A173089 - OEIS

%I #26 Sep 15 2024 13:57:27

%S 26,102,228,404,630,906,1232,1608,2034,2510,3036,3612,4238,4914,5640,

%T 6416,7242,8118,9044,10020,11046,12122,13248,14424,15650,16926,18252,

%U 19628,21054,22530,24056,25632,27258,28934,30660,32436,34262,36138,38064,40040,42066,44142,46268,48444,50670,52946,55272,57648

%N a(n) = 25*n^2 + n.

%C The identity (5000*n^2 + 200*n + 1)^2 - (25*n^2 + n)*(1000*n + 20)^2 = 1 can be written as A157511(n)^2 - a(n)*A157510(n)^2 = 1. This is the case s=5 of the identity (8*n^2*s^4 + 8*n*s^2 + 1)^2 -(n^2*s^2 + n)*(8*n*s^3 + 4*s)^2 = 1. - _Vincenzo Librandi_, Feb 04 2012

%H Vincenzo Librandi, <a href="/A173089/b173089.txt">Table of n, a(n) for n = 1..10000</a>

%H Vincenzo Librandi, <a href="http://mathforum.org/kb/message.jspa?messageID=5785989&amp;tstart=0">X^2-AY^2=1</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

%F G.f.: x*(-26 - 24*x)/(x-1)^3. - _Vincenzo Librandi_, Feb 04 2012

%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - _Vincenzo Librandi_, Feb 04 2012

%t LinearRecurrence[{3, -3, 1}, {26, 102, 228}, 50] (* _Vincenzo Librandi_, Feb 04 2012 *)

%t Table[25n^2+n,{n,50}] (* _Harvey P. Dale_, Sep 15 2024 *)

%o (Magma) [ 25*n^2+n: n in [1..50] ];

%o (PARI) for(n=1, 40, print1(25*n^2 + n", ")); \\ _Vincenzo Librandi_, Feb 04 2012

%Y Cf. A157510, A157511.

%K nonn,easy

%O 1,1

%A _Vincenzo Librandi_, Nov 22 2010