%I #39 Jun 29 2023 17:58:55
%S 15,161,495,-31679,-1093425,-23647519,-393425745,-4968639359,
%T -35359140465,375162560801,21473668418415,535788072480961,
%U 9867752001506895,141189807098209121,1383913884510780975,713562283940993281,-378544244105385903345
%N a(n) = 17^n*cos(2*n*arctan(1/4)) or denominator of tan(2*n*arctan(1/4)).
%C Note that A067360(n), A067361(n) and 17^n are primitive Pythagorean triples with hypotenuse 17^n.
%D Steven R. Finch, Mathematical Constants, Cambridge, 2003, pp. 430-433.
%H J. M. Borwein and R. Girgensohn, <a href="http://dx.doi.org/10.4153/CJM-1995-013-4">Addition theorems and binary expansions</a>, Canadian J. Math. 47 (1995) 262-273.
%H E. Eckert, <a href="http://www.jstor.org/stable/2690291">The group of primitive Pythagorean triangles</a>, Mathematics Magazine 57 (1984) 22-27.
%H Steven R. Finch, <a href="http://www.people.fas.harvard.edu/~sfinch/constant/plff/plff.html">Plouffe's Constant</a> [Broken link]
%H Steven R. Finch, <a href="http://web.archive.org/web/20010624104257/http://www.mathsoft.com/asolve/constant/plff/plff.html">Plouffe's Constant</a> [From the Wayback machine]
%H Simon Plouffe, <a href="https://cs.uwaterloo.ca/journals/JIS/compass.html">The Computation of Certain Numbers Using a Ruler and Compass</a>, J. Integer Seqs. Vol. 1 (1998), #98.1.3.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (30, -289).
%F a(n) = 17^n*cos(2*n*arctan(1/4)).
%F A recursive formula for T(n) = tan(2*n*arctan(1/4)) is T(n+1) = (8/15+T(n))/(1-8/15*T(n)). Unsigned a(n) is the absolute value of denominator of T(n). [And a(n) = 17^n*cos(n*arctan(8/15)). - _Peter Luschny_, Sep 29 2019]
%F From _Colin Barker_, Jul 25 2017: (Start)
%F G.f.: x*(15 - 289*x) / (1 - 30*x + 289*x^2).
%F a(n) = ((15 - 8*i)^n + (15 + 8*i)^n)/2 where i=sqrt(-1).
%F a(n) = 30*a(n-1) - 289*a(n-2) for n>2. (End)
%F a(n) = Re((8 + 15*i)^n) = Re((4 + i)^(2*n)) = (1/2)*V(2*n,P = 8,Q = 17), where V(n,P,Q) denotes the Lucas sequence of the second kind and i=sqrt(-1). - _Peter Bala_, Sep 24 2019
%p a[1] := 8/15; for n from 1 to 40 do a[n+1] := (8/15+a[n])/(1-8/15*a[n]):od: seq(abs(denom(a[n])), n=1..40);# a[n]=tan(2n arctan(1/4))
%t Table[t = Tan[2 n ArcTan[1/4]] // TrigToExp // Simplify; Sign[t] * Denominator[t], {n, 1, 17}] (* _Jean-François Alcover_, Jul 25 2017 *)
%Y Cf. A067360 (17^n sin(2n arctan(1/4))).
%Y Cf. A066770, A066771, A067358, A067359, A020888, A014498, A020892.
%K sign,easy,frac
%O 1,1
%A Barbara Haas Margolius, (b.margolius(AT)csuohio.edu), Jan 17 2002