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 A066771 5^n cos(2n arctan(1/2)) or denominator of tan(2n arctan(1/2)). 12

%I

%S 1,3,-7,-117,-527,-237,11753,76443,164833,-922077,-9653287,-34867797,

%T 32125393,1064447283,5583548873,6890111163,-98248054847,-761741108157,

%U -2114245277767,6358056037323,91004468168113,387075408075603,47340744250793,-9392840736385317

%N 5^n cos(2n arctan(1/2)) or denominator of tan(2n arctan(1/2)).

%C Let A =

%C [ -(3/5)-(2/5)i,-(2/5)i,-(2/5)i,-(2/5)i ]

%C [ -(2/5)i,-(3/5)+(2/5)i,-(2/5)i,(2/5)i ]

%C [ -(2/5)i,-(2/5)i,-(3/5)+(2/5)i,(2/5)i ]

%C [ -(2/5)i,(2/5)i,(2/5)i,-(3/5)-(2/5)i ]

%C be the Cayley transform of the matrix iH, where H =

%C [1,1,1,1]

%C [1,-1,1,-1]

%C [1,1,-1,-1]

%C [1,-1,-1,1]

%C is an Hadamard matrix of order 4 and i is the imaginary unit. Any diagonal entry of the matrix A^n is one of the four complex numbers (+ or -)(X/5^n)(+ or -)(Y/(5^n)i). Then a(n) is the X in [A^n]_(j,j), j=1,2,3,4. - _Simone Severini_, Apr 28 2004

%C Related to the (3,4,5) Pythagorean triple. Each unsigned term represents a leg in a Pythagorean triple in which the hypotenuse = 5^n. E.g. (3 + 4i)^3 = (-117 + 44i), considered as two legs of a triangle, hypotenuse = 125 = 5^3. - _Gary W. Adamson_, Aug 06 2006

%D S. 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 S. R. Finch, <a href="http://www.people.fas.harvard.edu/~sfinch/constant/plff/plff.html">Plouffe's Constant</a>

%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 (6,-25).

%F G.f.: ( 1-3*x ) / ( 1-6*x+25*x^2 ).

%F A recursive formula for T(n) = tan(2n arctan(1/2)) is T(n+1)=(4/3+T(n))/(1-4/3*T(n)). Unsigned A(n) is the absolute value of the denominator of T(n).

%F a(n) is the real part of (2+I)^(2n) = sum(k=0, n, 4^(n-k)*(-1)^k*C(2n, 2k) ). - _Benoit Cloitre_, Aug 03 2002

%F a(n) = real part of (3 + 4i)^n. - _Gary W. Adamson_, Aug 06 2006

%F a(n) = 6*a(n-1)-25*a(n-2). - _Gary Detlefs_, Jun 10 2010

%F a(n) = 5^n*cos(n*arccos(3/5)). - _Gary Detlefs_, Dec 11 2010

%p 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(denom(a[n])), n=1..40);# a[n]=tan(2n arctan(1/2))

%t CoefficientList[Series[(1-3x)/(1-6x+25x^2),{x,0,30}],x] (* or *) LinearRecurrence[{6,-25},{1,3},30] (* _Harvey P. Dale_, Jul 16 2011 *)

%o (PARI) a(n)=real((2+I)^(2*n))

%Y Cf. A066770 5^n sin(2n arctan(1/2)), A000351 powers of 5 and also hypotenuse of right triangle with legs given by A066770 and A066771.

%Y Note that A066770, A066771 and A000351 are primitive Pythagorean triples with hypotenuse 5^n. The offset of A000351 is 0, but the offset is 1 for A066770, A066771.

%Y Cf. A093378.

%Y Cf. A193410, A121622.

%K sign,easy,frac

%O 0,2

%A Barbara Haas Margolius, (b.margolius(AT)csuohio.edu), Jan 17 2002

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