OFFSET
0,4
COMMENTS
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
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
LINKS
Alois P. Heinz, Table of n, a(n) for n = 0..30 (terms n = 1..26 from Seiichi Manyama)
FORMULA
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
a(n) = [x^Fibonacci(n)] Product_{k>=1} 1/(1 - x^k). - Ilya Gutkovskiy, Jun 08 2017
EXAMPLE
MAPLE
F:= n-> (<<0|1>, <1|1>>^n)[1, 2]:
a:= n-> combinat[numbpart](F(n)):
seq(a(n), n=0..18); # Alois P. Heinz, Apr 06 2021
MATHEMATICA
Table[PartitionsP[Fibonacci[n]], {n, 1, 17}]
PROG
(Haskell)
a072214 = a000041 . a000045 . (+ 1) -- Reinhard Zumkeller, Dec 09 2015
(Magma) [NumberOfPartitions(Fibonacci(n)): n in [1..18]]; // Vincenzo Librandi May 09 2016
(PARI) a(n) = numbpart(fibonacci(n)); \\ Michel Marcus, May 09 2016
(Python)
from sympy import npartitions as p, fibonacci as f
def a(n): return p(f(n)) # Indranil Ghosh, Jun 08 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
Jeff Burch, Jul 03 2002
EXTENSIONS
Edited by Robert G. Wilson v, Jul 06 2002
a(18) by Vincenzo Librandi, May 09 2016
a(0)=1 prepended by Alois P. Heinz, Apr 06 2021
STATUS
approved