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A109061 To compute a(n) we first write down 9^n 1's in a row. Each row takes the rightmost 9th 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 9th part. The single element in the last row is a(n).

%I A109061 #11 Apr 01 2024 13:38:23
%S A109061 1,1,9,693,476121,2940705927,163444130390781,81756588582353417271,
%T A109061 368059416198072536171078649,14912674110246473369128526689667934,
%U A109061 5437955149300119215042866669813503145575607,17846712348533391270843269203829434120473501691723788
%N A109061 To compute a(n) we first write down 9^n 1's in a row. Each row takes the rightmost 9th 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 9th part. The single element in the last row is a(n).
%H A109061 Alois P. Heinz, <a href="/A109061/b109061.txt">Table of n, a(n) for n = 0..46</a>
%e A109061 For example, for n=3 the array, from 2nd row, follows:
%e A109061 1..2..3.....70..71..72..73..74..75..76..77..78..79..80..81
%e A109061 ........................73.147.222.298.375.453.532.612.693
%e A109061 .......................................................693
%e A109061 Therefore a(3)=693.
%p A109061 proc(n::nonnegint) local f,a; if n=0 or n=1 then return 1; end if; f:=L->[seq(add(L[i],i=8*nops(L)/9+1..j),j=8*nops(L)/9+1..nops(L))]; a:=f([seq(1,j=1..9^n)]); while nops(a)>9 do a:=f(a) end do; a[9]; end proc;
%t A109061 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 A109061 a[n_] := A[n, 9];
%t A109061 Table[a[n], {n, 0, 12}] (* _Jean-François Alcover_, Apr 01 2024, after _Alois P. Heinz_ in A355576 *)
%Y A109061 Cf. A107354, A109055, A109056, A109057, A109058, A109059, A109060, A109062.
%Y A109061 Column k=9 of A355576.
%K A109061 nonn
%O A109061 0,3
%A A109061 _Augustine O. Munagi_, Jun 17 2005
%E A109061 More terms from _Alois P. Heinz_, Jul 06 2022