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 A147557 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)... 2
 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; refs; listen; history; text; internal format)
 OFFSET 1,1 LINKS EXAMPLE 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. MATHEMATICA 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, Nov 10 2008] CROSSREFS Cf. A147541 Sequence in context: A135950 A202063 A200016 * A117025 A078021 A106342 Adjacent sequences:  A147554 A147555 A147556 * A147558 A147559 A147560 KEYWORD sign AUTHOR Neil Fernandez, Nov 07 2008 EXTENSIONS Corrected and extended by Zak Seidov, Nov 10 2008 STATUS approved

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