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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).

%I A109060 #11 Apr 01 2024 14:08:25
%S A109060 1,1,8,484,231736,886208954,27106585594040,6632714300472863716,
%T A109060 12983632019302863224103688,203325054125533158416534341556735,
%U A109060 25472733809776289439071490656049076425792,25529963965104465687252347321830255523307055463168
%N 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).
%H A109060 Alois P. Heinz, <a href="/A109060/b109060.txt">Table of n, a(n) for n = 0..47</a>
%e A109060 For example, for n=3 the array, from 2nd row, follows:
%e A109060 1..2..3.....53..54..55..56..57..58..59..60..61..62..63..64
%e A109060 ............................57.115.174.234.295.357.420.484
%e A109060 .......................................................484
%e A109060 Therefore a(3)=484.
%p A109060 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;
%t A109060 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}]];
%t A109060 a[n_] := A[n, 8];
%t A109060 Table[a[n], {n, 0, 12}] (* _Jean-François Alcover_, Apr 01 2024, after _Alois P. Heinz_ in A355576 *)
%Y A109060 Cf. A107354, A109055, A109056, A109057, A109058, A109059, A109061, A109062.
%Y A109060 Column k=8 of A355576.
%K A109060 nonn
%O A109060 0,3
%A A109060 _Augustine O. Munagi_, Jun 17 2005
%E A109060 More terms from _Alois P. Heinz_, Jul 06 2022