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A158488 a(n) = 64*n^2 + 8. 2

%I #30 Mar 05 2023 03:06:13

%S 72,264,584,1032,1608,2312,3144,4104,5192,6408,7752,9224,10824,12552,

%T 14408,16392,18504,20744,23112,25608,28232,30984,33864,36872,40008,

%U 43272,46664,50184,53832,57608,61512,65544,69704,73992,78408,82952,87624,92424,97352,102408

%N a(n) = 64*n^2 + 8.

%C The identity (16*n^2+1)^2 - (64*n^2+8)*(2*n)^2 = 1 can be written as A108211(n)^2 - a(n)*A005843(n)^2 = 1. - rewritten by _Bruno Berselli_, Nov 16 2011

%H Vincenzo Librandi, <a href="/A158488/b158488.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 a(1)=72, a(2)=264, a(3)=584, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - _Harvey P. Dale_, Nov 16 2011

%F G.f: x*(72 + 48*x + 8*x^2)/(1-x)^3. - _Vincenzo Librandi_, Feb 08 2012

%F From _Amiram Eldar_, Mar 05 2023: (Start)

%F Sum_{n>=1} 1/a(n) = (coth(Pi/(2*sqrt(2)))*Pi/(2*sqrt(2)) - 1)/16.

%F Sum_{n>=1} (-1)^(n+1)/a(n) = (1 - cosech(Pi/(2*sqrt(2)))*Pi/(2*sqrt(2)))/16. (End)

%p A158488:=n->64*n^2+8: seq(A158488(n), n=1..50); # _Wesley Ivan Hurt_, Apr 08 2017

%t 64Range[40]^2+8 (* or *) LinearRecurrence[{3,-3,1},{72,264,584},40] (* _Harvey P. Dale_, Nov 16 2011 *)

%o (Magma) I:=[72,264,584]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+Self(n-3): n in [1..50]]; // _Vincenzo Librandi_, Feb 08 2012

%o (PARI) for(n=1, 40, print1(64*n^2 + 8", ")); \\ _Vincenzo Librandi_, Feb 08 2012

%Y Cf. A005843, A108211.

%K nonn,easy

%O 1,1

%A _Vincenzo Librandi_, Mar 20 2009

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