A357674 a(n) = ( Sum_{k = 0..2*n} binomial(n+k-1,k) )^4 * ( Sum_{k = 0..2*n} binomial(n+k-1,k)^2 )^3.

1, 2187, 8422734375, 202402468703748096, 9223976224194016590174375, 587835594121137662072707812564687, 46157429480574073282465608886521546620928, 4181198339699286332943143923058721957212160000000, 420336565507755143573799144638372909582306681004894518439

Offset: 0

Author

Peter Bala, Oct 11 2022

Keywords

• nonn
• easy

Comments

Conjectures:

1) a(p) == a(1) (mod p^5) for all primes p >= 3 (checked up to p = 271).

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

3) Let m be a positive integer and set u(n) = ( Sum_{k = 0..m*n} binomial(n+k-1,k) )^(2*m) * ( Sum_{k = 0..m*n} binomial(n+k-1,k)^2 )^(m+1). Then the sequence {u(n)} satisfies the supercongruence u(p) == u(1) (mod p^5) for all primes p >= 7. This is the case m = 2. See A357672 for the case m = 1.

Examples

			Example of a supercongruence:
a(7) - a(1) = 4181198339699286332943143923058721957212160000000 - 2187 = (3^7)*(7^5)*211*298225180113209*1807736060307048120859243 == 0 (mod 7^5).
		

Crossrefs

Cf. A005809, A357565, A357566, A357671, A357672, A357673.

Programs

Maple

seq((add(binomial(n+k-1,k), k = 0..2*n))^4 * (add( binomial(n+k-1,k)^2, k = 0..2*n))^3, n = 0..20);

Mathematica

Table[Binomial[3*n,n]^4 * Sum[Binomial[n+k-1,k]^2, {k, 0, 2*n}]^3, {n, 0, 10}] (* Vaclav Kotesovec, May 31 2025 *)

PARI

a(n) = sum(k = 0, 2*n, binomial(n+k-1,k))^4 * sum(k = 0, 2*n, binomial(n+k-1,k)^2)^3; \\ Michel Marcus, Oct 24 2022

Formula

a(n) = ( A005809(n) )^4 * (Sum_{k = 0..2*n} binomial(n+k-1,k)^2 )^3.

a(n) ~ 3^(30*n+5) / (125 * Pi^5 * n^5 * 2^(20*n+10)). - Vaclav Kotesovec, May 31 2025