A recurrence that wiggles?

Question

Consider the following sequence $a_n$:

$a_1 = 0$

$a_n = 1 + \frac{1}{2^n-2} \sum_{i=1}^{n-1} \binom{n}{i} a_i$

The first few terms are $0,1,\frac{3}{2},\frac{13}{7},\frac{15}{7}$.

The sequence comes out of the analysis of a certain process, whose asymptotics we expect to be about $\log_2 n$. Numerical computations show that indeed $a_n \approx \log_2 n - C$, where $C \approx 0.318$. It should be easy to show that $a_n = \log_2 n + O(1)$.

Is it true that $a_n = \log_2 n - C + o(1)$ for some constant $C$?

There are some sequences defined by similar recurrence relations for which the "constant" part actually wiggles (unfortunately the reference escapes me).

Answer 1

This is the Erdos-Renyi "20 Questions" problem. You want to guess an integer in a set of N elements by selecting subsets and asking questions of the form "and is it in this subset?". For the question to provide information the subset cannot be empty or the entire set of uneliminated possibilities, and if an informative question (subset) is chosen at random one gets the above recursion for the expected length of the game.

Erdos and Renyi showed that the expected number of questions is $\log_2(n)$ plus a non-constant periodic function of $\log_2(n)$. I don't remember if they allowed uninformative questions, so the problem could be slightly different, but similar log-periodic behavior should appear in both versions of the problem. (added: the reason it should be the same is that the dominant $\log_2(n)$ term comes from the concentration of the size of the random subset around $k/2$ where $k$ is the number of remaining possibilities, and the log-periodic part comes from the $O(\sqrt{k})$ dispersion of the subset size near $k/2$. What happens at the very extremes of the distribution is much less influential.)

Answer 2

Not really an answer, but too long for a comment.

A recurrence with a wiggle

The following was proposed by Lennart Rade in a 1991 issue of the American Mathematical Monthly, and the solution appears in the January 1994 issue, on page 78.

Begin with $n$ fair coins all showing tails. Toss them all and pick up only the coins with tails. Repeat this procedure until all coins show heads.

The number of coins showing tails is a Markov chain that starts in state $n$ and ends up at 0. The transition probabilities are binomial: $p_{n,i}={n \choose i}\left({1\over 2}\right)^n \mbox{ for }0\leq i\leq n$.

The sequence $u_n=P_{n+1}(\mbox{chain hits }1)$ satisfies $u_0=1$ and for $n\geq1$ $$u_n=\sum_{j=0}^n{n+1\choose j}2^{-(n+1)} u_{n-j},$$ or $$u_n=\sum_{j=1}^n{n+1\choose j}{2^{-(n+1)}\over 1-2^{-(n+1)}} u_{n-j}.$$

In fact, this problem can be solved explicitly giving $$u_n={n+1\over2}\sum_{k=0}^\infty 2^{-k}(1-2^{-k})^n.$$

Analysis of this expression shows that, despite appearances, the sequence $u_n$ does not converge.

It slowly oscillates about the pseudo limit $${1\over2\ln(2)}=.72134$$ with an amplitude of about $${|\Gamma(1-2\pi i/\ln(2))|\over2\ln(2)}=.00000356.$$
It's a fun example to show students as the oscillations are invisible graphically.

For related examples see The asymptotic probability of a tie for first place, by B. Eisenberg, G. Stengle, and G. Strang, Annals of Applied Probability 3, 731-745 (1993).

Answer 3

Noam Elkies gives an example here.