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A072214
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Number of partitions of Fibonacci(n).
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3
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1, 1, 2, 3, 7, 22, 101, 792, 12310, 451276, 49995925, 22540654445, 60806135438329, 1596675274490756791, 758949605954969709105721, 14362612091531863067120268402228, 29498346711208035625096160181520548669694
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,3
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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
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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
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EXAMPLE
| 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)
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MATHEMATICA
| Table[ PartitionsP[ Fibonacci[n]], {n, 1, 17}]
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CROSSREFS
| Cf. A000041, A000045.
Sequence in context: A038159 A077210 A151908 * A007660 A158055 A156615
Adjacent sequences: A072211 A072212 A072213 * A072215 A072216 A072217
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KEYWORD
| nonn
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AUTHOR
| Jeff Burch (gburch(AT)erols.com), Jul 03 2002
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EXTENSIONS
| Edited by Robert G. Wilson v (rgwv(AT)rgwv.com), Jul 06 2002
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