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%I #28 Dec 28 2022 15:42:44
%S 1,4,5,11,-6,-22,-4,155,16,-182,-158,376,56,-1456,680,23155,-4966,
%T -28674,6132,117946,15792,-415426,-162814,512550,333904,-4231332,
%U 235968,15171332,-5259270,-68578566,15199212,736983115,-4403208,-1097465342
%N 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)...
%H Seiichi Manyama, <a href="/A147559/b147559.txt">Table of n, a(n) for n = 1..1000</a>
%F Product_{k>=1} (1+a(k)*x^k) = 1 + Sum_{k>=1} k^2*x^k. - _Seiichi Manyama_, Jun 24 2018
%e 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.
%t 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}];
%t Array[a, terms] /. sol (* _Jean-François Alcover_, Jun 20 2017 *)
%Y Cf. A000290, A147654, A253909.
%K sign,look
%O 1,2
%A _Neil Fernandez_, Nov 07 2008
%E Terms from a(11) on corrected by _R. J. Mathar_, Nov 11 2008
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