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)...

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

Offset: 1

Author

Neil Fernandez, Nov 07 2008

Keywords

• sign
• look

Examples

			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.
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A000290, A147654, A253909.

Programs

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-François Alcover, Jun 20 2017 *)

Formula

Product_{k>=1} (1+a(k)*x^k) = 1 + Sum_{k>=1} k^2*x^k. - Seiichi Manyama, Jun 24 2018

Extensions

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