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%I #29 Mar 20 2023 03:02:24
%S 1,57,225,505,897,1401,2017,2745,3585,4537,5601,6777,8065,9465,10977,
%T 12601,14337,16185,18145,20217,22401,24697,27105,29625,32257,35001,
%U 37857,40825,43905,47097,50401,53817,57345,60985,64737,68601,72577,76665,80865,85177,89601
%N a(n) = 56*n^2 + 1.
%C The identity (56*n^2 + 1)^2 - (784*n^2 + 28)*(2*n)^2 = 1 can be written as a(n)^2 - A158659(n)*A005843(n)^2 = 1.
%H Vincenzo Librandi, <a href="/A158660/b158660.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 + 54*x + 57*x^2)/(x-1)^3.
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F From _Amiram Eldar_, Mar 20 2023: (Start)
%F Sum_{n>=0} 1/a(n) = (coth(Pi/(2*sqrt(14)))*Pi/(2*sqrt(14)) + 1)/2.
%F Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/(2*sqrt(14)))*Pi/(2*sqrt(14)) + 1)/2. (End)
%t LinearRecurrence[{3, -3, 1}, {1, 57, 225}, 50] (* _Vincenzo Librandi_, Feb 17 2012 *)
%t 56 Range[0,40]^2+1 (* _Harvey P. Dale_, Jun 14 2022 *)
%o (Magma) I:=[1, 57, 225]; [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(56*n^2 + 1", ")); \\ _Vincenzo Librandi_, Feb 17 2012
%Y Cf. A158659, A005843.
%K nonn,easy
%O 0,2
%A _Vincenzo Librandi_, Mar 23 2009
%E Comment rephrased and redundant formula replaced by _R. J. Mathar_, Oct 19 2009