

A147559


Result of using the perfect squares 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, 4, 5, 11, 6, 22, 4, 155, 16, 182, 158, 376, 56, 1456, 680, 23155, 4966, 28674, 6132, 117946, 15792, 415426, 162814, 512550, 333904, 4231332, 235968, 15171332, 5259270, 68578566, 15199212, 736983115, 4403208, 1097465342
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OFFSET

1,2


LINKS

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


EXAMPLE

From the perfect squares, construct the series 1+x+4x^2+9x^3+16x^4+25x^5+... 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)=4. Then divide this quotient by (1+a(2)x^2), i.e. here (1+4x^2), to get (1+a(3)x^3+...), giving a(3)=5.


MATHEMATICA

terms = 34; sol = {a[1] > 1}; Do[sol = Append[sol, Solve[ SeriesCoefficient[ x*(1+x)/(1x)^3  (Product[1+a[k]*x^k, {k, 1, n}] /. sol), {x, 0, n}] == 0][[1, 1]]], {n, 2, terms}];
Array[a, terms] /. sol (* JeanFrançois Alcover, Jun 20 2017 *)


CROSSREFS

Cf. A000290.
Sequence in context: A074098 A196270 A126069 * A206028 A007429 A064945
Adjacent sequences: A147556 A147557 A147558 * A147560 A147561 A147562


KEYWORD

sign


AUTHOR

Neil Fernandez, Nov 07 2008


EXTENSIONS

Terms from a(11) on corrected by R. J. Mathar, Nov 11 2008


STATUS

approved



