A109058 To compute a(n) we first write down 6^n 1's in a row. Each row takes the rightmost 6th 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 6th part. The single element in the last row is a(n).

1, 1, 6, 201, 39656, 46769781, 330736663032, 14031372754200653, 3571582237574150514024, 5454701025672508908169570740, 49984143782624329482858175943128416, 2748177454593265010973723857947479180947553, 906585004703475512437226615670665677815744239819376

Offset: 0

Author

Augustine O. Munagi, Jun 17 2005

Keywords

• nonn

Examples

			For example, for n=3 the array, from 2nd row, follows:
1..2..3.....25..26..27..28..29..30..31..32..33..34..35..36
....................................31..63..96.130.165.201
.......................................................201
Therefore a(3)=201.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..51

Crossrefs

Cf. A107354, A109055, A109056, A109057, A109059, A109060, A109061, A109062.

Column k=6 of A355576.

Programs

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=5*nops(L)/6+1..j),j=5*nops(L)/6+1..nops(L))]; a:=f([seq(1,j=1..6^n)]); while nops(a)>6 do a:=f(a) end do; a[6]; 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, 6];
Table[a[n], {n, 0, 12}] (* Jean-François Alcover, Apr 01 2024, after Alois P. Heinz in A355576 *)

Extensions

More terms from Alois P. Heinz, Jul 06 2022