A121548 Triangle read by rows: T(n,k) is the number of compositions of n into k Fibonacci numbers (1 <= k <= n; only one 1 is considered as a Fibonacci number).

1, 1, 1, 1, 2, 1, 0, 3, 3, 1, 1, 2, 6, 4, 1, 0, 3, 7, 10, 5, 1, 0, 2, 9, 16, 15, 6, 1, 1, 2, 9, 23, 30, 21, 7, 1, 0, 2, 10, 28, 50, 50, 28, 8, 1, 0, 3, 9, 34, 71, 96, 77, 36, 9, 1, 0, 2, 12, 36, 95, 156, 168, 112, 45, 10, 1, 0, 0, 12, 43, 115, 231, 308, 274, 156, 55, 11, 1, 1, 2, 9, 48, 140, 312, 504, 560, 423, 210, 66, 12, 1

Offset: 1

Author

Emeric Deutsch, Aug 07 2006

Keywords

• nonn
• tabl

Examples

			T(5,3)=6 because we have [1,2,2], [2,1,2], [2,2,1], [1,1,3], [1,3,1] and [3,1,1].
Triangle starts:
  1;
  1,  1;
  1,  2,  1;
  0,  3,  3,  1;
  1,  2,  6,  4,  1;
  0,  3,  7, 10,  5,  1;
  0,  2,  9, 16, 15,  6,  1;
  ...
		

Links

• Alois P. Heinz, Rows n = 0..150, flattened

Crossrefs

Cf. A076739, A010056, A121549, A121550, A121551.

T(2n,n) gives A341072.

Programs

Maple

with(combinat): G:=1/(1-t*sum(z^fibonacci(i),i=2..40))-1: Gser:=simplify(series(G,z=0,25)): for n from 1 to 23 do P[n]:=sort(coeff(Gser,z,n)) od: for n from 1 to 15 do seq(coeff(P[n],t,j),j=1..n) od; # yields sequence in triangular form
# second Maple program:
g:= proc(n) g(n):= (t-> issqr(t+4) or issqr(t-4))(5*n^2) end:
T:= proc(n, t) option remember;
      `if`(n=0, `if`(t=0, 1, 0), `if`(t<1, 0, add(
      `if`(g(j), T(n-j, t-1), 0), j=1..n)))
    end:
seq(seq(T(n, k), k=1..n), n=1..14); # Alois P. Heinz, Oct 10 2022

Mathematica

nmax = 14;
T = Rest@CoefficientList[#, t]& /@ Rest@(1/(1 - t*Sum[z^Fibonacci[i],
     {i, 2, nmax}]) - 1 + O[z]^(nmax+1) // CoefficientList[#, z]&);
Table[T[[n, k]], {n, 1, nmax}, {k, 1, n}] // Flatten (* Jean-François Alcover, May 02 2022 *)

Formula

G.f.: G(t,z) = 1 / (1 - t*Sum_{i>=2} z^Fibonacci(i)) - 1.

Sum of terms in row n = A076739(n).

T(n,1) = A010056(n) (the characteristic function of the Fibonacci numbers);

T(n,2) = A121549(n);

T(n,3) = A121550(n);

Sum_{k=1..n} k*T(n,k) = A121551(n).