Nice idea! We can show directly that each term provides information about the next.
The density function of the distribution of the fractional part in the continued fractional algorithm converges to 1/[(1+x) ln(2)] (it seems this is also called the Gauss-Kuzmin distribution, since the two are so closely associated). So we can directly calculate the probability of getting a coefficient of n by integrating this from 1/(n+1) to 1/n, which gives -lg(1-1/(n+1)^2) as you say above. But we can also calculate the probability of getting an n followed by an m, by integrating this from 1/(n+1/m) to 1/(n+1/(m+1)), which gives -lg(1-1/(mn+1)(mn+m+n+2)). So dividing one by the other gives P(m|n) = lg(1-1/(mn+1)(mn+m+n+2))/lg(1-1/(n+1)^2), which is rather ugly, but the point is that it does depend on n.
This turns out to be an anticorrelation. High numbers are more likely to by followed by low numbers, and vice-versa. The probability of getting a 1 given you’ve just had a 1 is 36.6%, whereas if you’ve just had a very high number the probability of getting a 1 will be very close to 50% (since the distribution of the fractional part is tending to uniform).
I tend to view the golden ratio as the least irrational irrational number. It fills in the next gap after all the rational numbers. In the same way, 1⁄2 is the noninteger which shares the most algebraic properties with the integers, even though it’s furthest from them in a metric sense.
Nice idea! We can show directly that each term provides information about the next.
The density function of the distribution of the fractional part in the continued fractional algorithm converges to 1/[(1+x) ln(2)] (it seems this is also called the Gauss-Kuzmin distribution, since the two are so closely associated). So we can directly calculate the probability of getting a coefficient of n by integrating this from 1/(n+1) to 1/n, which gives -lg(1-1/(n+1)^2) as you say above. But we can also calculate the probability of getting an n followed by an m, by integrating this from 1/(n+1/m) to 1/(n+1/(m+1)), which gives -lg(1-1/(mn+1)(mn+m+n+2)). So dividing one by the other gives P(m|n) = lg(1-1/(mn+1)(mn+m+n+2))/lg(1-1/(n+1)^2), which is rather ugly, but the point is that it does depend on n.
This turns out to be an anticorrelation. High numbers are more likely to by followed by low numbers, and vice-versa. The probability of getting a 1 given you’ve just had a 1 is 36.6%, whereas if you’ve just had a very high number the probability of getting a 1 will be very close to 50% (since the distribution of the fractional part is tending to uniform).
Thanks! That’s surprisingly straightforward.
Huh, an extra reason why the golden ratio is the “most irrational”/most unusual irrational number.
I tend to view the golden ratio as the least irrational irrational number. It fills in the next gap after all the rational numbers. In the same way, 1⁄2 is the noninteger which shares the most algebraic properties with the integers, even though it’s furthest from them in a metric sense.