OFFSET
0,2
COMMENTS
Also, the number of alignments for 4 sequences of length n each (Slowinski 1998).
LINKS
Alois P. Heinz, Table of n, a(n) for n = 0..350
J. B. Slowinski, The Number of Multiple Alignments, Molecular Phylogenetics and Evolution 10:2 (1998), 264-266. doi:10.1006/mpev.1998.0522
FORMULA
Recurrence: (n-1)*n^3*(864*n^4 - 6480*n^3 + 17763*n^2 - 21015*n + 9059)*a(n) = 15*(n-1)*(44928*n^7 - 404352*n^6 + 1459788*n^5 - 2712556*n^4 + 2772389*n^3 - 1538829*n^2 + 423093*n - 43506)*a(n-1) + (188352*n^8 - 2166048*n^7 + 10541118*n^6 - 28166748*n^5 + 44769259*n^4 - 42719172*n^3 + 23364582*n^2 - 6470217*n + 671094)*a(n-2) + 3*(n-2)*(3456*n^7 - 38016*n^6 + 169116*n^5 - 388336*n^4 + 486619*n^3 - 322644*n^2 + 100014*n - 10989)*a(n-3) - (n-3)^3*(n-2)*(864*n^4 - 3024*n^3 + 3507*n^2 - 1473*n + 191)*a(n-4). - Vaclav Kotesovec, Mar 22 2016
a(n) ~ sqrt(8 + 6*sqrt(2) + sqrt(140 + 99*sqrt(2))) * (195 + 138*sqrt(2) + 4*sqrt(4756 + 3363*sqrt(2)))^n / (8 * Pi^(3/2) * n^(3/2)). - Vaclav Kotesovec, Mar 22 2016
MATHEMATICA
With[{k = 4}, Table[Sum[Sum[(-1)^i*Binomial[j, i]*Binomial[j - i, n]^k, {i, 0, j}], {j, 0, k*n}], {n, 0, 15}]] (* Vaclav Kotesovec, Mar 22 2016 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Alois P. Heinz, Oct 08 2015
STATUS
approved