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

%I #35 May 07 2024 04:56:14

%S 1,13,46,101,178,277,398,541,706,893,1102,1333,1586,1861,2158,2477,

%T 2818,3181,3566,3973,4402,4853,5326,5821,6338,6877,7438,8021,8626,

%U 9253,9902,10573,11266,11981,12718,13477,14258,15061,15886,16733,17602,18493,19406

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

%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=3, s=1. After 13, all terms are in A000408. - _Bruno Berselli_, Feb 06 2012

%H Bruno Berselli, <a href="/A010003/b010003.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+9*x+x^2)/(1-x)^3. - _Bruno Berselli_, Feb 06 2012

%F E.g.f.: (x*(x+1)*11+2)*e^x-1. - _Gopinath A. R._, Feb 14 2012

%F Sum_{n>=0} 1/a(n) = 3/4+sqrt(22)/44*Pi*coth( Pi*sqrt(22)/11) = 1.134242719070... - _R. J. Mathar_, May 07 2024

%t Join[{1}, 11 Range[41]^2 + 2] (* _Bruno Berselli_, Feb 06 2012 *)

%t Join[{1}, LinearRecurrence[{3, -3, 1}, {13, 46, 101}, 50]] (* _Vincenzo Librandi_, Aug 03 2015 *)

%o (PARI) A010003(n)=11*n^2+2-!n \\ _M. F. Hasler_, Feb 14 2012

%o (Magma) [1] cat [11*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_.

%E More terms from _Bruno Berselli_, Feb 06 2012