%I #20 Jul 15 2021 03:10:35
%S 1,1,3,199,249337,6062674201,3653786369479951,65709007885111803731947,
%T 40564683796482484146182142025377,
%U 969773549559254966290998252899999751714721,999999990999996719397362087568018696141879478712251051,49037072510879011742983689973641327840345400616866967292640434759551
%N a(n) = n! * [x^n] (Sum_{j>=0} n^(j*(j+1)/2) * x^j/j!)^(1/n) for n > 0, a(0) = 1.
%C All terms are odd.
%H Alois P. Heinz, <a href="/A342578/b342578.txt">Table of n, a(n) for n = 0..35</a>
%H Richard Stanley, <a href="https://mathoverflow.net/q/385402">Proof of the general conjecture</a>, MathOverflow, March 2021.
%F a(n) == 1 (mod n*(n-1)) for n >= 2 (see "general conjecture" in A178319 and link to proof by _Richard Stanley_ above).
%F a(n) ~ n^((n^2 + n - 2)/2). - _Vaclav Kotesovec_, Jul 15 2021
%p a:= n-> `if`(n>0, coeff(series(add(n^binomial(j+1, 2)*
%p x^j/j!, j=0..n)^(1/n), x, n+1), x, n)*n!, 1):
%p seq(a(n), n=0..12);
%Y Cf. A000217, A023813, A178315, A178319.
%Y Main diagonal of A346061.
%K nonn
%O 0,3
%A _Alois P. Heinz_, Mar 15 2021