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%I #3 Mar 31 2012 14:39:53
%S 2,3,-1,9,-4,0,-16,89,-52,60,-182,214,-620,966,-2142,10497,-7676,
%T 13684,-27530,48288,-98372,190928,-364464,619496,-1341508,2649990,
%U -4923220,9726940,-18510902,37055004,-69269976,213062855,-258284232,527143794
%N 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)...
%e 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.
%t 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] [From _Zak Seidov_, Nov 10 2008]
%Y Cf. A147541
%K sign
%O 1,1
%A _Neil Fernandez_, Nov 07 2008
%E Corrected and extended by _Zak Seidov_, Nov 10 2008
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