A133818 -id:A133818 - cp's OEIS frontend

15, 63, 143, 255, 399, 575, 783, 1023, 1295, 1599, 1935, 2303, 2703, 3135, 3599, 4095, 4623, 5183, 5775, 6399, 7055, 7743, 8463, 9215, 9999, 10815, 11663, 12543, 13455, 14399, 15375, 16383, 17423, 18495, 19599, 20735, 21903, 23103, 24335, 25599

Offset: 0

Miklos Kristof, Sep 15 2008

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.
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
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
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
Essentially the least common multiple of 4*n+1 and 4*n-1. - Colin Barker, Feb 11 2017

Bronstein-Semendjajew, Taschenbuch der Mathematik, 7th German ed., 1965, ch. 4.1.8.
Granino A. Korn and Theresa M. Korn, Mathematical Handbook for Scientists and Engineers, McGraw-Hill Book Company, New York (1968), pp. 980-981.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A005843, A074377, A093954, A133818, A158487.

[(4*n+3)*(4*n+5): n in [0..50]]; // Vincenzo Librandi, Sep 22 2011

A141759:=n->16*n^2 + 32*n + 15: seq(A141759(n), n=0..60);

LinearRecurrence[{3, -3, 1}, {15, 63, 143}, 50] (* Vincenzo Librandi, Feb 09 2012 *)

a(n)=n*(n+2)<<4+15 \\ Charles R Greathouse IV, Oct 27 2011

G.f.: (15+18*x-x^2)/(1-x)^3.
E.g.f.: (15+48*x+16*x^2)*exp(x).
a(n) = a(-n-2) = A016802(n+1) - 1. - Bruno Berselli, Sep 22 2011
From Amiram Eldar, Feb 04 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = Pi/(2*sqrt(2)) (A093954).
Product_{n>=0} (1 - 1/a(n)) = sin(Pi/(2*sqrt(2))). (End)

Formula indices corrected by R. J. Mathar, Jul 07 2009

cp's OEIS Frontend

A141759 a(n) = 16n^2 + 32n + 15.

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