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%I #76 Sep 08 2022 08:45:35
%S 15,63,143,255,399,575,783,1023,1295,1599,1935,2303,2703,3135,3599,
%T 4095,4623,5183,5775,6399,7055,7743,8463,9215,9999,10815,11663,12543,
%U 13455,14399,15375,16383,17423,18495,19599,20735,21903,23103,24335,25599
%N a(n) = 16n^2 + 32n + 15.
%C Via the partial fraction decomposition 1/((4n+3)*(4n+5)) = (1/2) *(1/(4n+3) -1/(4n+5)) we find 2*Sum_{n>=0} (-1)^n/a(n) = 2*Sum_{n>=0} (-1)^n/( (4*n+3)*(4*n+5) ) = 1/3 -1/5 -1/7 +1/9 +1/11 -1/13 -1/15 +1/17 +1/19 -- ++ ... = (1/1 + 1/3 -1/5 -1/7 +1/9 +1/11 -1/13 -1/15 +1/17 +1/19 -- ++ ..)-1 = Sum_{n>=0} (-1)^n/A016813(n) + Sum_{n>=0} (-1)^n/A004767(n) -1 = -1 + Sum_{n>=0} b(n)/n^1 where b(n) = 1, 0, 1, 0, -1, 0, -1, 0 is a sequence with period length 8, one of the Dirichlet L-series modulo 8. The alternating sum becomes -1 +L(m=8,r=4,s=1) = Pi*sqrt(2)/4-1 = A093954 - 1.
%C Pi = 4 - 8*Sum(1/a(n)) noted by Bronstein-Semendjajew for the variant a(n) = (4n-1)*(4n+1) starting at n=1. - _Frank Ellermann_, Sep 18 2011
%C The identity (16*n^2-1)^2 - (64*n^2-8)*(2*n)^2 = 1 can be written as a(n)^2 - A158487(n)*A005843(n)^2 = 1. - _Vincenzo Librandi_, Feb 09 2012
%C Sequence found by reading the line from 15, in the direction 15, 63,... in the square spiral whose vertices are the generalized decagonal numbers A074377. - _Omar E. Pol_, Nov 02 2012
%C Essentially the least common multiple of 4*n+1 and 4*n-1. - _Colin Barker_, Feb 11 2017
%D Bronstein-Semendjajew, Taschenbuch der Mathematik, 7th German ed., 1965, ch. 4.1.8.
%D Granino A. Korn and Theresa M. Korn, Mathematical Handbook for Scientists and Engineers, McGraw-Hill Book Company, New York (1968), pp. 980-981.
%H Vincenzo Librandi, <a href="/A141759/b141759.txt">Table of n, a(n) for n = 0..10000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F G.f.: (15+18*x-x^2)/(1-x)^3.
%F E.g.f.: (15+48*x+16*x^2)*exp(x).
%F a(n) = a(-n-2) = A016802(n+1) - 1. - _Bruno Berselli_, Sep 22 2011
%F From _Amiram Eldar_, Feb 04 2021: (Start)
%F Product_{n>=0} (1 + 1/a(n)) = Pi/(2*sqrt(2)) (A093954).
%F Product_{n>=0} (1 - 1/a(n)) = sin(Pi/(2*sqrt(2))). (End)
%p A141759:=n->16*n^2 + 32*n + 15: seq(A141759(n), n=0..60);
%t LinearRecurrence[{3, -3, 1}, {15, 63, 143}, 50] (* _Vincenzo Librandi_, Feb 09 2012 *)
%o (Magma) [(4*n+3)*(4*n+5): n in [0..50]]; // _Vincenzo Librandi_, Sep 22 2011
%o (PARI) a(n)=n*(n+2)<<4+15 \\ _Charles R Greathouse IV_, Oct 27 2011
%Y Cf. A005843, A074377, A093954, A133818, A158487.
%K nonn,easy
%O 0,1
%A _Miklos Kristof_, Sep 15 2008
%E Formula indices corrected by _R. J. Mathar_, Jul 07 2009
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