A158736 - OEIS

%I #22 Mar 22 2023 08:14:21

%S 69,279,629,1119,1749,2519,3429,4479,5669,6999,8469,10079,11829,13719,

%T 15749,17919,20229,22679,25269,27999,30869,33879,37029,40319,43749,

%U 47319,51029,54879,58869,62999,67269,71679,76229,80919,85749,90719,95829,101079,106469,111999

%N a(n) = 70*n^2 - 1.

%C The identity (70*n^2 - 1)^2 - (1225*n^2 - 35)*(2*n)^2 = 1 can be written as a(n)^2 - A158735(n)*A005843(n)^2 = 1.

%H Vincenzo Librandi, <a href="/A158736/b158736.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.: x*(-69 - 72*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(70))*Pi/sqrt(70))/2.

%F Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/sqrt(70))*Pi/sqrt(70) - 1)/2. (End)

%t LinearRecurrence[{3, -3, 1}, {69, 279, 629}, 50] (* _Vincenzo Librandi_, Feb 20 2012 *)

%o (Magma) I:=[69, 279, 629]; [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(70*n^2 - 1", ")); \\ _Vincenzo Librandi_, Feb 20 2012

%Y Cf. A005843, A158735.

%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