A364515 a(n) = (1/2) * (6*n)!*(2*n)!^2/((3*n)!*(4*n)!*n!^3) for n >= 1 with a(0) = 1.

1, 10, 594, 44200, 3640210, 317678760, 28782923400, 2677247862432, 253909524661650, 24443479528347880, 2381221777637046344, 234220654553738055840, 23223217841360039079304, 2318164676869886054980000, 232736024465276636260692000, 23482287467017571303962819200

Offset

0,2

Comments

Row 4 of A364513.

Links

Paolo Xausa, Table of n, a(n) for n = 0..400

Formula

a(n) = [x^n] (1 - x)^(2*n) * Legendre_P(4*n-1, (1 + x)/(1 - x)) for n >= 1.

a(n) = Sum_{k = 0..n} binomial(4*n-1, n-k)^2 * binomial(2*n+k-2, k).

a(n) = (1/2) * binomial(6*n, 2*n)*binomial(2*n, n)^2 / binomial(3*n, n).

a(n) = (4*n - 1)!*(3*n - 1/2)!*(n - 1/2)!/((2*n - 1)! * (2*n - 1/2)!^2 * n!^2) for n >= 1 (fractional factorials are defined in terms of the gamma function, for example, (3*n - 1/2)! = gamma(3*n + 1/2)).

a(n) ~ 108^n * sqrt(2)/(4*Pi*n).

P-recursive: a(0) = 1; for n >= 1, a(n) = 12*(6*n-1)*(6*n-5)*(2*n-1)^2 / ((4*n-1)*(4*n-3)*n^2) * a(n-1) with a(1) = 10.

Conjecture: 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.

Example

Examples of supercongruences:
a(7) - a(1) = 2677247862432 - 10 = 2*(7^3)*1999*1952323 == 0 (mod 7^3).
a(2*7) - a(2) = 232736024465276636260692000 - 594 = 2*(3^3)*(7^3)*1097* 11454314520615738059 == 0 (mod 7^3).

Maple

seq( (1/2) * (6*n)!*(2*n)!^2/((3*n)!*(4*n)!*n!^3), n = 1..15);

Mathematica

A364515[n_]:=If[n==0, 1, (1/2)(6n)!(2n)!^2/((3n)!(4n)!n!^3)]; Array[A364515, 15, 0] (* Paolo Xausa, Oct 05 2023 *)

Crossrefs

Cf. A364513.

Sequence in context: A323532 A273032 A185277 * A006441 A042751 A285372

Adjacent sequences: A364512 A364513 A364514 * A364516 A364517 A364518

Keyword

nonn,easy

Author

Peter Bala, Aug 01 2023

Status

approved