|
%I #21 Mar 05 2023 03:05:47
%S 30,105,230,405,630,905,1230,1605,2030,2505,3030,3605,4230,4905,5630,
%T 6405,7230,8105,9030,10005,11030,12105,13230,14405,15630,16905,18230,
%U 19605,21030,22505,24030,25605,27230,28905,30630,32405,34230,36105,38030,40005,42030
%N a(n) = 25*n^2 + 5.
%C The identity (10*n^2+1)^2 - (25*n^2+5) *(2*n)^2 = 1 can be written as A158187(n)^2 - a(n)*A005843(n)^2 = 1.
%H Vincenzo Librandi, <a href="/A158445/b158445.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 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F G.f: 5*x*(6+3*x+x^2)/(1-x)^3.
%F From _Amiram Eldar_, Mar 05 2023: (Start)
%F Sum_{n>=1} 1/a(n) = (coth(Pi/sqrt(5))*Pi/sqrt(5) - 1)/10.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = (1 - cosech(Pi/sqrt(5))*Pi/sqrt(5))/10. (End)
%t Table[25n^2+5,{n,50}]
%o (Magma) I:=[30, 105, 230]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+Self(n-3): n in [1..50]];
%o (PARI) a(n) = 25*n^2 + 5.
%Y Cf. A005843, A158187.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Mar 19 2009
|
