

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)...


3



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
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OFFSET

1,1


LINKS

Table of n, a(n) for n=1..34.


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 A300838
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



