A109060 To compute a(n) we first write down 8^n 1's in a row. Each row takes the rightmost 8th 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 8th part. The single element in the last row is a(n).
1, 1, 8, 484, 231736, 886208954, 27106585594040, 6632714300472863716, 12983632019302863224103688, 203325054125533158416534341556735, 25472733809776289439071490656049076425792, 25529963965104465687252347321830255523307055463168
Offset: 0
Keywords
Examples
For example, for n=3 the array, from 2nd row, follows: 1..2..3.....53..54..55..56..57..58..59..60..61..62..63..64 ............................57.115.174.234.295.357.420.484 .......................................................484 Therefore a(3)=484.
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..47
Crossrefs
Programs
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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=7*nops(L)/8+1..j),j=7*nops(L)/8+1..nops(L))]; a:=f([seq(1,j=1..8^n)]); while nops(a)>8 do a:=f(a) end do; a[8]; end proc;
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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, 8]; 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