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%I #28 Mar 22 2023 08:14:17
%S 71,287,647,1151,1799,2591,3527,4607,5831,7199,8711,10367,12167,14111,
%T 16199,18431,20807,23327,25991,28799,31751,34847,38087,41471,44999,
%U 48671,52487,56447,60551,64799,69191,73727,78407,83231,88199,93311,98567,103967,109511,115199
%N a(n) = 72*n^2 - 1.
%C The identity (72*n^2 - 1)^2 - (1296*n^2 - 36)*(2*n)^2 = 1 can be written as a(n)^2 - A158737(n)*A005843(n)^2 = 1.
%H Vincenzo Librandi, <a href="/A158738/b158738.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.: x*(-71 - 74*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 22 2023: (Start)
%F Sum_{n>=1} 1/a(n) = (1 - cot(Pi/(6*sqrt(2)))*Pi/(6*sqrt(2)))/2.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/(6*sqrt(2)))*Pi/(6*sqrt(2)) - 1)/2. (End)
%p A158738:=n->72*n^2 - 1; seq(A158738(n), n=1..40); # _Wesley Ivan Hurt_, Feb 01 2014
%t 72Range[40]^2-1 (* or *) LinearRecurrence[{3,-3,1},{71,287,647},40] (* _Harvey P. Dale_, May 01 2011 *)
%o (Magma) I:=[71,287,647]; [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(72*n^2 - 1", ")); \\ _Vincenzo Librandi_, Feb 21 2012
%Y Cf. A005843, A158737.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Mar 25 2009
%E Comment rewritten and formula replaced by _R. J. Mathar_, Oct 22 2009
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