A239003 Number of partitions of n into distinct Fibonacci numbers that are all greater than 2.

1, 0, 0, 1, 0, 1, 0, 0, 2, 0, 0, 1, 0, 2, 0, 0, 2, 0, 1, 0, 0, 3, 0, 0, 2, 0, 2, 0, 0, 3, 0, 0, 1, 0, 3, 0, 0, 3, 0, 2, 0, 0, 4, 0, 0, 2, 0, 3, 0, 0, 3, 0, 1, 0, 0, 4, 0, 0, 3, 0, 3, 0, 0, 5, 0, 0, 2, 0, 4, 0, 0, 4, 0, 2, 0, 0, 5, 0, 0, 3, 0, 3, 0, 0, 4, 0, 0

Offset: 0

Author

Clark Kimberling, Mar 08 2014

Keywords

• nonn
• easy
• look

Comments

a(n) > 0 if n+1 is a term of A003622; a(n) = 0 if n+1 is a term of A022342.

Examples

			There is one partition for n=0, the empty partition.  All parts are distinct, which means that there are no two parts that are equal. So a(0)=1.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..10946

Crossrefs

Cf. A000201, A001950, A000045, A000119, A239002, A000009.

Programs

Maple

F:= combinat[fibonacci]:
b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<4, 0,
       b(n, i-1)+`if`(F(i)>n, 0, b(n-F(i), i-1))))
    end:
a:= proc(n) local j; for j from ilog[(1+sqrt(5))/2](n+1)
       while F(j+1)<=n do od; b(n, j)
    end:
seq(a(n), n=0..100);  # Alois P. Heinz, Mar 15 2014

Mathematica

f = Table[Fibonacci[n], {n, 4, 75}];  b[n_] := SeriesCoefficient[Product[1 + x^f[[k]], {k, n}], {x, 0, n}]; u = Table[b[n], {n, 0, 60}]  (* A239003 *)
Flatten[Position[u, 0]]  (* A022342 *)

Formula

G.f.: product(1 + x^F(j), j=4..infinity). - Wolfdieter Lang, Mar 15 2014