

A147654


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)


4



1, 2, 1, 3, 0, 2, 0, 9, 0, 6, 0, 4, 0, 18, 0, 93, 0, 54, 0, 72, 0, 186, 0, 232
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OFFSET

1,2


LINKS

Table of n, a(n) for n=1..24.


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.


CROSSREFS

Cf. A147541
Sequence in context: A113288 A199580 A035215 * A071467 A125073 A281488
Adjacent sequences: A147651 A147652 A147653 * A147655 A147656 A147657


KEYWORD

sign


AUTHOR

Neil Fernandez, Nov 09 2008


STATUS

approved



