0,2

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 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 (ss54(AT)york.ac.uk), 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 (qntmpkt(AT)yahoo.com), Aug 06 2006

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.

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

S. R. Finch, Plouffe's Constant

S. Plouffe, The Computation of Certain Numbers Using a Ruler and Compass, J. Integer Seqs. Vol. 1 (1998), #98.1.3.

A(n)=5^n cos(2n arctan(1/2)). 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 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 (benoit7848c(AT)orange.fr), Aug 03 2002

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

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))

(PARI) a(n)=if(n<0, 0, real((2+I)^(2*n))

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 A0000351 are primitive Pythagorean triples with hypotenuse 5^n. The offset of A0000351 is zero, but the offset is 1 for A066770, A066771.

Cf. A093378.

Sequence in context: A014014 A015884 A156201 this_sequence A139159 A042329 A125956

Adjacent sequences: A066768 A066769 A066770 this_sequence A066772 A066773 A066774

sign,easy,frac

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