bibkey: hanna2025a378578 authors: Paul D. Hanna year: 2025 title: OEIS A378578, g.f. obtained by removing all factors of 3 from the coefficients of 1 + xA(x)^3 doi: null url: https://oeis.org/A378578 claim: “G.f. A(x) equals the series obtained by removing all factors of 3 from the coefficients in 1 + xA(x)^3. Conjecture: a(n) == binomial(3n,n)/(2n+1) (mod 2) for n >= 0.” strata_touched:

• D5/S1/Recurrence/Parity/StripThreeTernaryCatalanParity license: citation-only triage: anchor

OEIS A378578

Paul D. Hanna’s entry of January 3, 2025 defines the generating series by removing all factors of three from each coefficient of 1 + x*A(x)^3. The constant coefficient is one. For positive n, the coefficient a(n) is the three-free part of the coefficient of degree n-1 in A(x)^3. The entry conjectures that a(n) has the parity of the ternary Catalan number binomial(3*n,n)/(2*n+1) for every natural n.

The Lean module constructs the integer series by stabilizing successive approximations and proves its defining equation and uniqueness. Removing factors of three preserves parity. Frobenius and Lucas’s theorem identify the reduced coefficients with binomial(3*n,n) modulo two. Exact natural division by 2*n+1 preserves this parity. The subtraction formula for the quotient is stated only for positive n; the quotient at n=0 is one.

Verified locator

• URL: https://oeis.org/A378578
• NAME: “G.f. A(x) equals the series obtained by removing all factors of 3 from the coefficients in 1 + x*A(x)^3.”
• COMMENT: “Conjecture: a(n) == binomial(3n,n)/(2n+1) (mod 2) for n >= 0.”
• AUTHOR: Paul D. Hanna, Jan 03 2025.