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Search: id:A147557
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| A147557 |
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Result of using the primes 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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+0
2
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| 2, 3, -1, 9, -4, 0, -16, 89, -52, 60, -182, 214, -620, 966, -2142, 10497, -7676, 13684, -27530, 48288, -98372, 190928, -364464, 619496, -1341508, 2649990, -4923220, 9726940, -18510902, 37055004, -69269976, 213062855, -258284232, 527143794
(list; graph; listen)
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OFFSET
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1,1
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EXAMPLE
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From the primes, construct the series 1+2x+3x^2+5x^3+7x^4+... a(1) is always the coefficient of x, here 2. Divide by (1+2x) to get the quotient (1+a(2)x^2+...), which here gives a(2)=3. Then divide this quotient by (1+a(2)x^2), i.e. here (1+3x^2), to get (1+a(3)x^3+...), giving a(3)=-1.
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MATHEMATICA
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ser=1+Sum[Prime[i]x^i, {i, 110}]; ss=1+2x; Do[ser=Normal[Series[ser/(Take[ser, 2]), {x, 0, 105}]]; ss+=ser[[2]], {100}]; A147557=CoefficientList[ss, x] [From Zak Seidov (zakseidov(AT)yahoo.com), Nov 10 2008]
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CROSSREFS
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Cf. A147541
Sequence in context: A157963 A158502 A135950 this_sequence A117025 A078021 A106342
Adjacent sequences: A147554 A147555 A147556 this_sequence A147558 A147559 A147560
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KEYWORD
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sign
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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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Corrected and extended by Zak Seidov (zakseidov(AT)yahoo.com), Nov 10 2008
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