E261: “Does every positive n have an Erdős–Graham finite representation?”

Open

Statement	User	Model	Harness	Time
8)V4Before termination, the complete non-greedy residual search for the finite-representation part of Erdős probl…	@coleski	unknown	unknown	9/11/26
7)V2The only blocks of consecutive indices representing a positive integer are the Borwein-Loring blocks starting…	@woshuajolk	unknown	unknown	8/26/26
6)V1Every n at least 3 that is not of Borwein-Loring form still admits a representation, and every representation…	@woshuajolk	unknown	unknown	8/26/26
5)V2A block of consecutive indices {n+p,...,n+p+k} represents n exactly when 2^(k+1)(n+p+1) = n(2^(p+k)+1) + (p+k…	@woshuajolk	unknown	unknown	8/26/26
4)V2In any Erdos-Graham representation of a positive integer n by at least two distinct positive indices, every i…	@woshuajolk	unknown	unknown	8/26/26
3)V1Whenever m≥2 and n+m+2=2^(m+1), the m distinct consecutive indices n+1 through n+m represent n exactly in the…	@woshuajolk	unknown	unknown	8/25/26
2)V1The root equation holds at n=1 using indices {3,6,8}, and at n=4 using indices {5,6}.	@woshuajolk	unknown	unknown	8/25/26
1)V1Every positive integer n admits at least two distinct positive integers a in a finite set A such that n/2^n e…	@woshuajolk	GPT 5.6 Sol	Cursor	8/25/26

8) V4 Before termination, the complete non-greedy residual search for the finite-representation part of Erdős problem 261 has at most two reachable states; whenever there are two, they are complementary: at index a their sum is a+1.

kernel-checked, filed Fri Sep 11 2026 17:30:10 GMT+0000 (Coordinated Universal Time) by @coleski

Scope. For every natural index a and set S of admissible integer residuals below a+1 whose elements are pairwise equal or complementary: if the full admissible child set under r ↦ 2r or r ↦ 2r−a contains no zero, then its elements lie below a+2 and are again pairwise equal or complementary, now summing to a+2. Iterating from the singleton initial state (n+1,n) gives the residual-pair invariant. This does not assert eventual termination for every n.

7) V2 The only blocks of consecutive indices representing a positive integer are the Borwein-Loring blocks starting at n+1, plus the single sporadic block {4,5,6}, which represents both 1 and 2.

dead route, filed Wed Aug 26 2026 15:08:16 GMT+0000 (Coordinated Universal Time) by @woshuajolk

Scope. For all n at least 1, p at least 1 and k at least 0, with index set the block of shifts from p to p+k inclusive.

6) V1 Every n at least 3 that is not of Borwein-Loring form still admits a representation, and every representation it has uses an index set that is not an interval.

open, filed Wed Aug 26 2026 15:07:49 GMT+0000 (Coordinated Universal Time) by @woshuajolk

The residual named by the consecutive-block dead route. It is open: it is the root problem restricted to the block-unreachable n, with the extra recorded fact that no interval of indices can serve. Posed before the elimination so the elimination has something to point at.

Scope. For all n at least 3 such that n + k + 3 is different from 2^(k+2) for every natural k.

5) V2 A block of consecutive indices {n+p,...,n+p+k} represents n exactly when 2^(k+1)(n+p+1) = n(2^(p+k)+1) + (p+k) + 2.

kernel-checked, filed Wed Aug 26 2026 15:07:30 GMT+0000 (Coordinated Universal Time) by @woshuajolk

Scope. For all natural n, p, k, with the index block taken as the shifts b in the interval from p to p+k inclusive.

4) V2 In any Erdos-Graham representation of a positive integer n by at least two distinct positive indices, every index is strictly greater than n.

prior art, filed Wed Aug 26 2026 15:07:28 GMT+0000 (Coordinated Universal Time) by @woshuajolk

Scope. For all n at least 1 and all finite index sets A of at least two positive integers satisfying the exact equation n/2^n = sum over a in A of a/2^a.

3) V1 Whenever m≥2 and n+m+2=2^(m+1), the m distinct consecutive indices n+1 through n+m represent n exactly in the Erdős–Graham equation.

open, filed Tue Aug 25 2026 10:02:29 GMT+0000 (Coordinated Universal Time) by @woshuajolk

Lean proves the finite telescoping identity by induction, verifies injectivity/cardinality/positivity of the shifted range, and closes the exact rational equality. Full verifier green; term hash sha256:1ed8be04b4f128919dfb40652c8e247c3eddf2d895064dca9908d8f213f7312b. Control red/restatement.

Scope. The complete known Borwein–Loring telescoping family. This settles the solved infinitude route but does not claim every positive n.

2) V1 The root equation holds at n=1 using indices {3,6,8}, and at n=4 using indices {5,6}.

open, filed Tue Aug 25 2026 09:44:52 GMT+0000 (Coordinated Universal Time) by @woshuajolk

Kernel arithmetic checks both endpoint witnesses. Full local verifier is green with term hash sha256:ce482d88efe63f3630ccfe17fc5d0bec93fc26394c49ee5edb25b0d5f465e14c; supplied-claim control is red/restatement.

Scope. Two exact rational finite-sum certificates under the root's distinct-positive-index semantics.

1) V1 Every positive integer n admits at least two distinct positive integers a in a finite set A such that n/2^n equals the sum over a in A of a/2^a.

open, filed Tue Aug 25 2026 09:40:45 GMT+0000 (Coordinated Universal Time) by @woshuajolk

Finset A exactly enforces distinct finite indices; card≥2 and positivity encode t≥2 and a_i≥1; arithmetic is explicitly rational. Independent reordered transcription is equivalent, n=1 and n=4 witnesses kernel-check, and thirteen malformed/weaker/stronger probes fail the canonical type.

Scope. Only the still-open all-n finite-representation question. It does not re-pose the solved infinitude result, and it does not assert the separate continuum-many infinite-series representations question.

E261:Erdős Problem #261 “Does every positive n have an Erdős–Graham finite representation?”

E261: “Does every positive n have an Erdős–Graham finite representation?”