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A195913 The denominator in a fraction expansion of log(2)-Pi/8.

%I A195913 #40 Aug 12 2019 02:41:26
%S A195913 2,3,12,30,35,56,90,99,132,182,195,240,306,323,380,462,483,552,650,
%T A195913 675,756,870,899,992,1122,1155,1260,1406,1443,1560,1722,1763,1892,
%U A195913 2070,2115,2256,2450,2499,2652,2862,2915
%N A195913 The denominator in a fraction expansion of log(2)-Pi/8.
%C A195913 The minus sign in front of a fraction is considered the sign of the numerator and hence the sign of the fraction does not appear in this sequence.
%D A195913 Granino A. Korn and Theresa M. Korn, Mathematical Handbook for Scientists and Engineers, McGraw-Hill Book Company, New York (1968).
%H A195913 Mohammad K. Azarian, <a href="https://www.jstor.org/stable/24340810">Problem 1218</a>, Pi Mu Epsilon Journal, Vol. 13, No. 2, Spring 2010, p. 116. <a href="https://www.jstor.org/stable/24337914">Solution</a> published in Vol. 13, No. 3, Fall 2010, pp. 183-185.
%F A195913 log(2) - Pi/8 = Sum_{n>=1} (-1)^(n+1)*(1/n) + (-1/2)*Sum_{n>=0} (-1)^n*(1/(2*n+1)).
%F A195913 Empirical g.f.: x*(2+x+9*x^2+14*x^3+3*x^4+3*x^5) / ((1-x)^3*(1+x+x^2)^2). - _Colin Barker_, Dec 17 2015
%F A195913 From _Bernard Schott_, Aug 11 2019: (Start)
%F A195913 k >= 1, a(3*k) = (4*k-1) * 4*k,
%F A195913 k >= 0, a(3*k+1) = (4*k+1) * (4*k+2),
%F A195913 k >= 0, a(3*k+2) = (4*k+1) * (4*k+3).
%F A195913 The even terms a(3*k) and a(3*k+1) come from log(2) and the odd terms a(3*k+2) come from - Pi/8. (End)
%e A195913 1/2 - 1/3 + 1/12 + 1/30 - 1/35 + 1/56 + 1/90 - 1/99 + 1/132 + 1/182 - 1/195 + 1/240 + ... = [(1 - 1/2) + (1/3 - 1/4) + (1/5 - 1/6) + (1/7 - 1/8) + (1/9 - 1/10) + (1/11 - 1/12) + ...] - (1/2)*[(1 - 1/3) + (1/5 - 1/7) + (1/9 - 1/11) + (1/13 - 1/15) + ... ] = log(2) - Pi/8.
%Y A195913 Cf. A195909, A195697, A195947, A164833, A118324, A098289, A075549, A016655, A019675, A161685, A144981, A168056, A004772.
%K A195913 nonn,frac
%O A195913 1,1
%A A195913 _Mohammad K. Azarian_, Sep 25 2011