A109059 To compute a(n) we first write down 7^n 1's in a row. Each row takes the rightmost 7th 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 7th part. The single element in the last row is a(n).

1, 1, 7, 322, 102249, 226742516, 3518406903403, 382149784071841422, 290546585470549214822793, 1546306129153609960601346281449, 57606719909341067627899562630623352149, 15022729501707009545842655841005666468590455864, 27423481304702360472157221630747597794702587610760693525

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.....38..39..40..41..42..43..44..45..46..47..48..49
................................43..87.132.178.225.273.322
.......................................................322
Therefore a(3)=322.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..49

Crossrefs

Cf. A107354, A109055, A109056, A109057, A109058, A109060, A109061, A109062.

Column k=7 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=6*nops(L)/7+1..j),j=6*nops(L)/7+1..nops(L))]; a:=f([seq(1,j=1..7^n)]); while nops(a)>7 do a:=f(a) end do; a[7]; 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, 7];
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