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A141759
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a(n) = 16n^2 + 32n + 15.
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7
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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
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OFFSET
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0,1
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COMMENTS
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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
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REFERENCES
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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.
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LINKS
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Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
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FORMULA
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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)
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MAPLE
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A141759:=n->16*n^2 + 32*n + 15: seq(A141759(n), n=0..60);
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MATHEMATICA
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LinearRecurrence[{3, -3, 1}, {15, 63, 143}, 50] (* Vincenzo Librandi, Feb 09 2012 *)
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PROG
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(Magma) [(4*n+3)*(4*n+5): n in [0..50]]; // Vincenzo Librandi, Sep 22 2011
(PARI) a(n)=n*(n+2)<<4+15 \\ Charles R Greathouse IV, Oct 27 2011
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CROSSREFS
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Cf. A005843, A074377, A093954, A133818, A158487.
Sequence in context: A065915 A062965 A157968 * A335574 A305616 A104473
Adjacent sequences: A141756 A141757 A141758 * A141760 A141761 A141762
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KEYWORD
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nonn,easy
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AUTHOR
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Miklos Kristof, Sep 15 2008
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EXTENSIONS
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Formula indices corrected by R. J. Mathar, Jul 07 2009
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STATUS
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approved
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