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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).
9
1, 1, 3, 24, 541, 35649, 6979689, 4085743032, 7166723910237, 37698139930450365, 594816080266215640710, 28154472624850002001979592, 3997853576535778666975681355079, 1703042427700923785323670557504832751, 2176429411666209822350337722381643148477248
OFFSET
0,3
COMMENTS
Comment from Franklin T. Adams-Watters, Jul 13 2006: This is the number of subpartitions of the sequence 3^n-1. As such it can also be computed adding forward, with 3^n terms in the n-th line:
1...........................................................................
1.1 1.......................................................................
1.2.3.3..3..3..3..3..3......................................................
1.3.6.9.12.15.18.21.24.24.24.24.24.24.24.24.24.24.24.24.24.24.24.24.24.24.24
LINKS
EXAMPLE
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.
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;
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 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Augustine O. Munagi, Jun 17 2005
EXTENSIONS
More terms from Paul D. Hanna
STATUS
approved