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%I #24 Mar 15 2014 22:07:55
%S 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,
%T 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,
%U 4,0,0,4,0,2,0,0,5,0,0,3,0,3,0,0,4,0,0
%N Number of partitions of n into distinct Fibonacci numbers that are all greater than 2.
%C a(n) > 0 if n+1 is a term of A003622; a(n) = 0 if n+1 is a term of A022342.
%H Alois P. Heinz, <a href="/A239003/b239003.txt">Table of n, a(n) for n = 0..10946</a>
%F G.f.: product(1 + x^F(j), j=4..infinity). - _Wolfdieter Lang_, Mar 15 2014
%e 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.
%p F:= combinat[fibonacci]:
%p b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<4, 0,
%p b(n, i-1)+`if`(F(i)>n, 0, b(n-F(i), i-1))))
%p end:
%p a:= proc(n) local j; for j from ilog[(1+sqrt(5))/2](n+1)
%p while F(j+1)<=n do od; b(n, j)
%p end:
%p seq(a(n), n=0..100); # _Alois P. Heinz_, Mar 15 2014
%t 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 *)
%t Flatten[Position[u, 0]] (* A022342 *)
%Y Cf. A000201, A001950, A000045, A000119, A239002, A000009.
%K nonn,easy,look
%O 0,9
%A _Clark Kimberling_, Mar 08 2014
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