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%I #13 Mar 28 2023 07:37:49
%S 1,2,11,124,2225,56546,1928707,85029596,4687436609,314255427490,
%T 25077179715131,2343489559096412,253185531592066801,
%U 31279831940279656514,4376923336721600128115,687815536092999747916156,120491486068612766739548417,23378730923206887237941740226
%N a(n) = n! * [x^n] (1 + x)^n * exp(x*(1 + x)^n).
%C We conjecture that a(n+k) == a(n) (mod k) for all n and k. If true, then for each k, the sequence a(n) taken modulo k is a periodic sequence and the period divides k. For example, modulo 7 the sequence becomes [1, 2, 4, 5, 6, 0, 4, 1, 2, 4, 5, 6, 0, 4, 1, 2, 4, 5, 6, 0, 4, ...], apparently a periodic sequence of period 7.
%C More generally, let F(x) and G(x) denote power series with integer coefficients with F(0) = G(0) = 1. Define b(n) = n! * [x^n] exp(x*G(x)^n)*F(x)^n. Then we conjecture that b(n+k) == b(n) (mod k) for all n and k.
%H Vaclav Kotesovec, <a href="/A361036/b361036.txt">Table of n, a(n) for n = 0..270</a>
%F a(n) = n!*Sum_{i = 0..n} Sum_{j = 0..n-i} binomial(n,i+j)*binomial(j*n,i)/j!.
%F a(n) ~ n! * exp(r*(1+r)^n) * (1+r)^(n/2 + 1) / (sqrt(2*Pi*n*(3 + n*r)) * r^(n+1)), where r = 2*LambertW(n/2)/n - (n + 2*LambertW(n/2)) * (n - 4*LambertW(n/2)^3) / (n^3 * (3 + 2*LambertW(n/2))). - _Vaclav Kotesovec_, Mar 28 2023
%p seq( n!*add(add(binomial(n,i+j)*binomial(j*n,i)/j!, j = 0..n-i), i = 0..n), n = 0..20);
%t Table[n! * Sum[Sum[Binomial[n, i + j]*Binomial[j*n, i]/j!, {j, 0, n - i}], {i, 0, n}], {n, 0, 20}] (* _Vaclav Kotesovec_, Mar 27 2023 *)
%Y Cf. A047974, A278070, A361281.
%K nonn,easy
%O 0,2
%A _Peter Bala_, Mar 13 2023
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