Search: id:A147559
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%I A147559
%S A147559 1,4,5,11,6,22,4,155,16,182,158,376,56,1456,680,23155,4966,
%T A147559 28674,6132,117946,15792,415426,162814,512550,333904,4231332,235968,
%U A147559 15171332,5259270,68578566,15199212,736983115,4403208,1097465342
%V A147559 1,4,5,11,-6,-22,-4,155,16,-182,-158,376,56,-1456,680,23155,-4966,
%W A147559 -28674,6132,117946,15792,-415426,-162814,512550,333904,-4231332,235968,
%X A147559 15171332,-5259270,-68578566,15199212,736983115,-4403208,-1097465342
%N 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(3x^3)...
%e A147559 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.
%Y A147559 Cf. A000290
%Y A147559 Sequence in context: A052508 A074098 A126069 this_sequence A007429 A064945 
               A069820
%Y A147559 Adjacent sequences: A147556 A147557 A147558 this_sequence A147560 A147561 
               A147562
%K A147559 sign
%O A147559 1,2
%A A147559 N. Fernandez (primeness(AT)borve.org), Nov 07 2008
%E A147559 Terms from a(11) on corrected by R. J. Mathar (mathar(AT)strw.leidenuniv.nl), 
               Nov 11 2008

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