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%I #7 Feb 25 2017 02:48:19
%S 1,-8,-20,-24,144,-384,-832,-896,4352,-10240,-21504,-22528,102400,
%T -229376,-475136,-491520,2162688,-4718592,-9699328,-9961472,42991616,
%U -92274688,-188743680,-192937984,822083584,-1744830464,-3556769792
%N Denominators of a BBP series for Pi/4.
%C From the BBP formula for Pi, the following expression for Pi/4 in unit numerators is obtained
%C Pi/4 = Sum((1/(8k+1)+1/(-2*(8k+4))+1/(-4*(8k+5))+1/(-4*(8k+6)))/16^k, k>=0)
%C Therefore a(n) such that
%C a(4*n) = (8*n+1)*16^n.
%C a(4*n+1) = -2*(8*n+4)*16^n.
%C a(4*n+2) = -4*(8*n+5)*16^n.
%C a(4*n+3) = -4*(8*n+6)*16^n.
%C has
%C Sum_{n >= 0} (1/a(n)) = Pi/4.
%C Using PARI/GP suminf(n=0,1/(2^(n-2)*(2*(-1+(-1)^n+(1-I)*(-I)^n+(1+I)*I^n)+(-3+3*(-1)^n+(4-I)*(-I)^n+(4+I)*I^n)*n)))= 0.7853981633974483096156608454...=Pi/4. - _Alexander R. Povolotsky_, Sep 01 2009
%H G. C. Greubel, <a href="/A164916/b164916.txt">Table of n, a(n) for n = 0..1000</a>
%F G.f.: (1-8*x-20*x^2-24*x^3+112*x^4-128*x^5-192*x^6-128*x^7)/(1-16*x^4)^2.
%F a(n)= 2^(n-2)*(2*(-1+(-1)^n+(1-I)*(-I)^n+(1+I)*I^n)+(-3+3*(-1)^n+(4-I)*(-I)^n+(4+I)*I^n)*n). - _Alexander R. Povolotsky_, Sep 01 2009
%t CoefficientList[Series[(1 - 8*x - 20*x^2 - 24*x^3 + 112*x^4 - 128*x^5 - 192*x^6 - 128*x^7)/(1 - 16*x^4)^2, {x,0,50}], x] (* _G. C. Greubel_, Feb 25 2017 *)
%o (PARI) x='x + O('x^50); Vec((1 - 8*x - 20*x^2 - 24*x^3 + 112*x^4 - 128*x^5 - 192*x^6 - 128*x^7)/(1 - 16*x^4)^2) \\ _G. C. Greubel_, Feb 25 2017
%Y Cf. A048581, A066968, A154925, A154962, A156269, A157142.
%K frac,sign
%O 0,2
%A _Jaume Oliver Lafont_, Aug 31 2009
%E Comment section corrected by _Jaume Oliver Lafont_, Sep 03 2009
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