A062205 Number of alignments of n strings of length 4.

1, 1, 321, 699121, 5552351121, 117029959485121, 5402040231378569121, 480086443888959812703121, 74896283763383392805211587121, 19133358944433370977791260580721121, 7581761490297442738124283591348762605121, 4461925444770180839552702516305804230194739121

Offset: 0

Author

Angelo Dalli, Jun 13 2001

Keywords

• nonn

Comments

Conjectures: a(n) == 1 (mod 80); for fixed k, the sequence a(n) (mod k) eventually becomes periodic. - Peter Bala, Dec 19 2019

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..100

Crossrefs

See A062204 for references, formulas and comments.

Row n=4 of A262809.

Programs

Mathematica

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 *)

Formula

From Vaclav Kotesovec, Mar 22 2016: (Start)

a(n) ~ 2^(5*n-3) * n!^4 / (Pi^(3/2) * n^(3/2) * 3^n * (log(2))^(4*n+1)).

a(n) ~ sqrt(Pi) * 2^(5*n-1) * n^(4*n+1/2) / (3^n * exp(4*n) * (log(2))^(4*n+1)).

(End)

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

Extensions

Revised by Max Alekseyev, Mar 13 2009