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).
1, 1, 5, 115, 12885, 7173370, 19940684251, 277078842941900, 19249144351745111125, 6686277384080730564862875, 11612516024884420913314995604000, 100841213012622614260440382077516990500, 4378443591626306255827149380635713364079323075
Offset: 0
Keywords
Examples
For example, for n=3 the array, from 2nd row, follows: 1..2..3.....14..15..16..17..18..19..20..21..22..23..24..25 ........................................21..43..66..90.115 .......................................................115 Therefore a(3)=115.
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..54
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=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;
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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, 5]; Table[a[n], {n, 0, 12}] (* Jean-François Alcover, Apr 02 2024, after Alois P. Heinz in A355576 *)
Extensions
More terms from Alois P. Heinz, Jul 06 2022