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a(0) = 1, a(n) = 18*n^2 + 2 for n>0.
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%I #45 May 07 2024 05:55:53

%S 1,20,74,164,290,452,650,884,1154,1460,1802,2180,2594,3044,3530,4052,

%T 4610,5204,5834,6500,7202,7940,8714,9524,10370,11252,12170,13124,

%U 14114,15140,16202,17300,18434,19604,20810,22052,23330,24644,25994,27380,28802,30260

%N a(0) = 1, a(n) = 18*n^2 + 2 for n>0.

%C The identity (18*n^2+2)^2-(9*n^2+2)*(6*n)^2=4 can be written as a(n+1)^2-A010002(n+1)*A008588(n+1)^2=4. - _Vincenzo Librandi_, Feb 07 2012

%H Bruno Berselli, <a href="/A010008/b010008.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+16*x+x^2)/(1-x)^3. - _Bruno Berselli_, Feb 06 2012

%F a(n) = (3*n-1)^2+(3*n+1)^2 = (n-1)^2+(n+1)^2+(4*n)^2 for n>0. - _Bruno Berselli_, Feb 06 2012

%F E.g.f.: (x*(x+1)*18+2)*e^x-1. - _Gopinath A. R._, Feb 14 2012

%F Sum_{n>=0} 1/a(n) = 3/4+ (1/12)*Pi*coth(Pi/3) = 1.0853330948... - _R. J. Mathar_, May 07 2024

%F a(n) = 2*A247792(n), n>0. - _R. J. Mathar_, May 07 2024

%F a(n) = A069131(n)+A069131(n+1). - _R. J. Mathar_, May 07 2024

%t Join[{1}, 18 Range[41]^2 + 2] (* _Bruno Berselli_, Feb 06 2012 *)

%t Join[{1}, LinearRecurrence[{3, -3, 1}, {20, 74, 164}, 50]] (* _Vincenzo Librandi_, Aug 03 2015 *)

%o (Magma) [1] cat [18*n^2+2: n in [1..50]]; // _Vincenzo Librandi_, Aug 03 2015

%Y After 20, all terms are in A000408.

%Y Cf. A206399.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_.

%E More terms from _Bruno Berselli_, Feb 06 2012