A109056 To compute a(n) we first write down 4^n 1's in a row. Each row takes the rightmost 4th 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 4th part. The single element in the last row is a(n).

1, 1, 4, 58, 3236, 713727, 627642640, 2205897096672, 31004442653082720, 1743005531132374350208, 391947224244531572312436328, 352545281714327012273215572739472, 1268416358395092955994185170741834144224, 18254446075150458724007419019753847268167282688

Offset

0,3

Links

Alois P. Heinz, Table of n, a(n) for n = 0..58

Example

For example, for n=3 the array looks like this:
1..1.....1..1..1..1..1..1..1..1..1..1..1..1..1..1..1..1..1
............1..2..3..4..5..6..7..8..9.10.11.12.13.14.15.16
...............................................13.27.42.58
........................................................58
Therefore a(4)=58.

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=3*nops(L)/4+1..j), j=3*nops(L)/4+1..nops(L))]; a:=f([seq(1, j=1..4^n)]); while nops(a)>4 do a:=f(a) end do; a[4]; 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, 4];
Table[a[n], {n, 0, 13}] (* Jean-François Alcover, Apr 01 2024, after Alois P. Heinz in A355576 *)

Crossrefs

Cf. A107354, A109055, A109057, A109058, A109059, A109060, A109061.

Column k=4 of A355576.

Sequence in context: A216352 A295406 A229528 * A326944 A155204 A290765

Adjacent sequences: A109053 A109054 A109055 * A109057 A109058 A109059

Keyword

nonn

Author

Augustine O. Munagi, Jun 17 2005

Extensions

More terms from Alois P. Heinz, Jul 06 2022

Status

approved