%I #13 May 11 2023 16:57:15
%S 1,4,68,1336,27972,607004,13478072,304083224,6941422916,159882680452,
%T 3708781743068,86526900550864,2028273983776440,47733938489878528,
%U 1127187050415921304,26694934151138897336,633813403549444601156,15082008687681962081088,359592614152718921447108
%N a(n) = [x^n] E(x)^n where E(x) = exp( Sum_{k >= 1} binomial(2*k,k)^2*x^k/k ).
%C Compare with A359108(n) = [x^n] F(x)^n where F(x) = exp( Sum_{k >= 1} binomial(2*k,k)*x^k/k ).
%C More generally, we inductively define a family of sequences {a(i,n) : n >= 0}, i >= 0, by setting a(0,n) = binomial(2*n,n)^2 and, for i >= 1, a(i,n) = [x^n] ( exp(Sum_{k >= 1} a(i-1,k)*x^k/k) )^n. In this notation the present sequence is {a(1,n)}.
%C We conjecture that the sequences {a(i,n) : n >= 0}, i >= 1, satisfy the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5, and positive integers n and r.
%F Conjecture: the supercongruence a(n*p^r) == a(n(p^(r-1)) (mod p^(3*r)) holds for all primes p >= 5 and positive integers n and r.
%p E(n,x) := series( exp(n*add(binomial(2*k,k)^2*x^k/k, k = 1..20)), x, 21 ):
%p seq(coeftayl(E(n,x), x = 0, n), n = 0..20);
%Y Cf. A000984, A002894, A359108, A362722 - A362733.
%K nonn,easy
%O 0,2
%A _Peter Bala_, May 05 2023