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A109057 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).

%I A109057 #12 Apr 02 2024 02:59:44
%S A109057 1,1,5,115,12885,7173370,19940684251,277078842941900,
%T A109057 19249144351745111125,6686277384080730564862875,
%U A109057 11612516024884420913314995604000,100841213012622614260440382077516990500,4378443591626306255827149380635713364079323075
%N A109057 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 A109057 Alois P. Heinz, <a href="/A109057/b109057.txt">Table of n, a(n) for n = 0..54</a>
%e A109057 For example, for n=3 the array, from 2nd row, follows:
%e A109057 1..2..3.....14..15..16..17..18..19..20..21..22..23..24..25
%e A109057 ........................................21..43..66..90.115
%e A109057 .......................................................115
%e A109057 Therefore a(3)=115.
%p A109057 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 A109057 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 A109057 a[n_] := A[n, 5];
%t A109057 Table[a[n], {n, 0, 12}] (* _Jean-François Alcover_, Apr 02 2024, after _Alois P. Heinz_ in A355576 *)
%Y A109057 Cf. A107354, A109055, A109056, A109058, A109059, A109060, A109061, A109062.
%Y A109057 Column k=5 of A355576.
%K A109057 nonn
%O A109057 0,3
%A A109057 _Augustine O. Munagi_, Jun 17 2005
%E A109057 More terms from _Alois P. Heinz_, Jul 06 2022