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%I #25 Mar 23 2023 03:31:07
%S 1,79,313,703,1249,1951,2809,3823,4993,6319,7801,9439,11233,13183,
%T 15289,17551,19969,22543,25273,28159,31201,34399,37753,41263,44929,
%U 48751,52729,56863,61153,65599,70201,74959,79873,84943,90169,95551,101089,106783,112633,118639
%N a(n) = 78*n^2 + 1.
%C The identity (78*n^2 + 1)^2 - (1521*n^2 + 39)*(2*n)^2 = 1 can be written as a(n)^2 - A158768(n)*A005843(n)^2 = 1.
%H Vincenzo Librandi, <a href="/A158769/b158769.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&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 G.f.: -(1 + 76*x + 79*x^2)/(x-1)^3.
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F From _Amiram Eldar_, Mar 23 2023: (Start)
%F Sum_{n>=0} 1/a(n) = (coth(Pi/sqrt(78))*Pi/sqrt(78) + 1)/2.
%F Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/sqrt(78))*Pi/sqrt(78) + 1)/2. (End)
%t LinearRecurrence[{3, -3, 1}, {1, 79, 313}, 50] (* _Vincenzo Librandi_, Feb 21 2012 *)
%t 78*Range[0,40]^2+1 (* _Harvey P. Dale_, Dec 06 2018 *)
%o (Magma) I:=[1, 79, 313]; [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 21 2012
%o (PARI) for(n=0, 40, print1(78*n^2 + 1", ")); \\ _Vincenzo Librandi_, Feb 21 2012
%Y Cf. A005843, A158768.
%K nonn,easy
%O 0,2
%A _Vincenzo Librandi_, Mar 26 2009
%E Comment rewritten, a(0) added, and formula replaced by _R. J. Mathar_, Oct 22 2009
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