Continued Fractions - The HAKMEM Constant

The HAKMEM Constant

HAKMEM features the expression:

\frac{\sqrt{3 / π^{2} + e}}{\tanh (\sqrt{5}) - \sin (69)}

and states it is easily computed using continued fractions.

Observe $\tanh \sqrt{5}$ can be found with one division, one square root, and one application of the continued fraction of Gauss:

\tanh \sqrt{5} = \frac{\sqrt{5}}{1 + \frac{5}{3 + \frac{5}{5 + . . .}}}

The most troublesome subexpression is $\sin (69)$ , for the $\tan$ continued fraction expansion behaves poorly, and the terms in the Taylor series for $\sin (69)$ about 0 grow for quite some time before shrinking to acceptable levels for continued fractions.

As previously suggested, one possibility is to compute the continued fraction expansion of $\cos 1$ using the Taylor series, then evaluate Chebyshev polynomials and other trigonometric identities to find $\sin 69$ .

I followed this route in a test program for the frac library. Starting from $\cos 1$ , I used double-angle formulas to find $\cos 64$ , and the fifth Chebyshev polynomial (of the first kind) to find $\cos 5$ . Then since we may square root continued fractions, we can use $\sin x = \sqrt{1 - \cos^{2} x}$ to find $\sin 64$ and $\sin 5$ . Lastly, the angle addition formula yields $\sin 69$ .

Thus using techniques Gosper describes, and the Taylor series, we find the first 100 decimal places of the HAKMEM constant:

1
59170 96974 31217 53554 22849 04695 38245 87294 24160 11857
90051 76587 45519 76440 69814 05924 94706 86439 45987 93518

and the first 100 terms of its continued fraction expansion:

[0; 1, 1, 1, 2, 4, 2, 2, 1, 4, 1, 6, 2, 9, 13, 1, 1, 8, 3, 7, 1, 10, 6, 11, 2, 2, 2, 3, 3, 6, 12, 2, 1, 12, 1, 2, 2, 5, 1, 2, 1, 1, 24, 6, 1, 1, 2, 11, 1, 8, 50, 1, 1, 2, 3, 1, 2, 1, 130, 1, 29, 3, 1, 9, 1, 1, 1, 1, 16, 1, 2, 1, 1, 4, 1, 3, 1, 1, 7, 1, 1, 5, 1, 1168, 1, 6, 1, 10, 1, 3, 1, 8, 1, 12, 21, 1, 1, 7, 1, 1, 1, . . .]

For computing 1000 digits of the HAKMEM constant, my program was twice as fast as running these bc -l commands:

scale=1000
pi=4*a(1)
x=2*sqrt(5)
(sqrt(3/pi^2+e(1)))/((e(x)-1)/(e(x)+1)-s(69))

This is gratifying, especially as I could easily optimize further. (For example, I should be able to remove three integer divisions in my quadratic algorithm implementation; better ways of finding $\sin 69$ exist).