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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.

%I A357674 #19 May 31 2025 03:41:17
%S A357674 1,2187,8422734375,202402468703748096,9223976224194016590174375,
%T A357674 587835594121137662072707812564687,
%U A357674 46157429480574073282465608886521546620928,4181198339699286332943143923058721957212160000000,420336565507755143573799144638372909582306681004894518439
%N 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.
%C A357674 Conjectures:
%C A357674 1) a(p) == a(1) (mod p^5) for all primes p >= 3 (checked up to p = 271).
%C A357674 2) For r >= 2, and all primes p >= 3, a(p^r) == a(p^(r-1)) ( mod p^(3*r+3) ).
%C A357674 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.
%F A357674 a(n) = ( A005809(n) )^4 * (Sum_{k = 0..2*n} binomial(n+k-1,k)^2 )^3.
%F A357674 a(n) ~ 3^(30*n+5) / (125 * Pi^5 * n^5 * 2^(20*n+10)). - _Vaclav Kotesovec_, May 31 2025
%e A357674 Example of a supercongruence:
%e A357674 a(7) - a(1) = 4181198339699286332943143923058721957212160000000 - 2187 = (3^7)*(7^5)*211*298225180113209*1807736060307048120859243 == 0 (mod 7^5).
%p A357674 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);
%t A357674 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 *)
%o A357674 (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
%Y A357674 Cf. A005809, A357565, A357566, A357671, A357672, A357673.
%K A357674 nonn,easy
%O A357674 0,2
%A A357674 _Peter Bala_, Oct 11 2022