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%I #26 Mar 23 2023 03:30:58
%S 1482,6045,13650,24297,37986,54717,74490,97305,123162,152061,184002,
%T 218985,257010,298077,342186,389337,439530,492765,549042,608361,
%U 670722,736125,804570,876057,950586,1028157,1108770,1192425,1279122,1368861,1461642,1557465,1656330
%N a(n) = 1521*n^2 - 39.
%C The identity (78*n^2-1)^2-(1521*n^2-39)*(2*n)^2 = 1 can be written as A158771(n)^2-a(n)*A005843(n)^2 = 1.
%H Vincenzo Librandi, <a href="/A158770/b158770.txt">Table of n, a(n) for n = 1..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.: 39*x*(-38-41*x+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>=1} 1/a(n) = (1 - cot(Pi/sqrt(39))*Pi/sqrt(39))/78.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/sqrt(39))*Pi/sqrt(39) - 1)/78. (End)
%p A158770:=n->1521*n^2-39: seq(A158770(n), n=1..50); # _Wesley Ivan Hurt_, May 03 2017
%t LinearRecurrence[{3, -3, 1}, {1482, 6045, 13650}, 50] (* _Vincenzo Librandi_, Feb 21 2012 *)
%t Table[1521n^2-39,{n,30}] (* _Harvey P. Dale_, Aug 23 2019 *)
%o (Magma) I:=[1482, 6045, 13650]; [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=1, 40, print1(1521*n^2 - 39", ")); \\ _Vincenzo Librandi_, Feb 21 2012
%Y Cf. A005843, A158771.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Mar 26 2009
%E Comment rewritten and formula replaced by _R. J. Mathar_, Oct 22 2009
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