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a(0) = 0 and a(n) = 2*n^2 + 2*n - 1, for n>=1.
11

%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