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A147654
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Result of using the positive integers 1,2,3,... 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(3)x^3)
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4
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1, 2, 1, 3, 0, -2, 0, 9, 0, -6, 0, 4, 0, -18, 0, 93, 0, -54, 0, 72, 0, -186, 0, 232
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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EXAMPLE
| From the positive integers 1,2,3,..., construct the series 1+x+2x^2+3x^3+4x^4+... 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)=2. Then divide this quotient by (1+a(2)x^2), i.e. here (1+2x^2), to get (1+a(3)x^3+...), giving a(3)=1.
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CROSSREFS
| Cf. A147541
Sequence in context: A113288 A199580 A035215 * A071467 A125073 A071461
Adjacent sequences: A147651 A147652 A147653 * A147655 A147656 A147657
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KEYWORD
| sign
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AUTHOR
| N. Fernandez (primeness(AT)borve.org), Nov 09 2008
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