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a(n) = 62*n^2 + 1.
2

%I #27 Feb 27 2024 12:07:03

%S 1,63,249,559,993,1551,2233,3039,3969,5023,6201,7503,8929,10479,12153,

%T 13951,15873,17919,20089,22383,24801,27343,30009,32799,35713,38751,

%U 41913,45199,48609,52143,55801,59583,63489,67519,71673,75951,80353,84879,89529,94303,99201

%N a(n) = 62*n^2 + 1.

%C The identity (62*n^2 + 1)^2 - (961*n^2 + 31)*(2*n)^2 = 1 can be written as a(n)^2 - A158675(n)*A005843(n)^2 = 1.

%H Vincenzo Librandi, <a href="/A158676/b158676.txt">Table of n, a(n) for n = 0..10000</a>

%H Vincenzo Librandi, <a href="https://web.archive.org/web/20090309225914/http://mathforum.org/kb/message.jspa?messageID=5785989&amp;tstart=0">X^2-AY^2=1</a>, Math Forum, 2007. [Wayback Machine link]

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).

%F G.f.: -(1 + 60*x + 63*x^2)/(x-1)^3.

%F From _Amiram Eldar_, Mar 21 2023: (Start)

%F Sum_{n>=0} 1/a(n) = (coth(Pi/sqrt(62))*Pi/sqrt(62) + 1)/2.

%F Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/sqrt(62))*Pi/sqrt(62) + 1)/2. (End)

%t LinearRecurrence[{3, -3, 1}, {1, 63, 249}, 50] (* _Vincenzo Librandi_, Feb 18 2012 *)

%t 62*Range[0,40]^2+1 (* _Harvey P. Dale_, Mar 26 2022 *)

%o (Magma) I:=[1, 63, 249]; [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 18 2012

%o (PARI) for(n=0, 40, print1(62*n^2 + 1", ")); \\ _Vincenzo Librandi_, Feb 18 2012

%Y Cf. A005843, A158676.

%K nonn,easy

%O 0,2

%A _Vincenzo Librandi_, Mar 24 2009

%E Comment rewritten, a(0) added and formula replaced by _R. J. Mathar_, Oct 22 2009