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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 (list; graph; refs; listen; history; text; internal format)
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)/(1-x)^3 - (Product[1+a[k]*x^k, {k, 1, n}] /. sol), {x, 0, n}] == 0][[1, 1]]], {n, 2, terms}];

Array[a, terms] /. sol (* Jean-Fran├žois 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

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Last modified November 19 11:04 EST 2017. Contains 294936 sequences.