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

0,9

Comments

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

Links

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

Formula

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

Example

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.

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 *)

Crossrefs

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

Sequence in context: A084888 A091400 A129448 * A123759 A072453 A307303

Adjacent sequences: A239000 A239001 A239002 * A239004 A239005 A239006

Keyword

nonn,easy,look

Author

Clark Kimberling, Mar 08 2014

Status

approved