1. FormalConjectures
  2. OEIS
  3. «323557»
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Pronic indices for odd coefficients of $\sum_{n \ge 0} x^n \frac{(1+x^n)^n}{(1+x^{n+1})^{n+1}}$

Coefficients of G.f. $\sum_{n\ge 0} x^n \cdot \frac{(1 + x^n)^n}{(1 + x^{n+1})^{n+1}}$. The $m$-th term $a(m)$ is the coefficient of $x^m$, which is explicitly given by the sum: $$ a(m) = \sum_{n=0}^m \sum_{k=0}^n \binom{n}{k} (-1)^j \binom{n+j}{j},$$ where $j = \frac{m - n(k+1)}{n+1}$, and the term is zero unless $j$ is a natural number.

References:

    A323557

    arxiv/2605.22763 Advancing Mathematics Research with AI-Driven Formal Proof Search by George Tsoukalas et al.

@[expose] public sectionnamespace OeisA323557open Nat

Coefficients of G.f. $\sum_{n\ge 0} x^n \cdot \frac{(1 + x^n)^n}{(1 + x^{n+1})^{n+1}}$. The $m$-th term $a(m)$ is the coefficient of $x^m$, which is explicitly given by the sum: $$ a(m) = \sum_{n=0}^m \sum_{k=0}^n \binom{n}{k} (-1)^j \binom{n+j}{j},$$ where $j = \frac{m - n(k+1)}{n+1}$, and the term is zero unless $j$ is a natural number.

def a (m : ℕ) : ℤ := Finset.sum (Finset.range (m + 1)) fun n => Finset.sum (Finset.range (n + 1)) fun k => let exp_x_num := n * (k + 1) if exp_x_num ≤ m then let remainder := m - exp_x_num if (n + 1) ∣ remainder then let j : ℕ := remainder / (n + 1) let c₁ : ℤ := (n.choose k) let c₂ : ℤ := (choose (n + j) j) let sign : ℤ := if Even j then 1 else -1 sign * c₁ * c₂ else 0 else 0@[category test, AMS 11] lemma a_0 : a 0 = 1 := ⊢ a 0 = 1 All goals completed! 🐙@[category test, AMS 11] lemma a_1 : a 1 = 0 := ⊢ a 1 = 0 All goals completed! 🐙@[category test, AMS 11] lemma a_2 : a 2 = 3 := ⊢ a 2 = 3 All goals completed! 🐙@[category test, AMS 11] lemma a_3 : a 3 = -2 := ⊢ a 3 = -2 All goals completed! 🐙@[category test, AMS 11] lemma a_4 : a 4 = 2 := ⊢ a 4 = 2 All goals completed! 🐙

Conjecture: Odd terms occur only at positions $n(n+1)$ for $n \ge 0$ (verified for initial 32600 terms).

A formal proof has been found with the methods described in arxiv/2605.22763.

This statement is equivalent to OeisA325046.odd_a_implies_pronic. The two sequences differ only by the sign $(-1)^j$ on each summand, so they agree modulo $2$ and have the same odd indices.

@[category research solved, AMS 11, formal_proof using formal_conjectures at "https://github.com/mo271/formal-conjectures/blob/a32396489dcb8f86c3549b93aa358ac6a10a3a1f/FormalConjectures/OEIS/323557.wip.lean#L193"] theorem odd_a_implies_pronic (m : ℕ) : Odd (a m) → ∃ n : ℕ, m = n * (n + 1) := m:ℕ⊢ Odd (a m) → ∃ n, m = n * (n + 1) All goals completed! 🐙end OeisA323557