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%I #15 Dec 27 2019 09:57:24
%S 1,1,321,699121,5552351121,117029959485121,5402040231378569121,
%T 480086443888959812703121,74896283763383392805211587121,
%U 19133358944433370977791260580721121,7581761490297442738124283591348762605121,4461925444770180839552702516305804230194739121
%N Number of alignments of n strings of length 4.
%C Conjectures: a(n) == 1 (mod 80); for fixed k, the sequence a(n) (mod k) eventually becomes periodic. - _Peter Bala_, Dec 19 2019
%H Alois P. Heinz, <a href="/A062205/b062205.txt">Table of n, a(n) for n = 0..100</a>
%F From _Vaclav Kotesovec_, Mar 22 2016: (Start)
%F a(n) ~ 2^(5*n-3) * n!^4 / (Pi^(3/2) * n^(3/2) * 3^n * (log(2))^(4*n+1)).
%F a(n) ~ sqrt(Pi) * 2^(5*n-1) * n^(4*n+1/2) / (3^n * exp(4*n) * (log(2))^(4*n+1)).
%F (End)
%F It appears that a(n) = (1/(2*6^n))*Sum_{k = 0..n} (-1)^(n-k)*binomial(n,k) *A055203(n+k) for n >= 1. - _Peter Bala_, Dec 19 2019
%t With[{r = 4}, Flatten[{1, Table[Sum[Sum[(-1)^i*Binomial[j, i]*Binomial[j - i, r]^k, {i, 0, j}], {j, 0, k*r}], {k, 1, 15}]}]] (* _Vaclav Kotesovec_, Mar 22 2016 *)
%Y See A062204 for references, formulas and comments.
%Y Row n=4 of A262809.
%K nonn
%O 0,3
%A _Angelo Dalli_, Jun 13 2001
%E Revised by _Max Alekseyev_, Mar 13 2009
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