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%I #12 Apr 02 2024 02:59:44
%S 1,1,5,115,12885,7173370,19940684251,277078842941900,
%T 19249144351745111125,6686277384080730564862875,
%U 11612516024884420913314995604000,100841213012622614260440382077516990500,4378443591626306255827149380635713364079323075
%N To compute a(n) we first write down 5^n 1's in a row. Each row takes the rightmost 5th 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 5th part. The single element in the last row is a(n).
%H Alois P. Heinz, <a href="/A109057/b109057.txt">Table of n, a(n) for n = 0..54</a>
%e For example, for n=3 the array, from 2nd row, follows:
%e 1..2..3.....14..15..16..17..18..19..20..21..22..23..24..25
%e ........................................21..43..66..90.115
%e .......................................................115
%e Therefore a(3)=115.
%p proc(n::nonnegint) local f,a; if n=0 or n=1 then return 1; end if; f:=L->[seq(add(L[i],i=4*nops(L)/5+1..j),j=4*nops(L)/5+1..nops(L))]; a:=f([seq(1,j=1..5^n)]); while nops(a)>5 do a:=f(a) end do; a[5]; end proc;
%t 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 a[n_] := A[n, 5];
%t Table[a[n], {n, 0, 12}] (* _Jean-François Alcover_, Apr 02 2024, after _Alois P. Heinz_ in A355576 *)
%Y Cf. A107354, A109055, A109056, A109058, A109059, A109060, A109061, A109062.
%Y Column k=5 of A355576.
%K nonn
%O 0,3
%A _Augustine O. Munagi_, Jun 17 2005
%E More terms from _Alois P. Heinz_, Jul 06 2022
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