The Gauss-Kuzmin distribution gives the probability of an integer appearing as a partial denominator in the continued fraction of a real number $x$ as $$ P(a_k = k) = -\log_2\left(1 - \frac{1}{(k+1)^2} \right) $$

The entropy (in bits) of this distribution is \begin{align*} H = \sum_{k=1}^{\infty} \log_2\left(1 - \frac{1}{(k+1)^2} \right)\log_{2}\left(\log_2\left(1 - \frac{1}{(k+1)^2} \right) \right) \end{align*} The sequence is logarithmically convergent with only positive terms. I used 1000000 terms at 2000 digits of precision using MPFR, and was only able to recover 4 decimal digits. Others have indicated that this constant is very difficult to compute.

In the spirit of the SIAM 100 digit challenge, are there sequence acceleration tools that could make the sequence converge faster?

(Hope you have fun with this one!)


1 Answer 1


It's fairly easy to evaluate, to do this expand the logs in Taylor series in $x=(k+1)^{-2}$: $$ \log_2(1-x) = \frac{-1}{\log 2}\sum_{m\geq1}\frac{x^{m}}{m}$$ $$ \log_2(-\log_2(1-x)) = \frac{\log x}{\log 2} - \frac{\log\log 2}{\log 2} + \sum_{n\geq 1}a_n x^n, $$ where $a_n$ are Taylor series coefficients of the l.h.s. after the log-singularity is subtracted. These can be easily calculated directly using numerical quadrature (code below).

Using the identities (here, as above, $x=(k+1)^{-2}$) $$ \sum_{k\geq 1}x^s = \zeta(2s)-1, $$ $$ \sum_{k\geq 1}x^s\log x = 2\zeta'(2s), $$ we can rewrite the goal sum as a sum of three terms: $$ \frac{-1}{(\log 2)^2}\sum_{k,m\geq 1}\frac{x^m\log x}{m} = \frac{-1}{(\log 2)^2} \sum_{m\geq 1}\frac{2\zeta'(2s)}{m} = 4.067782509260337209451548799089911685899767925643908, $$ the second $$ \frac{\log\log 2}{(\log 2)^2}\sum_{k,m\geq 1}\frac{x^m}{m} = \frac{\log\log 2}{\log 2} = -0.52876637294489761424749777977881481518723706368332, $$ and a triple sum with separately calculated coefficients: $$ \frac{-1}{\log 2}\sum_{k,m,n\geq1} \frac{a_n}{m}x^{m+n} = \frac{-1}{\log 2}\sum_{m,n\geq1}\frac{a_n}{m}(\zeta(2m+2n)-1) = -0.1064886215397004957702703848332558106111776391548. $$ Because $\zeta(a)-1 = O(2^{-a})$, these sums converge extremely quickly, and can be evaluated directly without numerical extrapolation techniques, yielding the result $$ 3.4325275147757390994337806344778410601013532228057895 $$ It looks like it matches the number your link gives ($3.432527514776$) to at least the digits given there. I used fifty digits, and the most time-consuming portion of this is calculating the derivatives.

from mpmath import mp
from mpmath import *
import mpmath
mp.dps = 50

lg = lambda x: mp.log(x, b=2)

def A(n):
    if n == 0:
        return 0
    return (mp.diff(lambda x: lg(-lg(1-x))-(log(x)-log(log(2)))/log(2), 0, n, method="quad")/mp.gamma(n+1)).real

s1 = -log(2)**(-2)*nsum(lambda m: 2*zeta(2*m, derivative=1)/m, (1, mp.inf))
s2 = log(log(2))/log(2)**2 * nsum(lambda m: (zeta(2*m)-1)/m, (1, mp.inf))
s3 = -1/log(2) * nsum(monitor(lambda m, n: A(n)/m * (zeta(2*m+2*n)-1)), (1, mp.inf), (1, mp.inf), method="direct")

print(s1 + s2 + s3)
  • 1
    $\begingroup$ This is awesome. There is so much good stuff in this answer! $\endgroup$ Jun 29, 2020 at 1:26
  • $\begingroup$ This is also my first exposure to 'memoize' and 'monitor', which are awesome tools. $\endgroup$
    – user14717
    Jun 29, 2020 at 13:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.