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%I #35 Sep 08 2022 08:45:06
%S 1,1,1,2,3,7,22,101,792,12310,451276,49995925,22540654445,
%T 60806135438329,1596675274490756791,758949605954969709105721,
%U 14362612091531863067120268402228,29498346711208035625096160181520548669694,23537552807178094028466621551669121053281242290608650
%N Number of partitions of Fibonacci(n).
%C Also number of partitions of F(n+2) whose highest term is F(n+1) ( or, which is the same, whose number of terms is F(n+1)). - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Sep 14 2007
%C Divide the set of partitions P(i,j) in two subsets : 1) Partitions containing at least one term 1; Deleting a term 1, we prove that their number is P(i-1,j-1) 2). Subtracting 1 from each term of the other partitions we prove that their number is P(i-j,j) Hence P(i,j) - P(i-1,j-1) = P(i-j,j) Replacing successively in this formula i by i-1 and j by j-1 and summing all these equalities we get, if j>= floor((i+1)/2) P(i,j)=sum ({k,1,j}P(i-j;k))= A000041(i-j) As for i=F(n+2) and j=F(n+1) the condition is satisfied : P(F(n+2),F(n+1)) = P (F(n+2),F(n+1)= A000041(n) = 1072214(n). - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Sep 14 2007
%H Alois P. Heinz, <a href="/A072214/b072214.txt">Table of n, a(n) for n = 0..30</a> (terms n = 1..26 from Seiichi Manyama)
%F Let P(i,j) denote the number of partitions of i whose highest term is j A072214(n) = A000041(F(n)) = P(F(n+2),F(n+1)) - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Sep 14 2007
%F a(n) = A000041(A000045(n)). - _Michel Marcus_, May 09 2016
%F a(n) = [x^Fibonacci(n)] Product_{k>=1} 1/(1 - x^k). - _Ilya Gutkovskiy_, Jun 08 2017
%e F(5) = 5, F(4) = 3: 5 = 3+2 = 3+1+1 (or 5 = 3+1+1 = 2+2+1), then P(5,3) = 2 = A000041(2) = A000041(F(3)) = A072214(3).
%p F:= n-> (<<0|1>, <1|1>>^n)[1, 2]:
%p a:= n-> combinat[numbpart](F(n)):
%p seq(a(n), n=0..18); # _Alois P. Heinz_, Apr 06 2021
%t Table[PartitionsP[Fibonacci[n]], {n, 1, 17}]
%o (Haskell)
%o a072214 = a000041 . a000045 . (+ 1) -- _Reinhard Zumkeller_, Dec 09 2015
%o (Magma) [NumberOfPartitions(Fibonacci(n)): n in [1..18]]; // _Vincenzo Librandi_ May 09 2016
%o (PARI) a(n) = numbpart(fibonacci(n)); \\ _Michel Marcus_, May 09 2016
%o (Python)
%o from sympy import npartitions as p, fibonacci as f
%o def a(n): return p(f(n)) # _Indranil Ghosh_, Jun 08 2017
%Y Cf. A000041, A000045, A072241.
%K nonn
%O 0,4
%A _Jeff Burch_, Jul 03 2002
%E Edited by _Robert G. Wilson v_, Jul 06 2002
%E a(18) by _Vincenzo Librandi_, May 09 2016
%E a(0)=1 prepended by _Alois P. Heinz_, Apr 06 2021
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