What's the value of this game (rebalancing counters)?

Question

Suppose you have the following game:

There are infinitely many counters $\{c_1,c_2,\ldots\}$, all initialized to 0.

In each step, you may choose a counter $c_i$ and increase it's value by 1.

Unfortunately, every $T$ steps, each counter that has a positive value is decremented by 1.

Also, the values of the counters are bounded by $M$, so you can't increment a counter any further.

Given as many steps as you like, can many positive valued counters can you reach?

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EDIT: As this question is unanswered for two weeks, I have posted it in CSTheory as well.

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Detailed buildup for $\approx T\log(M)$ positive counters:

1. While you have less than $T-1$ counters at value $M$:

   • Increment the minimal index counter whose value is strictly less than $M$. This has to converge as the sum of the counters are bound to increase every $T$ steps.

2. Let $r = T$.

3. While ($c_0>0$)

   a. while ($c_0>c_r$)

   • Increment $c_r$

   b. $r = r + 1$

Now for the analysis: first observation is that the number of positive counters is $r$.

Now let $m_r$ be the maximal value $c_{r}$ have reached. For $r=T$ we get $M(1-\frac{1}{T})$. For $r=T+1$ we get $m_r(1-\frac{1}{T})=M(1-\frac{1}{T})^2$, or in general $$\forall r\geq T:m_r = M(1-\frac{1}{T})^{r-T+1}$$

Next we notice that when $\forall m_r$ is achieved, $c_0=m_r$. This means the loop will halt when $m_r < 1$ (give or take integrality and end-of-game-strategies).

This gives us $$M(1-\frac{1}{T})^{r-T+1} < 1$$ $$(1-\frac{1}{T})^{r-T+1} < M^{-1}$$ $$({r-T+1}) \log (1-\frac{1}{T}) < -\log M$$ $${r-T+1} < \frac{-\log M}{\log (1-\frac{1}{T})}$$ $$r < \frac{-\log M}{\log (1-\frac{1}{T})} + T - 1$$ $$r < \frac{\log M}{\sum_{k\geq 1}^{\infty} {\frac{1}{kT^k}}} + T - 1 < T(\log M + 1) -1$$

Does this look right?

Is it possible to do better? Can anyone prove this is optimal?