A109055 To compute a(n) we first write down 3^n 1's in a row. Each row takes the rightmost 3rd 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 3rd part. The single element in the last row is a(n).
1, 1, 3, 24, 541, 35649, 6979689, 4085743032, 7166723910237, 37698139930450365, 594816080266215640710, 28154472624850002001979592, 3997853576535778666975681355079, 1703042427700923785323670557504832751, 2176429411666209822350337722381643148477248
Offset: 0
Keywords
Examples
For example, for n=3 the array looks like this: 1..1..1..1..1........1..1..1..1..1..1..1..1..1..1 ........................1..2..3..4..5..6..7..8..9 ..........................................7.15.24 ...............................................24 Therefore a(3)=24.
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..65
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=2*nops(L)/3+1..j),j=2*nops(L)/3+1..nops(L))]; a:=f([seq(1,j=1..3^n)]); while nops(a)>3 do a:=f(a) end do; a[3]; 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, 3]; Table[a[n], {n, 0, 14}] (* Jean-François Alcover, Apr 01 2024, after Alois P. Heinz in A355576 *)
Extensions
More terms from Paul D. Hanna
Comments