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A066770 a(n) = 5^n*sin(2n*arctan(1/2)) or numerator of tan(2n*arctan(1/2)).

%I A066770 #39 May 24 2021 13:18:06
%S A066770 4,24,44,-336,-3116,-10296,16124,354144,1721764,1476984,-34182196,
%T A066770 -242017776,-597551756,2465133864,29729597084,116749235904,
%U A066770 -42744511676,-3175197967656,-17982575014036,-28515500892816,278471369994004,2383715742284424,7340510203856444
%N A066770 a(n) = 5^n*sin(2n*arctan(1/2)) or numerator of tan(2n*arctan(1/2)).
%D A066770 Steven R. Finch, Mathematical Constants, Cambridge, 2003, pp. 430-433.
%H A066770 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 A066770 E. Eckert, <a href="http://www.jstor.org/stable/2690291">The group of primitive Pythagorean triangles</a>, Mathematics Magazine 57 (1984) 22-27.
%H A066770 Steven R. Finch, <a href="http://www.people.fas.harvard.edu/~sfinch/constant/plff/plff.html">Plouffe's Constant</a> [Broken link]
%H A066770 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 A066770 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 A066770 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (6,-25).
%F A066770 G.f.: 4*x/(1-6*x+25*x^2). - _Ralf Stephan_, Jun 12 2003
%F A066770 a(n) = 5^n*sin(2*n*arctan(1/2)). A recursive formula for T(n) = tan(2*n*arctan(1/2)) is T(n+1) = (4/3+T(n))/(1-4/3*T(n)). Unsigned a(n) is the absolute value of numerator of T(n).
%F A066770 a(n) is the imaginary part of (2+I)^(2*n) = Sum_{k=0..n} 2^(2*n-2*k-1)*(-1)^k*binomial(2*n, 2*k+1). - _Benoit Cloitre_, Aug 03 2002
%F A066770 a(n) = 6*a(n-1)-25*a(n-2), n>2. - _Gary Detlefs_, Dec 11 2010
%F A066770 a(n) = 5^n*sin(n*x), where x = arcsin(4/5) = 0.927295218.. . - _Gary Detlefs_, Dec 11 2010
%p A066770 a[1] := 4/3; for n from 1 to 40 do a[n+1] := (4/3+a[n])/(1-4/3*a[n]):od: seq(abs(numer(a[n])), n=1..40);# a[n]=tan(2n arctan(1/2))
%t A066770 Table[ 5^n*Sin[2*n*ArcCot[2]] // Simplify, {n, 1, 23}] (* _Jean-François Alcover_, Mar 04 2013 *)
%o A066770 (PARI) a(n)=imag((2+I)^(2*n))
%Y A066770 Cf. A066771, A000351 powers of 5 and also hypotenuse of right triangle with legs given by A066770 and A066771.
%Y A066770 Note that A066770, A066771 and A000351 are primitive Pythagorean triples with hypotenuse 5^n. The offset of A000351 is zero, but the offset is 1 for A066770, A066771.
%K A066770 sign,easy,frac
%O A066770 1,1
%A A066770 Barbara Haas Margolius, (b.margolius(AT)csuohio.edu), Jan 17 2002