A147557 Result of using the primes 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)...

2, 3, -1, 9, -4, 0, -16, 89, -52, 60, -182, 214, -620, 966, -2142, 10497, -7676, 13684, -27530, 48288, -98372, 190928, -364464, 619496, -1341508, 2649990, -4923220, 9726940, -18510902, 37055004, -69269976, 213062855, -258284232, 527143794

Offset: 1

Author

Neil Fernandez, Nov 07 2008

Keywords

• sign

Examples

			From the primes, construct the series 1+2x+3x^2+5x^3+7x^4+... a(1) is always the coefficient of x, here 2. Divide by (1+2x) to get the quotient (1+a(2)x^2+...), which here gives a(2)=3. Then divide this quotient by (1+a(2)x^2), i.e. here (1+3x^2), to get (1+a(3)x^3+...), giving a(3)=-1.
		

Crossrefs

Cf. A147541

Programs

Mathematica

ser=1+Sum[Prime[i]x^i,{i,110}];ss=1+2x;Do[ser=Normal[Series[ser/(Take[ser,2]),{x,0,105}]];ss+=ser[[2]],{100}];A147557=CoefficientList[ss,x] (* Zak Seidov, Nov 10 2008 *)

Extensions

Corrected and extended by Zak Seidov, Nov 10 2008