A375177 a(n) = 1/(n + 1)^2 * Sum_{k = 1..n+1} (k^4)*binomial(n+1, k)^2.

1, 5, 26, 134, 670, 3262, 15540, 72732, 335478, 1528670, 6894316, 30820660, 136736236, 602610764, 2640266600, 11508115320, 49928451750, 215717144670, 928515985980, 3983029119300, 17032882625220, 72631992447300, 308911087394520, 1310670689270280, 5548646191175100

Offset

1,2

Comments

Compare with the identity 1/(n+1)^2 * Sum_{k = 1..n+1} (k^2)*binomial(n+1, k)^2 = binomial(2*n, n) = A000984(n).

The central binomial coefficients satisfy the supercongruence binomial(2*p, p) == 2 (mod p^3) for all primes p >= 5 (Wolstenholme's theorem).

Links

Table of n, a(n) for n=1..25.

Romeo Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2012), arXiv:1111.3057 [math.NT], (2011).

Formula

a(n) = hypergeom([2, 2, -n, -n], [1, 1, 1], 1).

a(n) = (1/2) * P(n)/(2*n - 1) * binomial(2*n, n), where P(n) = n^3 + 4*n^2 + 2*n - 2.

P-recursive: n*P(n-1)*a(n) = 2*(2*n - 3)*P(n)*a(n-1) with a(1) = 1.

a(p) == 2*p + 2 (mod p^3) for all primes p >= 5.

Maple

seq( 1/(n+1)^2 *add( (k^4)*binomial(n+1, k)^2, k = 1..n+1), n = 0..25);
# faster program for large n
a := proc (n) option remember; if n = 0 then 1 else (4*n-6)*(n^3+4*n^2+2*n-2)*a(n-1)/(n*(n^3+n^2-3*n-1)) end if; end:
seq(a(n), n = 0..25);

Crossrefs

Cf. A000984, A092443.

Sequence in context: A047757 A047755 A047768 * A022032 A255118 A052918

Adjacent sequences: A375174 A375175 A375176 * A375178 A375179 A375180

Keyword

nonn,easy

Author

Peter Bala, Aug 02 2024

Status

approved