A357566 a(n) = ( Sum_{k = 0..n} binomial(n+k-1,k)^2 )^3 * ( Sum_{k = 0..n} binomial(n+k-1,k)^3 )^2.

1, 32, 3556224, 4816142496896, 14260946236464636800, 62923492736113950202540032, 355372959542696519903013302282592, 2376354966106399942850054560101358877184, 17973185649572984869873798116070605084766512000, 149319509846904520286037745483655872001727895961600000

Offset: 0

Author

Peter Bala, Oct 16 2022

Keywords

• nonn
• easy

Comments

Conjectures:

1) a(p) == a(1) (mod p^5) for all odd primes p except p = 5 (checked up to p = 271).

2) a(p^r) == a(p^(r-1)) (mod p^(3*r+3)) for r >= 2 and all primes p >= 5.

3) More generally, let m be a positive integer and set u(n) = ( Sum_{k = 0..m*n} binomial(n+k-1,k)^2 )^(m+2) * ( Sum_{k = 0..m*n} binomial(n+k-1,k)^3 )^(2*m). Then the supercongruences u(p) == u(1) (mod p^5) hold for all primes p >= 5.

4) u(p^r) == u(p^(r-1)) (mod p^(3*r+3)) for r >= 2 and all primes p >= 5.

Examples

			a(7) - a(1) = 2376354966106399942850054560101358877184 - 32 = (2^5)*(7^5)*19*31*317*339247*25170329*2771351868561767 == 0 (mod 7^5).
		

Crossrefs

Cf. A357565, A357671, A357672, A357673, A357674.

Cf. A333592.

Programs

Maple

seq((add(binomial(n+k-1,k)^2, k = 0..n))^3 * (add( binomial(n+k-1,k)^3, k = 0..n))^2, n = 0..20);

Mathematica

Table[Sum[Binomial[n+k-1,k]^2, {k, 0, n}]^3 * Sum[Binomial[n+k-1,k]^3, {k, 0, n}]^2, {n, 0, 10}] (* Vaclav Kotesovec, May 31 2025 *)

PARI

a(n) = sum(k = 0, n, binomial(n+k-1,k)^2)^3 * sum(k = 0, n, binomial(n+k-1,k)^3)^2; \\ Michel Marcus, Oct 25 2022

Formula

a(n) ~ 2^(24*n) / (1323*Pi^6*n^6). - Vaclav Kotesovec, May 31 2025