Let $a(n)$ for $n \geq 1$ be the number of powers of two $2^m$ that lie between $3^{n-1}$ and $3^{n}$:

$$a(n) = 1, 2, 1, 2, 1, 2, 2, 1, ...$$

It represents the increments between successive terms of allowable dropping times in the Collatz ($3x+1$) problem (see oeis A022921).

Is there an easy way to prove that the sequence never becomes periodic? i.e. do not exist $n_0, k \geq 1$ such that for all $n \geq n_0$, $a(n) = a(n_0 + (n - n_0)\bmod k))$ ?

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    $\begingroup$ It has an irrational Cesàro sum: $\log_23$. $\endgroup$ – Emil Jeřábek supports Monica Feb 28 '14 at 15:10
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    $\begingroup$ even more interesting: there are two types of partial sums, say $a=1,2$ and $b=1,2,2$. Then you can rewrite it $a_1(n)=a,a,a,b,a,a,a,b,a,a,b,a,a,a,b,...$ ,say. THis has again two types of "words",$c=a,a,a,b$ and $d=a,a,b$, say. Then one can rewrite this as $a_2(n) = c,d,d,d,c,d,d,c,d,d,d,....$. Then this has two types of "words"... and so on. It is a nice game to find the ever more compressed expression, and is also related to the continued fraction of log(3)/log(2) (or log(3/2), dan't have it at hand at the moment) $\endgroup$ – Gottfried Helms Mar 1 '14 at 0:42
  • $\begingroup$ (cont) I forgot to mention: the a-periodicity guarantees, that this does not run into a repeated sequence of only one letter but can be continued forever to sharper compressions. If we write the sequence instead as infinite product, beginning with (3/2)*(3/2)*(3/4)*... where we take care that we never fall below 1, then the compression as words provide the "best approximations" of powers of 3 to powers of 2. $\endgroup$ – Gottfried Helms Mar 1 '14 at 1:01

$a(n)=\lfloor \alpha n\rfloor-\lfloor \alpha(n-1)\rfloor$, where $\alpha=\log_23$. Sequences of this type are called Sturmian and are well known (and easily shown) not to be periodic. Basically: for any $x,x'$ whose difference is not an integer, there is a multiple of $\alpha$ such that $\lfloor x+n\alpha\rfloor-\lfloor x\rfloor\ne \lfloor x'+n\alpha\rfloor-\lfloor x'\rfloor$ by density of multiples of $\alpha$ in the circle.

  • $\begingroup$ Thanks! I noticed that Chapter 2 of Lothaire's "Algebraic combinatorics on words" is entirely devoted to "Sturmian words"; I'll try to read it (my mathematical skills are poor). $\endgroup$ – Marzio De Biasi Feb 28 '14 at 15:43

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