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a(0) = 1, a(n) = 22*n^2 + 2 for n>0.
1

%I #37 May 07 2024 06:00:37

%S 1,24,90,200,354,552,794,1080,1410,1784,2202,2664,3170,3720,4314,4952,

%T 5634,6360,7130,7944,8802,9704,10650,11640,12674,13752,14874,16040,

%U 17250,18504,19802,21144,22530,23960,25434,26952,28514,30120,31770,33464,35202,36984

%N a(0) = 1, a(n) = 22*n^2 + 2 for n>0.

%C From _Bruno Berselli_, Feb 06 2012: (Start)

%C First trisection of A008259.

%C Apart from the first term, numbers of the form (r^2+2*s^2)*n^2+2 = (r*n)^2+(s*n-1)^2+(s*n+1)^2: in this case is r=2, s=3. After 1, all terms are in A000408. (End)

%H Bruno Berselli, <a href="/A010012/b010012.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 G.f.: (1+x)*(1+20*x+x^2)/(1-x)^3. - _Bruno Berselli_, Feb 06 2012

%F E.g.f.: (x*(x+1)*22+2)*e^x-1. - _Gopinath A. R._, Feb 14 2012

%F Sum_{n>=0} 1/a(n) = 3/4 + sqrt(11)/44*Pi*coth( Pi/sqrt(11)) = 1.0706480516966... - _R. J. Mathar_, May 07 2024

%F a(n) = A069173(n)+A069173(n+1). - _R. J. Mathar_, May 07 2024

%t Join[{1}, 22 Range[41]^2 + 2] (* _Bruno Berselli_, Feb 06 2012 *)

%t Join[{1},LinearRecurrence[{3,-3,1},{24,90,200},50]] (* _Harvey P. Dale_, Jul 20 2013 *)

%t CoefficientList[Series[(1 + x) (1 + 20 x + x^2)/(1 - x)^3, {x, 0, 50}], x] (* _Vincenzo Librandi_, Aug 03 2015 *)

%o (Magma) [1] cat [22*n^2+2: n in [1..50]]; // _Vincenzo Librandi_, Aug 03 2015

%Y Cf. A206399.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_.