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Search: id:A019274
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| A019274 |
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Number of recursive calls needed to compute the n-th Fibonacci number F(n), starting with F(1) = F(2) = 1. |
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+0
2
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| 0, 0, 2, 4, 8, 14, 24, 40, 66, 108, 176, 286, 464, 752, 1218, 1972, 3192, 5166, 8360, 13528, 21890, 35420, 57312, 92734, 150048, 242784, 392834, 635620, 1028456, 1664078, 2692536, 4356616, 7049154, 11405772, 18454928, 29860702, 48315632
(list; graph; listen)
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OFFSET
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1,3
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COMMENT
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Let g=F(2)+F(3)+...+F(n)=F(n+2)-2. Some numbers in the range [0,g] have unique representations of the form sum_{i=1..n} a(i)*F(i) where each a(i) is 1 or -1. These numbers have the form g-k for k in the sequence. - Louis ten Bosch (louis_ten_bosch(AT)hotmail.com), Jan 01 2003
a(n+2) = Sum[k=0..n, Fib(n-k) + k*Fib(n-k) ].
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LINKS
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Kim Trammell, sci.math article
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FORMULA
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a(n)=a(n-1)+a(n-2)+2. a(n)=2*F(n)-2.
a(n)=sum{k=0..n, (2-2*0^(n-k))*F(k)}; - Paul Barry (pbarry(AT)wit.ie), Oct 24 2007
a(n)=F(n)+F(n+3)-2 n>=-1 {where F(n) is the n-th Fibonacci number} - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jan 31 2008
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MAPLE
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with(combinat): seq(fibonacci(n)+fibonacci(n+3)-2, n=-1..35); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jan 31 2008
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CROSSREFS
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Cf. A000045.
Antidiagonal sums of array A017125.
Adjacent sequences: A019271 A019272 A019273 this_sequence A019275 A019276 A019277
Sequence in context: A123655 A084683 A118544 this_sequence A066456 A066342 A020956
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KEYWORD
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nonn,easy,nice
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AUTHOR
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Kim Trammell (kim(AT)coc.com) and others.
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