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%I #25 Mar 19 2023 02:46:43
%S 26,702,2730,6110,10842,16926,24362,33150,43290,54782,67626,81822,
%T 97370,114270,132522,152126,173082,195390,219050,244062,270426,298142,
%U 327210,357630,389402,422526,457002,492830,530010,568542,608426,649662,692250,736190,781482,828126
%N a(n) = 676*n^2 + 26.
%C The identity (52*n^2 + 1)^2 - (676*n^2 + 26)*(2*n)^2 = 1 can be written as A158644(n)^2 - a(n)*A005843(n)^2 = 1.
%H Vincenzo Librandi, <a href="/A158643/b158643.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.: -26*(1 + 24*x + 27*x^2)/(x-1)^3.
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F From _Amiram Eldar_, Mar 19 2023: (Start)
%F Sum_{n>=0} 1/a(n) = (coth(Pi/sqrt(26))*Pi/sqrt(26) + 1)/52.
%F Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/sqrt(26))*Pi/sqrt(26) + 1)/52. (End)
%t 26(26Range[0,40]^2+1) (* _Harvey P. Dale_, Mar 30 2011 *)
%t LinearRecurrence[{3, -3, 1}, {26, 702, 2730}, 50] (* _Vincenzo Librandi_, Feb 17 2012 *)
%o (Magma) I:=[26, 702, 2730]; [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 17 2012
%o (PARI) for(n=0, 40, print1(676*n^2 + 26", ")); \\ _Vincenzo Librandi_, Feb 17 2012
%Y Cf. A005843, A158644.
%K nonn,easy
%O 0,1
%A _Vincenzo Librandi_, Mar 23 2009
%E Comment rephrased and redundant formula replaced by _R. J. Mathar_, Oct 19 2009
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