A364514 a(0) = 1 and a(n) = [x^n] (1 - x)^(2*n) * Legendre_P(n-1, (1 + x)/(1 - x)) for n >= 1.

1, -2, 0, 16, 0, -252, 0, 4800, 0, -100100, 0, 2201472, 0, -50139936, 0, 1170614016, 0, -27839740500, 0, 671626956000, 0, -16388657193480, 0, 403645030064640, 0, -10018806017062752, 0, 250305475771456000, 0, -6288594802355952000, 0, 158759294846918261760

Offset

0,2

Comments

Row 1 of A364513.

Links

Paolo Xausa, Table of n, a(n) for n = 0..1000

Formula

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

a(2*n) = 0 for n >= 1; a(2*n+1) = (-1)^(n+1) * 2/(2*n + 1) * (3*n + 1)!/n!^3.

a(2*n+1) ~ (-1)^(n+1) * 3^(3*n) * 3*sqrt(3)/(2*Pi*n).

P-recursive: a(0) = 1, a(1) = -2 and for n >= 2, a(n) = -(3*n - 1)*(3*n - 5)*(3*n - 6)/(n*(n - 1)^2) * a(n-2).

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.

a(n) = Sum_{k = 0..n} (-1)^k * binomial(n+k-1, k) * binomial(n-1, k) * binomial(2*n-k, n-k). - Peter Bala, Aug 13 2023

Maple

a := proc(n) option remember; if n = 0 then 1 elif n = 1 then -2 else -(3*n - 1)*(3*n - 5)*(3*n - 6)/(n*(n - 1)^2) * a(n-2) end if; end:
seq(a(n), n = 0..15);

Mathematica

A364514[n_]:=A364514[n]=Which[n==0, 1, n==1, -2, True, -(3n-1)(3n-5)(3n-6)/(n(n-1)^2)A364514[n-2]]; Array[A364514, 40, 0] (* Paolo Xausa, Oct 05 2023 *)

Crossrefs

Cf. A364513.

Sequence in context: A086261 A111978 A146558 * A025600 A009006 A155585

Adjacent sequences: A364511 A364512 A364513 * A364515 A364516 A364517

Keyword

sign,easy

Author

Peter Bala, Aug 02 2023

Status

approved