A109057 To compute a(n) we first write down 5^n 1's in a row. Each row takes the rightmost 5th part of the previous row and each element in it equals sum of the elements of the previous row starting with the first of the rightmost 5th part. The single element in the last row is a(n).

1, 1, 5, 115, 12885, 7173370, 19940684251, 277078842941900, 19249144351745111125, 6686277384080730564862875, 11612516024884420913314995604000, 100841213012622614260440382077516990500, 4378443591626306255827149380635713364079323075

Offset

0,3

Links

Alois P. Heinz, Table of n, a(n) for n = 0..54

Example

For example, for n=3 the array, from 2nd row, follows:
1..2..3.....14..15..16..17..18..19..20..21..22..23..24..25
........................................21..43..66..90.115
.......................................................115
Therefore a(3)=115.

Maple

proc(n::nonnegint) local f, a; if n=0 or n=1 then return 1; end if; f:=L->[seq(add(L[i], i=4*nops(L)/5+1..j), j=4*nops(L)/5+1..nops(L))]; a:=f([seq(1, j=1..5^n)]); while nops(a)>5 do a:=f(a) end do; a[5]; end proc;

Mathematica

A[n_, k_] := A[n, k] = If[n == 0, 1, -Sum[A[j, k]*(-1)^(n - j)* Binomial[If[j == 0, 1, k^j], n - j], {j, 0, n - 1}]];
a[n_] := A[n, 5];
Table[a[n], {n, 0, 12}] (* Jean-François Alcover, Apr 02 2024, after Alois P. Heinz in A355576 *)

Crossrefs

Cf. A107354, A109055, A109056, A109058, A109059, A109060, A109061, A109062.

Column k=5 of A355576.

Sequence in context: A208959 A209053 A255884 * A245105 A080988 A230338

Adjacent sequences: A109054 A109055 A109056 * A109058 A109059 A109060

Keyword

nonn

Author

Augustine O. Munagi, Jun 17 2005

Extensions

More terms from Alois P. Heinz, Jul 06 2022

Status

approved