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Search: id:A147558
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| A147558 |
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Result of using the Fibonacci numbers as coefficients in an infinite polynomial series in x and then expressing this series as (1+a(1)x)(1+a(2)x^2)(1+a(3x^3)... |
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+0
1
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| 1, 1, 1, 2, 2, 3, 4, 8, 8, 14, 18, 29, 40, 68, 88, 174, 210, 344, 492, 852, 1144, 1962, 2786, 4601, 6704, 11240, 16096, 27738, 39650, 64936, 97108, 168408, 236880, 397110, 589298, 979496, 1459960, 2421132, 3604880, 6086790
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OFFSET
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1,4
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EXAMPLE
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From the Fibonacci numbers, beginning 1,1, construct the series 1+x+x^2+2x^3+3x^4+5x^5+... a(1) is always the coefficient of x, here 1. Divide by (1+a(1)x), i.e. here (1+x), to get the quotient (1+a(2)x^2+...), which here gives a(2)=1. Then divide this quotient by (1+a(2)x^2), i.e. here (1+x^2), to get (1+a(3)x^3+...), giving a(3)=1.
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CROSSREFS
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Cf. A000045, A147558
Cf. A147542. [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Mar 12 2009]
Sequence in context: A011784 A032252 A112708 this_sequence A032243 A153922 A153943
Adjacent sequences: A147555 A147556 A147557 this_sequence A147559 A147560 A147561
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KEYWORD
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nonn
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AUTHOR
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N. Fernandez (primeness(AT)borve.org), Nov 07 2008
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EXTENSIONS
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More terms from R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Mar 12 2009
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