slug: oeis-a204061-companion-pell-exp-base-five-residue bibkey: hanna2012a204061 doi: null url: https://oeis.org/A204061 triage: theorem motivation_gids:

• D5/S1/Recurrence/Invariants/CompanionPellExpBaseFiveResidue

Hanna’s A204061 coefficients modulo five

Problem

OEIS A204061, NAME (%N, verbatim):

G.f.: exp( Sum_{n>=1} A001333(n)^2 * x^n/n ) where A001333(n) = A002203(n)/2, one-half the companion Pell numbers.

COMMENT conjecture (%C, verbatim):

a(n) == 1 (mod 5) iff n has no 2’s in its base 5 expansion (A023729), otherwise a(n) == 0 (mod 5); this is a conjecture needing proof.

FORMULA (%F, verbatim):

G.f.: 1 / ( sqrt(1+x) * (1-6*x+x^2)^(1/4) ).

AUTHOR (%A, verbatim):

_Paul D. Hanna_, Jan 10 2012

The exact formal target is, for every n : Nat, exists z : Z_5, (z : Q_5) = a(n) and PadicInt.toZMod z = (if 2 in Nat.digits 5 n then 0 else 1). Here a(n) is coefficient n of the defining exponential in Q_5. Its exponent uses the frozen sequence Q : Nat → Nat from D5/S1/Recurrence/PellCompanionGcd: A001333, one-half the companion Pell numbers, canonically embedded into Q_5. This proves the residue conjecture in the 5-adic reading.

Motivation

The conjecture links a recurrence-defined exponential to a base-five digit condition at every natural index. Its unconditional residue law also constructs a 5-adic integral representative for each coefficient.

Gap

The orchestrator’s readings dated 2026-09-15 report that the OEIS revision still says “conjecture needing proof”, with no settlement in %C, %F, or %H. The supplied search results were: arXiv API all:"A204061", zero entries; OpenAlex "A204061", zero; MathOverflow API, zero; GitHub code search, only joeis/loda/OEIS-Spider mirrors and no Lean proof; formal-conjectures, zero. The repository prior-art.sh search reported zero A204061 hits and only PellCompanionGcd for A001333. That module supplies the frozen defining sequence and proves a gcd statement. Pre-registration: issue #7889.

Route

1. For the defining Pell-square exponential f, the order-three recurrence of the squared frozen sequence Q gives a differential equation. Formal differentiation proves constant coefficient one and f^4 * ((1+X)^2 * (1-6X+X^2)) = 1.
2. Let R=(1+X)^2*(1-6X+X^2). Substitute R-1 into the binomial series with exponent -4.inv over Z_5 to construct g with g^4*R=1 and constant coefficient one. Uniqueness of the fourth root with that constant coefficient in Q_5 identifies its image with f.
3. Reduce g modulo five. Frobenius and the quartic equation give B=(1+X+X^3+X^4)*B(X^5) over ZMod 5, so coefficient n is zero when n mod 5=2 and otherwise equals coefficient n div 5.
4. Inside result’s proof, a have performs strong induction through n div 5 and identifies the coefficients with the indicator that the base-five expansion has no digit two. Apply this to the residue of coefficient n of g.

Falsifier

A natural index whose exponential coefficient is not 5-adic integral, or whose residue is neither the specified zero nor the specified one, would contradict the theorem. At index zero the digit list is empty and the coefficient is one. Finite tests alone cannot establish the universal law.

Evidence

• lake env lean D5/S1/Recurrence/Invariants/CompanionPellExpBaseFiveResidue.lean: exit 0. The public declarations are exactly a and result.
• #print axioms D5.S1.Recurrence.Invariants.CompanionPellExpBaseFiveResidue.result: [propext, Classical.choice, Quot.sound]. The source contains no sorry.
• tools/scripts/agent/header-check.sh D5/S1/Recurrence/Invariants/CompanionPellExpBaseFiveResidue.lean: exit 0; seven-line header, generality I, utility: none.
• The orchestrator reports: measured by the orchestrator outside the sandbox: lean-report 0, emit 0 ×2, deposit-uncovered 0.

Single-import deletion checks, supplied to Lean by process substitution:

Triage

theorem, resolution proved for the exact kernel-checked 5-adic statement above.

ASSUMED-UNVERIFIED

Exhaustive literature coverage, priority, and integrality over ℤ are not established by this result.