A364510 a(n) = binomial(4*n, n)^2.

1, 16, 784, 48400, 3312400, 240374016, 18116083216, 1401950721600, 110634634890000, 8862957169158400, 718528370729238784, 58818762721626513424, 4853704694918904043024, 403242220875862752160000, 33694913171561404510440000, 2829611125043050701300998400

Offset

0,2

Links

Paolo Xausa, Table of n, a(n) for n = 0..500

Wikipedia, Dixon's identity

Formula

a(n) = Sum_{i = -n..n} (-1)^i * binomial(2*n, n+i)^2 * binomial(4*n, 2*n+i).

Compare with Dixon's identity: Sum_{i = -n..n} (-1)^i * binomial(2*n, n+i)^3 = (3*n)!/n!^3.

a(n) = A005810(n)^2.

P-recursive: a(n) = 16 * ( (4*n - 1)*(4*n - 2)*(4*n - 3)/(3*n*(3*n - 1)*(3*n - 2)) )^2 * a(n-1) with a(0) = 1.

a(n) ~ c^n * 2/(3*Pi*n), where c = (2^16)/(3^6).

a(n) = [x^n] G(x)^(16*n), where the power series G(x) = 1 + x + 9*x^2 + 225*x^3 + 7525*x^4 + 295228*x^5 + 12787152*x^6 + 592477457*x^7 + 28827755219*x^8 + ... appears to have integer coefficients.

exp( Sum_{n > = 1} a(n)*x^n/n ) = F(x)^16, where the power series F(x) = 1 + x + 25*x^2 + 1033*x^3 + 53077*x^4 + 3081944*x^5 + 193543624*x^6 + 12835533333*x^7 + 886092805699*x^8 + ... appears to have integer coefficients.

The supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3^r)) hold for all primes p >= 5 and all positive integers n and r.

a(n) = [x^(3*n)] ( (1 - x)^(6*n)*Legendre_P(2*n, (1 + x)/(1 - x)) ). - Peter Bala, Aug 14 2023

Maple

seq( binomial(4*n, n)^2, n = 0..15);

Mathematica

A364510[n_]:=Binomial[4n, n]^2; Array[A364510, 15, 0] (* Paolo Xausa, Oct 05 2023 *)

Crossrefs

Cf. A002894, A005810, A364506, A364511, A364512.

Row 2 of A364509.

Sequence in context: A232519 A231083 A173984 * A186855 A159872 A220750

Adjacent sequences: A364507 A364508 A364509 * A364511 A364512 A364513

Keyword

nonn,easy

Author

Peter Bala, Jul 28 2023

Status

approved