A067360 - OEIS

A067360

a(n) = 17^n sin(2n arctan(1/4)) or numerator of tan(2n arctan(1/4)).

8, 240, 4888, 77280, 905768, 4839120, -116593352, -4896306240, -113193708472, -1980778750800, -26710380775592, -228866364286560, 853309115549288, 91741652745294480, 2505643247965090168, 48655959795562600320, 735547895204966951048

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OFFSET

1,1

COMMENTS

Note that A067360(n), A067361(n) and 17^n are primitive Pythagorean triples with hypotenuse 17^n.

REFERENCES

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

LINKS

Table of n, a(n) for n=1..17.

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.

Index entries for linear recurrences with constant coefficients, signature (30, -289).

FORMULA

a(n) = 17^n sin(2n arctan(1/4)). A recursive formula for T(n) = tan(2n arctan(1/4)) is T(n+1)=(8/15+T(n))/(1-8/15*T(n)). Unsigned a(n) is the absolute value of numerator of T(n).

Conjectures from Colin Barker, Jul 25 2017: (Start)

G.f.: 8*x / (1 - 30*x + 289*x^2).

a(n) = i*((15 - 8*i)^n - (15 + 8*i)^n)/2 where i=sqrt(-1).

a(n) = 30*a(n-1) - 289*a(n-2) for n>2.

(End)

MAPLE

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(numer(a[n])), n=1..40); # a[n]=tan(2n arctan(1/4))

MATHEMATICA

Table[Tan[2n ArcTan[1/4]] // TrigToExp // Simplify // Numerator, {n, 1, 17} ] (* Jean-François Alcover, Jul 25 2017 *)

CROSSREFS

Cf. A067361 (17^n cos(2n arctan(1/4))).

Cf. A066770, A066771, A067358, A067359, A020888, A014498, A020892.