slug: hanna-nested-square-odd-coefficients
bibkey: hanna2026a392210
doi: null
url: https://oeis.org/A392210
triage: theorem
motivation_gids:
  - D5/S1/Recurrence/Parity/NestedSquareOddEven.result

The Nested-Square Generating Function Has Even Coefficients at Odd Indices

Problem

OEIS A392210, by Paul D. Hanna, Feb 02 2026:

G.f.: x + sq(x/(1-x) + sq(x/(1-2x) + sq(x/(1-3x) + sq(x/(1-4*x) + …)))) where sq(x) = x^2, an infinitely nested square.

The comment carrying the question:

It appears that a(2*n-1) is even for n > 1.

Motivation

The frozen theorem D5/S1/Recurrence/Parity/NestedSquareOddEven.result settles the comment for every solution of the entry’s functional equation A(x) = x + A(x/(1-x))^2, and shows that a solution exists.

Gap

Issue 9657 records the screen carried out before the work. The A-number appears nowhere under Problems/, D5/, Library/ or Blueprint/, nor in the screening records, nor in google-deepmind/formal-conjectures. The entry records no proof and no reference; its cross-references state no parity result for this sequence, and a catalogue search by the terms returns only this entry. This is a bounded negative finding.

Route

One. Write F = A(x/(1-x)). The coefficient of x^(2m+1) in F^2 is Σ_{i=0}^{2m+1} f_i f_(2m+1-i). Split the range into i ≤ m and i ≥ m+1; reflecting the second half by i ↦ 2m+1-i turns it into the first, so the sum is twice Σ_{i=0}^{m} f_i f_(2m+1-i).

Two. For n > 1 put 2n-1 = 2m+1 with m ≥ 1. The term x contributes nothing at that index, so the coefficient of A there equals that of F^2, which is even by step one.

Three. A solution exists. Define a(0) = 0, a(1) = 1 and, for n ≥ 2, a(n) = Σ_{k=1}^{n-1} b(k) b(n-k) with b(m) = Σ_{k=1}^{m} C(m-1, k-1) a(k), the entry’s own recurrence. Since (x/(1-x))^k has coefficient C(m-1, k-1) at x^m, substituting x/(1-x) into Σ a(k) x^k gives Σ b(m) x^m, and comparing coefficients of x + (Σ b(m) x^m)^2 with Σ a(n) x^n gives the equation.

Falsifier

The statement would fail if some odd-index coefficient were odd. Reading the index as 2n+1 with n ≥ 1 changes nothing; reading it as 2n-1 with n ≥ 1 would include a(1) = 1, which the comment excludes by requiring n > 1.

Evidence

The entry’s recurrence reproduces its first twelve terms, and iterating the functional equation on truncated power series reproduces the same twelve terms independently. The coefficients a(3) through a(39) at odd indices are all even.

Triage

theorem; Tier 1 named external open question from 2026, preregistered in issue 9657 before the work. The computational use is none: the delivered statement quantifies over every solution and every index.

ASSUMED-UNVERIFIED

The literature screen is bounded. The entry was read in full and records no proof and no reference to one. Citation indices, printed sources and the resolved set of arXiv:2608.11941 were not exhaustively compared, so no worldwide priority claim is made. The weight is stated plainly: the argument is a symmetry of the square, and what is settled is that the sentence sat on the entry unjudged.