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%I #37 Jul 03 2023 08:42:47
%S 0,3,11,23,39,59,83,111,143,179,219,263,311,363,419,479,543,611,683,
%T 759,839,923,1011,1103,1199,1299,1403,1511,1623,1739,1859,1983,2111,
%U 2243,2379,2519,2663,2811,2963,3119,3279,3443,3611,3783,3959,4139,4323,4511
%N a(0) = 0 and a(n) = 2*n^2 + 2*n - 1, for n>=1.
%C Previous name was: Sequence gives X values that satisfy the integer equation 2*X^3 + 3*X^2 = Y^2.
%C To find Y values: b(n) = (2*n^2 + 2*n - 1)*(2*n - 1).
%H G. C. Greubel, <a href="/A132209/b132209.txt">Table of n, a(n) for n = 0..5000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3, -3, 1).
%F a(n) = 2*n^2 + 2*n - 1 for n>=1.
%F G.f.: x*(1+x)*(3-x)/(1-x)^3. - _R. J. Mathar_, Nov 14 2007
%F E.g.f.: 1 + (2*x^2 + 4*x -1)*exp(x). - _G. C. Greubel_, Jul 13 2017
%F From _Amiram Eldar_, Mar 07 2021: (Start)
%F Sum_{n>=1} 1/a(n) = 1 + sqrt(3)*Pi*tan(sqrt(3)*Pi/2)/6.
%F Product_{n>=1} (1 + 1/a(n)) = -Pi*sec(sqrt(3)*Pi/2)/2.
%F Product_{n>=1} (1 - 1/a(n)) = cos(sqrt(5)*Pi/2)*sec(sqrt(3)*Pi/2)/2. (End)
%t Join[{0}, LinearRecurrence[{3, -3, 1}, {3, 11, 23}, 40]] (* _Vincenzo Librandi_, Sep 22 2015 *)
%o (Magma) [0] cat [2*n^2+2*n-1: n in [1..50]]; // _Vincenzo Librandi_, Sep 22 2015
%o (PARI) for(n=0,50, print1(if(n==0, 0, 2*n^2 + 2*n -1), ", ")) \\ _G. C. Greubel_, Jul 13 2017
%Y Cf. A005563, A046092, A001082, A002378, A036666, A062717, A028347, A087475, A000217.
%K nonn
%O 0,2
%A _Mohamed Bouhamida_, Nov 06 2007
%E Edited by the Associate Editors of the OEIS, Nov 15 2009
%E More terms from _Vincenzo Librandi_, Sep 22 2015
%E Shorter name (using formula given) from _Joerg Arndt_, Sep 27 2015
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