A158735 - OEIS

%I #24 Mar 22 2023 08:14:41

%S 1190,4865,10990,19565,30590,44065,59990,78365,99190,122465,148190,

%T 176365,206990,240065,275590,313565,353990,396865,442190,489965,

%U 540190,592865,647990,705565,765590,828065,892990,960365,1030190,1102465,1177190,1254365,1333990,1416065

%N a(n) = 1225*n^2 - 35.

%C The identity (70*n^2 - 1)^2 - (1225*n^2 - 35)*(2*n)^2 = 1 can be written as A158736(n)^2 - a(n)*A005843(n)^2 = 1.

%H Vincenzo Librandi, <a href="/A158735/b158735.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&amp;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.: 35*x*(-34 - 37*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/sqrt(35))*Pi/sqrt(35))/70.

%F Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/sqrt(35))*Pi/sqrt(35) - 1)/70. (End)

%t LinearRecurrence[{3, -3, 1}, {1190, 4865, 10990}, 50] (* _Vincenzo Librandi_, Feb 20 2012 *)

%t 1225Range[30]^2-35 (* _Harvey P. Dale_, May 08 2021 *)

%o (Magma) I:=[1190, 4865, 10990]; [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 20 2012

%o (PARI) for(n=1, 40, print1(1225*n^2 - 35", ")); \\ _Vincenzo Librandi_, Feb 20 2012

%Y Cf. A005843, A158736.

%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