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a(n) = 2*n^2 + 6*n + 2.
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%I #58 Dec 23 2022 07:38:52

%S 2,10,22,38,58,82,110,142,178,218,262,310,362,418,478,542,610,682,758,

%T 838,922,1010,1102,1198,1298,1402,1510,1622,1738,1858,1982,2110,2242,

%U 2378,2518,2662,2810,2962,3118,3278,3442,3610,3782,3958,4138,4322,4510

%N a(n) = 2*n^2 + 6*n + 2.

%C Values of polynomial K_2 related to A090285: a(n) = K_2(n) = Sum_{k>=0} A090285(2,k)*2^k*binomial(n,k).

%C Numbers k such that 2*k+5 is a square. - _Vincenzo Librandi_, Oct 10 2013

%C a(n) is the area of a triangle with vertices at (b(n-2),b(n-1)), (b(n),b(n+1)), and (b(n+2),B(n+3)) for b(k)=A000292(k) with n>1. - _J. M. Bergot_, Mar 23 2017

%H Vincenzo Librandi, <a href="/A090288/b090288.txt">Table of n, a(n) for n = 0..1000</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*A028387(n).

%F G.f.: 2*(1 +2*x -x^2)/(1-x)^3. - _R. J. Mathar_, Apr 02 2008

%F E.g.f.: 2*(1 +4*x +x^2)*exp(x). - _G. C. Greubel_, Jul 13 2017

%F Sum_{n>=0} 1/a(n) = 1/2 + Pi*tan(sqrt(5)*Pi/2)/(2*sqrt(5)). - _Amiram Eldar_, Dec 23 2022

%t Table[2*(n^2 +3*n +1), {n, 0, 50}] (* _Vincenzo Librandi_, Oct 10 2013 *)

%t LinearRecurrence[{3,-3,1},{2,10,22},50] (* _Harvey P. Dale_, May 04 2017 *)

%o (PARI) a(n)=2*n^2+6*n+2 \\ _Charles R Greathouse IV_, Sep 24 2015

%o (Magma) [2*(1+3*n+n^2): n in [0..50]]; // _G. C. Greubel_, May 31 2019

%o (Sage) [2*(1+3*n+n^2) for n in (0..50)] # _G. C. Greubel_, May 31 2019

%o (GAP) List([0..50], n-> 2*(1+3*n+n^2)) # _G. C. Greubel_, May 31 2019

%Y Cf. A028387, A090285.

%K nonn,easy

%O 0,1

%A _Philippe Deléham_, Jan 25 2004

%E Corrected by _T. D. Noe_, Nov 12 2006