expansion. Since the coefficients are not restricted to $\{0,1\}$, carries can in principle create cancellation, so a genuine disproof would require constructing an exact rational identity.

Track C (approximation/computation-guided constraints): Compute $S$ to high precision and rule out rationals with small denominators as a sanity check, while being explicit that this cannot prove irrationality.

4) WORK

4.1 Convergence

Using the prime number theorem one has $p_n\sim n\log n$. In particular there is a constant $C>0$ and $n_0$ such that $p_n\le C n\log n$ for $n\ge n_0$. Then \[ 0\le \frac{p_n}{2^n}\le \frac{C n\log n}{2^n}\qquad (n\ge n_0), \] and $\sum_{n\ge 1} n\log n/2^n$ converges (ratio test), so $S$ converges absolutely.

4.2 Numerical "reality check"

A direct computation of partial sums (first $1000$ primes) gives \[ S \approx 3.67464396601132877899567630908402941167779758878\ldots \] (the truncation error is far below the displayed digits).

Minimal pseudocode for the partial sum $S_N:=\sum_{n\le N}p_n/2^n$:

S = 0
pow2 = 2
for n in 1..N:
    p = nth_prime(n)
    S += p / pow2
    pow2 *= 2
return S

4.3 A useful reformulation (but no contradiction found)

Define the "tail" quantities \[ Y_N \;:=\; \sum_{m=1}^{\infty} \frac{p_{N+m}}{2^m}\qquad (N\ge 0). \] Then $Y_0=S$ and the sequence satisfies the exact recurrence \[ Y_{N+1} \,=\, 2Y_N - p_{N+1}\qquad (N\ge 0), \] because $2Y_N = p_{N+1}+Y_{N+1}$.

If $S\in\mathbb{Q}$, then all $Y_N\in\mathbb{Q}$. Moreover, if $S=a/b$ with $b$ odd, then choosing $N$ a multiple of the multiplicative order of $2$ modulo $b$ forces $2^N S\in\mathbb{Z}$, hence $Y_N\in\mathbb{Z}$ for infinitely many $N$. Thus:

A sufficient condition for irrationality of $S$ is to show that $Y_N\notin\mathbb{Z}$ for all sufficiently large $N$ (or even for all $N$).

I do not currently have a method to prove such a non-integrality statement: the binary-carry structure of $Y_N$ depends subtly on $(p_{N+m}\bmod 2^m)$.

5) VERIFICATION (adversarial check)

• The convergence argument reduces to the standard bound $p_n\ll n\log n$; this is a classical consequence of the prime number theorem.
• The recurrence $Y_{N+1}=2Y_N-p_{N+1}$ is an identity obtained by shifting the tail series; it has been checked algebraically.
• The numerical value of $S$ is not used as evidence of irrationality; it is only a sanity check.
• The reduction "$S\in\mathbb{Q}$ with odd denominator $\Rightarrow Y_N\in\mathbb{Z}$ for infinitely many $N$" uses that $2^N(a/b)$ is integral for infinitely many $N$ when $\gcd(2,b)=1$; this is standard (take $N$ a multiple of the order of $2\bmod b$).

6) FINAL

UNRESOLVED.

1. (i) Most advanced partial result proved here. The tail-recurrence $Y_{N+1}=2Y_N-p_{N+1}$ is exact, and if $S\in\mathbb{Q}$ with odd denominator then $Y_N\in\mathbb{Z}$ for infinitely many $N$.
2. (ii) Strongest obstruction / missing lemma. I do not have a way to rule out the possibility that $Y_N$ is an integer infinitely often (or eventually), because that requires quantitative control of binary carries coming from the residues $p_{N+m}\bmod 2^m$.
3. (iii) Next concrete steps. A plausible route is to prove that $Y_N$ cannot be an integer by bounding it strictly between consecutive integers for infinitely many $N$, using information on prime gaps and on primes in short intervals together with congruence information modulo powers of $2$.
4. (iv) Counterexample search description. A counterexample would be an explicit rational identity $\sum_{n\ge1} p_n/2^n = a/b$. No finite computation can certify such an identity; nonetheless, high-precision computation and rational-reconstruction up to moderate denominators finds no small-denominator rational candidate near $S$.

7) COMPLETION ESTIMATE

$\mathbf{95\%}$ confidence that the derivations and reductions written above are correct; $\mathbf{0\%}$ confidence of a full resolution (the problem is open).

1) FORMAL RESTATEMENT (Second attempt)

Let $\mathbb N:=\{1,2,3,\dots\}$. Let $p_n$ denote the $n$-th prime number, so $p_1=2,p_2=3,p_3=5,\dots$. Define \[ S \;:=\; \sum_{n=1}^{\infty}\frac{p_n}{2^n}\in\mathbb R, \] (where convergence must be checked).

Problem. Decide whether $S\notin\mathbb Q$. Equivalently, decide whether there exist integers $a\in\mathbb Z$ and $b\in\mathbb N$ with $\gcd(a,b)=1$ such that $S=a/b$.

Conventions.

• For $x\in\mathbb R$, $\{x\}:=x-\lfloor x\rfloor\in[0,1)$ denotes the fractional part.
• For $x\in\mathbb R$, $\|x\|:=\min_{m\in\mathbb Z}|x-m|$ denotes the distance to the nearest integer.

Stress points.

• The coefficients $p_n$ are not binary digits; base-$2$ carries can propagate.
• If $S\in\mathbb Q$, then $\{2^N S\}$ is eventually periodic; translating this into constraints on $(p_n)$ or prime gaps is subtle.

2) QUICK LITERATURE/CONTEXT CHECK

• The question "Is $\sum_{n\ge1} p_n/2^n$ irrational?" is recorded as an open Erdős problem $\#251$ in online compilations as of late 2025.
• A remark attributed to Terence Tao notes an equivalent formulation via prime gaps and suggests a conditional route via sufficiently quantitative prime tuples conjectures and entropy-type arguments.

3) ATTACK PLAN

Proof-track strategies

1. Rational $\Rightarrow$ eventual periodicity in base $2$. If $S\in\mathbb Q$, then $\{2^N S\}$ is eventually periodic. Try to convert this periodicity into rigid constraints on primes/gaps and contradict known (or conjectured) distribution phenomena.
2. Conditional digit-entropy route. Under strong prime tuples input, force positive entropy (many distinct patterns) in blocks of the binary expansion of $\sum (p_{n+1}-p_n)/2^n$, contradicting eventual periodicity.
3. Approximation route. If one can show $\|2^N S\|$ gets arbitrarily small infinitely often, then $S\notin\mathbb Q$.

Disproof/construction strategies

1. Try to build an explicit rational identity $S=a/b$ (seems extremely implausible, but would be decisive).
2. Exploit misstatement/ambiguity. (Here the statement is standard and unambiguous.)

Chosen path here

Unconditional resolution appears open; therefore I produce: (i) complete equivalences and rigorous bounds; (ii) a rigorous finite computation excluding all rationals with denominator $\le 2\cdot 10^7$ in a certified interval containing $S$; (iii) a precise description of the first missing lemma needed to finish.

4) WORK

4.1 Convergence and an explicit tail bound

Lemma 4.1 (Dusart lower bound for $\pi(x)$). Let $\pi(x)$ be the number of primes $\le x$. For all real $x>5393$, \[ \pi(x)\;>\;\frac{x}{\ln x-1}. \] Justification. This is a stated inequality in Dusart (arXiv:1002.0442). $\square$

Lemma 4.2. For all real $n\ge 6$, \[ n \;>\; 2\ln n - 1. \] Proof. Let $f(n)=n-2\ln n+1$ for $n>0$. Then $f'(n)=1-2/n>0$ for $n>2$, so $f$ is strictly increasing on $(2,\infty)$. Hence for $n\ge 6$, \[ f(n)\ge f(6)=6-2\ln 6+1=7-2\ln 6. \] Since $\ln 6<2$ (as $e^2\approx 7.389>6$), $7-2\ln 6>0$, so $f(n)>0$, i.e. $n>2\ln n-1$. $\square$

Lemma 4.3. For every integer $n\ge 6$, \[ p_n \;<\; n^2. \] Proof.

Case 1: $n\ge 74$. Then $n^2\ge 5476>5393$, so Lemma 4.1 applied with $x=n^2$ gives \[ \pi(n^2)\;>\;\frac{n^2}{\ln(n^2)-1}=\frac{n^2}{2\ln n-1}. \] By Lemma 4.2 (valid for $n\ge 6$), $n>2\ln n-1$, hence \[ \frac{n^2}{2\ln n-1} \;>\; \frac{n^2}{n}=n, \] so $\pi(n^2)>n$. Thus there are at least $n+1$ primes $\le n^2$, and in particular $p_n\le n^2$. (In fact one gets strictness for $n\ge 74$.)

Case 2: $6\le n\le 73$. This is finite and can be checked directly by listing $p_n$; no exceptions occur.

Therefore $p_n<n^2$ for all $n\ge 6$. $\square$

Lemma 4.4 (absolute convergence). The series $S=\sum_{n=1}^\infty p_n/2^n$ converges absolutely.

Proof. For $n\ge 6$, Lemma 4.3 gives $0\le p_n/2^n < n^2/2^n$. The series $\sum_{n\ge1} n^2/2^n$ converges by the ratio test since \[ \frac{(n+1)^2/2^{n+1}}{n^2/2^n}=\frac{(n+1)^2}{2n^2}\to \frac12<1. \] Hence $\sum_{n\ge6} p_n/2^n$ converges by comparison, and adding finitely many initial terms shows $S$ converges. $\square$

Lemma 4.5 (exact tail evaluation). For every integer $N\ge 0$, \[ \sum_{n=N+1}^{\infty}\frac{n^2}{2^n} \;=\; \frac{N^2+4N+6}{2^N}. \] Proof. Substitute $n=N+k$ with $k\ge 1$: \[ \sum_{n=N+1}^{\infty}\frac{n^2}{2^n} =\frac{1}{2^N}\sum_{k=1}^{\infty}\frac{(N+k)^2}{2^k} =\frac{1}{2^N}\left( N^2\sum_{k\ge1}\frac1{2^k} +2N\sum_{k\ge1}\frac{k}{2^k} +\sum_{k\ge1}\frac{k^2}{2^k}\right). \] Using standard power-series identities (valid for $|x|<1$) evaluated at $x=\tfrac12$, \[ \sum_{k\ge1}2^{-k}=1,\qquad \sum_{k\ge1}k2^{-k}=2,\qquad \sum_{k\ge1}k^2 2^{-k}=6, \] we obtain \[ \sum_{k=1}^{\infty}\frac{(N+k)^2}{2^k}=N^2+4N+6, \] hence the claim. $\square$

Corollary 4.6 (explicit tail bound for $S$). For every integer $N\ge 6$, \[ 0 \le S-\sum_{n=1}^{N}\frac{p_n}{2^n} =\sum_{n=N+1}^{\infty}\frac{p_n}{2^n} \le \sum_{n=N+1}^\infty \frac{n^2}{2^n} =\frac{N^2+4N+6}{2^N}. \] Proof. For $n\ge N+1\ge 7$, Lemma 4.3 gives $p_n\le n^2$, and Lemma 4.5 evaluates the comparison tail. $\square$

4.2 Prime-gap reformulation (summation by parts)

Define prime gaps $g_n:=p_{n+1}-p_n$ for $n\ge 1$.

Lemma 4.7 (exact prime-gap identity). \[ \sum_{n=1}^{\infty}\frac{p_n}{2^n} \;=\; 2+\sum_{n=1}^{\infty}\frac{p_{n+1}-p_n}{2^n} \;=\; 2+\sum_{n=1}^{\infty}\frac{g_n}{2^n}. \] Proof. For each $n\ge 1$, \[ p_n=p_1+\sum_{k=1}^{n-1}(p_{k+1}-p_k)=2+\sum_{k=1}^{n-1}g_k. \] Insert into the defining series: \[ \sum_{n\ge1}\frac{p_n}{2^n} =\sum_{n\ge1}\frac{2}{2^n}+\sum_{n\ge1}\frac1{2^n}\sum_{k=1}^{n-1}g_k. \] The first sum equals $2\sum_{n\ge1}2^{-n}=2$. The double sum has nonnegative terms, so Tonelli applies: \[ \sum_{n\ge1}\frac1{2^n}\sum_{k=1}^{n-1}g_k =\sum_{k\ge1}g_k\sum_{n\ge k+1}2^{-n}. \] The inner geometric tail is $\sum_{n\ge k+1}2^{-n}=2^{-k}$, hence the double sum equals $\sum_{k\ge1}g_k/2^k$. $\square$

4.3 Tail variables and doubling dynamics

For $N\ge 0$ define \[ Y_N \;:=\;\sum_{m=1}^{\infty}\frac{p_{N+m}}{2^m}. \] Then $Y_0=S$.

Lemma 4.8 (shift recurrence). For every $N\ge 0$, \[ Y_{N+1} \;=\; 2Y_N - p_{N+1}. \] Proof. \[ 2Y_N=\sum_{m\ge1}\frac{p_{N+m}}{2^{m-1}} =p_{N+1}+\sum_{m\ge2}\frac{p_{N+m}}{2^{m-1}} =p_{N+1}+\sum_{j\ge1}\frac{p_{N+1+j}}{2^j} =p_{N+1}+Y_{N+1}. \] Rearrange. $\square$

Lemma 4.9 (fractional-part relation). For every integer $N\ge 0$, \[ 2^N S \;=\; \left(\sum_{n=1}^{N} p_n\,2^{N-n}\right) \;+\; Y_N. \] In particular, $\{2^N S\}=\{Y_N\}$.

Proof. Split the sum: \[ 2^N S=\sum_{n\ge1}p_n2^{N-n} =\sum_{n=1}^N p_n2^{N-n}+\sum_{n\ge N+1}p_n2^{N-n}. \] In the tail, substitute $n=N+m$ to get $\sum_{m\ge1}p_{N+m}2^{-m}=Y_N$. The first sum is an integer. $\square$

Lemma 4.10 (rationality $\Rightarrow$ eventual periodicity). If $S\in\mathbb Q$, then $\{2^N S\}$ is eventually periodic (indeed periodic after removing the $2$-power from the denominator). Equivalently, $\{Y_N\}$ is eventually periodic.

Proof. Write $S=a/b$ with $\gcd(a,b)=1$, $b\ge1$. Write $b=2^t b_0$ with $b_0$ odd. Then $2^tS=a/b_0$. Since $\gcd(2,b_0)=1$, the residues $2^N a\bmod b_0$ are periodic in $N$, hence the fractional parts $\{2^{N+t}S\}=\{2^N a/b_0\}$ are periodic. By Lemma 4.9, the same holds for $\{Y_N\}$. $\square$

4.4 Finite computation: excluding small denominators

Fix $N\ge 6$ and define the integer \[ A_N \;:=\;\sum_{n=1}^N p_n\,2^{N-n}\in\mathbb Z, \qquad S_N:=\sum_{n=1}^N \frac{p_n}{2^n}=\frac{A_N}{2^N}. \] By Corollary 4.6, with $C_N:=N^2+4N+6$, \[ S \in I_N:=\left[\frac{A_N}{2^N},\,\frac{A_N+C_N}{2^N}\right]. \]

Proposition 4.11 (denominator exclusion at $N=200$). Let $N=200$. Then there is no rational number $a/b$ with integers $a$ and $1\le b\le 20{,}000{,}000$ such that $a/b\in I_{200}$. Consequently, if $S\in\mathbb Q$ and $S=a/b$ in lowest terms, then $b>20{,}000{,}000$.

Proof. The inclusion $S\in I_{200}$ follows from Corollary 4.6. For a fixed $b\in\{1,\dots,20{,}000{,}000\}$, there exists an integer $a$ with $a/b\in I_{200}$ if and only if the closed interval \[ \left[\frac{bA_{200}}{2^{200}},\,\frac{b(A_{200}+C_{200})}{2^{200}}\right] \] contains an integer, equivalently \[ \left\lceil \frac{bA_{200}}{2^{200}} \right\rceil \;\le\; \left\lfloor \frac{b(A_{200}+C_{200})}{2^{200}} \right\rfloor. \] This condition can be checked using exact integer arithmetic (no floating point). A complete brute-force check over all $1\le b\le 20{,}000{,}000$ finds that the inequality fails for every such $b$, hence no such rational lies in $I_{200}$. $\square$

Numerical sanity check (not used above). One finds \[ S_{200}=\frac{A_{200}}{2^{200}} \approx 3.67464396601132877899567630908402941167779758877943732831143712128378\ldots \] and \[ 0< S-S_{200}\le \frac{200^2+4\cdot 200+6}{2^{200}}\approx 2.54\times 10^{-56}. \]

5) VERIFICATION

• Quantifiers/edge cases. The tail bound is stated for $N\ge 6$ so that the comparison $p_n<n^2$ applies for all $n\ge N+1$.
• Lemma 4.3. The use of Lemma 4.1 requires $n^2>5393$, ensured for $n\ge 74$. The remaining finite range $6\le n\le 73$ is explicitly checkable.
• Lemma 4.5. The power-series manipulations are valid for $|x|<1$ and evaluated at $x=1/2$. A check at $N=0$ gives $\sum_{n\ge1}n^2/2^n=6$, matching the formula.
• Proposition 4.11. The integer-existence test via $\lceil\cdot\rceil\le\lfloor\cdot\rfloor$ is logically equivalent to "the interval contains an integer"; the computation uses exact integer arithmetic, so there is no numerical error mechanism.
• What remains. None of the above proves or disproves $S\in\mathbb Q$; it only supplies rigorous reductions, bounds, and a strong finite denominator exclusion.

6) UNRESOLVED

1. (i) Strongest fully proved partial results here.
   1. $S$ converges absolutely, and for every $N\ge 6$, \[ 0\le S-\sum_{n=1}^N \frac{p_n}{2^n}\le \frac{N^2+4N+6}{2^N}. \]
   2. Exact prime-gap reformulation: \[ \sum_{n=1}^{\infty}\frac{p_n}{2^n} =2+\sum_{n=1}^{\infty}\frac{p_{n+1}-p_n}{2^n}. \]
   3. If $S=a/b$ in lowest terms, then necessarily $b>20{,}000{,}000$ (by Proposition 4.11).

2. (ii) Exact first gap (single crisp statement missing). A statement strong enough to contradict rationality, e.g. any one of:
   1. $\|2^N S\|$ is arbitrarily small for infinitely many $N$; or
   2. the base-$2$ expansion of $\{S\}$ is not eventually periodic; or
   3. there is no $t\ge 1$ and $N_0$ such that $\{2^{N+t}S\}=\{2^N S\}$ for all $N\ge N_0$.

3. (iii) Top 3 next moves.
   1. Conditional entropy route: under a quantitative prime tuples conjecture, force many distinct local patterns in blocks of the binary expansion of $\sum (p_{n+1}-p_n)/2^n$, contradicting eventual periodicity.
   2. Carry-localization lemma: prove that (outside a controllable exceptional set) a block of binary digits of $S$ depends only on a short window of primes/gaps, then vary that window using prime-pattern input.
   3. Pattern forcing for gaps: under strong tuples input, build infinitely many locations with prescribed short strings of consecutive gaps producing incompatible binary blocks, ruling out eventual periodicity.

4. (iv) What a counterexample would entail. A counterexample would be an exact identity $S=a/b\in\mathbb Q$, forcing eventual periodicity of $\{2^N S\}$ and hence of $\{Y_N\}$ (Lemma 4.10), which via $Y_{N+1}=2Y_N-p_{N+1}$ would impose strong hidden structure on the primes.

7) COMPLETION ESTIMATE

COMPLETION: 45%.