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import FormalConjecturesUtilCoefficient of $(xyz)^n$ in $((x+y)(y+z)(z+x))^n$ equaling $\binom{2n}{n}^3$
The sequence $a(n)$ is defined by $a(n) = \binom{2n}{n}^3$.
References:
arxiv/2605.22763 Advancing Mathematics Research with AI-Driven Formal Proof Search by George Tsoukalas et al.
namespace OeisA2897open Polynomialopen Nat MvPolynomial
The sequence $a(n)$ is defined by $a(n) = \binom{2n}{n}^3$.
We use Nat.choose (2 * n) n for the central binomial coefficient.
def a (n : ℕ) : ℕ := (Nat.choose (2 * n) n) ^ 3abbrev Vars := Fin 3
The finsupp corresponding to the monomial $x^n y^n z^n$.
This is the map $\lambda i. n$. Since Fin 3 is finite, this function is finitely supported.
noncomputable def xyzPowN (n : ℕ) : Finsupp Vars ℕ :=
Finsupp.ofSupportFinite (fun _ : Vars => n) (Set.toFinite _)local notation "P" => MvPolynomial Vars ℤThe polynomial $pPoly(X, Y, Z) = (1 + X + Y + Z)^{2n} (1 + X + Y - Z)^n (1 + X - Y + Z)^n$. We identify $X_0, X_1, X_2$ with $X, Y, Z$.
noncomputable def pPoly (n : ℕ) : P :=
let X := MvPolynomial.X 0
let Y := MvPolynomial.X 1
let Z := MvPolynomial.X 2
let p1 : P := 1 + X + Y + Z
let p2 : P := 1 + X + Y - Z
let p3 : P := 1 + X - Y + Z
p1 ^ (2 * n) * p2 ^ n * p3 ^ n@[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 = 8 := ⊢ a 1 = 8 All goals completed! 🐙@[category test, AMS 11]
lemma a_2 : a 2 = 216 := ⊢ a 2 = 216 All goals completed! 🐙@[category test, AMS 11]
lemma a_3 : a 3 = 8000 := ⊢ a 3 = 8000 All goals completed! 🐙@[category test, AMS 11]
lemma a_4 : a 4 = 343000 := ⊢ a 4 = 343000 All goals completed! 🐙Conjecture: $a(n) = [x^n y^n z^n] (1+x+y+z)^{2n} (1+x+y-z)^n (1+x-y+z)^n$. - Peter Bala, Apr 10 2022
A formal proof has been found with the methods described in arxiv/2605.22763.
@[category research solved, AMS 11, formal_proof using formal_conjectures at
"https://github.com/mo271/formal-conjectures/blob/a32396489dcb8f86c3549b93aa358ac6a10a3a1f/FormalConjectures/OEIS/2897.wip.lean#L408"]
theorem a_eq_coeff (n : ℕ) : (a n : ℤ) = MvPolynomial.coeff (xyzPowN n) (pPoly n) := n:ℕ⊢ ↑(a n) = MvPolynomial.coeff (xyzPowN n) (pPoly n)
All goals completed! 🐙end OeisA2897