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

1, 3, -7, -117, -527, -237, 11753, 76443, 164833, -922077, -9653287, -34867797, 32125393, 1064447283, 5583548873, 6890111163, -98248054847, -761741108157, -2114245277767, 6358056037323, 91004468168113, 387075408075603, 47340744250793, -9392840736385317

Offset: 0

Author

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

Keywords

• sign
• easy
• frac

Comments

Let A =

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

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

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

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

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

[1, 1, 1, 1]

[1, -1, 1, -1]

[1, 1, -1, -1]

[1, -1, -1, 1]

is a 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

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

a(n) = 5^n*cos(nC-nA), where C is the angle opposite side AB and A is the angle opposite side BC in a triangle ABC having sidelengths |BC|=3, |CA|=4, |AB|=5 - Clark Kimberling, Oct 02 2024

If a prime p divides a term, then the indices n such that p divides a(n) comprise an arithmetic sequence; e.g., 7 divides a(4n+2) for n >= 0; 13 divides a(6n+3) for n>= 0. See the Renault paper in References. - Clark Kimberling, Oct 03 2024

References

• Steven R. Finch, Mathematical Constants, Cambridge, 2003, pp. 430-433.

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..1000
• J. M. Borwein and R. Girgensohn, Addition theorems and binary expansions, Canadian J. Math. 47 (1995) 262-273.
• E. Eckert, The group of primitive Pythagorean triangles, Mathematics Magazine 57 (1984) 22-27.
• Steven R. Finch, Plouffe's Constant [Broken link]
• Steven R. Finch, Plouffe's Constant [From the Wayback machine]
• Simon Plouffe, The Computation of Certain Numbers Using a Ruler and Compass, J. Integer Seqs. Vol. 1 (1998), #98.1.3.
• Marc Renault, The Period, Rank, and Order of the (a,b)-Fibonacci Sequence mod m, Math. Mag. 86 (2013) pp. 372-380.
• Index entries for linear recurrences with constant coefficients, signature (6,-25).

Crossrefs

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.

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.

Cf. A093378.

Cf. A193410, A121622.

Cf. A139030.

Programs

Maple

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))

Mathematica

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 *)

PARI

a(n)=real((2+I)^(2*n))

Formula

G.f.: ( 1-3*x ) / ( 1-6*x+25*x^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 the denominator of T(n).

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

a(n) = real part of (3 + 4i)^n. - Gary W. Adamson, Aug 06 2006

a(n) = 6*a(n-1) - 25*a(n-2). - Gary Detlefs, Jun 10 2010

a(n) = 5^n*cos(n*arccos(3/5)). - Gary Detlefs, Dec 11 2010

a(n) = (-1)^n * hypergeom([1,-n,1/2-n],[1/2,1],-4). - Gerry Martens, Jul 28 2023