A375177 - OEIS

%I #13 Mar 12 2026 02:11:00

%S 1,5,26,134,670,3262,15540,72732,335478,1528670,6894316,30820660,

%T 136736236,602610764,2640266600,11508115320,49928451750,215717144670,

%U 928515985980,3983029119300,17032882625220,72631992447300,308911087394520,1310670689270280,5548646191175100

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

%C 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).

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

%H Romeo Meštrović, <a href="https://arxiv.org/abs/1111.3057">Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2012)</a>, arXiv:1111.3057 [math.NT], (2011).

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

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

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

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

%p seq( 1/(n+1)^2 *add( (k^4)*binomial(n+1, k)^2, k = 1..n+1), n = 0..25);

%p # Alternative: faster program for large n

%p 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:

%p seq(a(n), n = 0..25);

%Y Cf. A000984, A092443.

%K nonn,easy

%O 1,2

%A _Peter Bala_, Aug 02 2024