%I #33 Mar 06 2023 02:20:43
%S 1,23,89,199,353,551,793,1079,1409,1783,2201,2663,3169,3719,4313,4951,
%T 5633,6359,7129,7943,8801,9703,10649,11639,12673,13751,14873,16039,
%U 17249,18503,19801,21143,22529,23959,25433,26951,28513,30119,31769,33463,35201,36983
%N a(n) = 22*n^2 + 1.
%C From the identity (22*n^2 + 1)^2 - (121*n^2 + 11)*(2*n)^2 = 1 we derive a(n)^2 - A158536(n) * A005843(n)^2 = 1.
%H Vincenzo Librandi, <a href="/A158537/b158537.txt">Table of n, a(n) for n = 0..10000</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 + 20*x + 23*x^2)/(1-x)^3.
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F From _Amiram Eldar_, Mar 06 2023: (Start)
%F Sum_{n>=0} 1/a(n) = (coth(Pi/sqrt(22))*Pi/sqrt(22) + 1)/2.
%F Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/sqrt(22))*Pi/sqrt(22) + 1)/2. (End)
%t LinearRecurrence[{3, -3, 1}, {1, 23, 89}, 50] (* _Vincenzo Librandi_, Feb 12 2012 *)
%t 22*Range[0,40]^2+1 (* _Harvey P. Dale_, May 04 2019 *)
%o (Magma) I:=[1, 23, 89]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]]; // _Vincenzo Librandi_, Feb 10 2012
%o (PARI) for(n=1, 40, print1(22*n^2 + 1", ")); \\ _Vincenzo Librandi_, Feb 10 2012
%Y Cf. A005843, A158536.
%K nonn,easy
%O 0,2
%A _Vincenzo Librandi_, Mar 21 2009
%E Comment rewritten, a(0) added by _R. J. Mathar_, Oct 16 2009