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The problem, the result, and the mechanism of the proof.

Published · 2026 · Sole author

An Explicit Formula and a Recurrence for OEIS A265233

Juntao Jin

Journal of Integer Sequences 29 (2026), Article 26.5.3

I express the array count A265233 as a binomial transform of Franel numbers, prove Mathar’s recurrence, and derive its generating function and asymptotic growth.

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@article{Jin2026A265233,
  author = {Juntao Jin},
  title = {An Explicit Formula and a Recurrence for OEIS A265233},
  journal = {Journal of Integer Sequences},
  volume = {29},
  year = {2026},
  note = {Article 26.5.3},
  url = {https://cs.uwaterloo.ca/journals/JIS/VOL29/Jin/jin48.html}
}

What is counted

Let aₙ count 3 × n arrays on {0,1,2}. Each symbol occurs n times; vertically adjacent entries must differ, while the top and bottom entry of a column may agree. Read the first row from left to right, then the second and third: the symbols first appear in the order 0,1,2. Set a₀ = 1 for the empty array.

(a0,a1,a2,a3,…)=(1,1,7,56,…).(a_0,a_1,a_2,a_3,\ldots)=(1,1,7,56,\ldots).

Theorem 1: formula and recurrence

Write fₘ for the mth Franel number. The explicit formula holds for n ≥ 1; the empty array is treated separately. The recurrence is valid for n ≥ 4, with the four initial values above.

fm=∑k=0m(mk)3,an=16∑m=0n(nm)4 n−mfm(n≥1).\begin{aligned}f_m&=\sum_{k=0}^{m}\binom{m}{k}^{3},\\a_n&=\frac16\sum_{m=0}^{n}\binom nm4^{\,n-m}f_m\quad(n\ge1).\end{aligned}
n2an=(19n2−19n+6)an−1−96(n−1)2an−2+144(n−1)(n−2)an−3(n≥4).\begin{aligned}n^2a_n={}&(19n^2-19n+6)a_{n-1}\\&-96(n-1)^2a_{n-2}\\&+144(n-1)(n-2)a_{n-3}\quad(n\ge4).\end{aligned}

This proves the recurrence conjectured by Mathar in 2015. The displayed statements are Theorem 1 of the published paper, not a new claim added for this website.

The mechanism of the proof

First ignore the first-occurrence convention and count labeled arrays bₙ. For n ≥ 1 the six permutations of the three symbols act freely, so bₙ = 6aₙ. The twelve allowed columns have a polynomial that splits into a product and a single extra term:

P(x,y,z)=(x+y)(y+z)(z+x)+4xyz,bn=[xnynzn]P(x,y,z)n.\begin{aligned}P(x,y,z)&=(x+y)(y+z)(z+x)+4xyz,\\b_n&=[x^ny^nz^n]P(x,y,z)^n.\end{aligned}

The central coefficient of ((x+y)(y+z)(z+x))ᵐ is fₘ. Expanding the extra 4xyz term yields the binomial transform. Transforming the classical differential equation for the Franel generating function and extracting coefficients then gives the recurrence. The restriction n ≥ 4 keeps the exceptional empty-array normalization out of that conversion.

Corollary 2: generating function and growth

A(x)=∑n≥0anxn=2F1 ⁣(13,23;1;27x(1−4x)2)+56,\begin{aligned}A(x)&=\sum_{n\ge0}a_nx^n\\&=\frac{{}_2F_1\!\left(\frac13,\frac23;1;27x(1-4x)^2\right)+5}{6},\end{aligned}
an∼36π 12nn(n→∞).a_n\sim\frac{\sqrt3}{6\pi}\,\frac{12^n}{n}\qquad(n\to\infty).

Here ₂F₁ is the Gauss hypergeometric function. The paper derives van Hoeij’s recorded generating function and Kotesovec’s conjectured leading asymptotic. The proof uses a logarithmic singular expansion and an explicit error estimate; the full analytic conditions and error term are in §4–5.

Statement and proof summary checked against the official PDF and TeX source, Theorem 1, Corollary 2 and §§2–5. Full proof and conditions