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Fix any integer $n\geq 1$, and let $a_0,\dots,a_n\in\mathbb{N}$ be positive integers. Define $a\stackrel{\rm def}{=} \sum_{k=0}^n a_k$, and $$ \Delta \stackrel{\rm def}{=} \min_{0\leq k\leq n-1} a_{k+1}\left(2^{a_k}-1\right). $$ What is the best upper bound (as a function of $a$ and $n$) than can be established on $\Delta$?

By an averaging argument and a coarse majoration, I can show that $\Delta \leq a\left(2^{\frac{a}{n}}-1\right)$; but that seems "highly non-optimal." In the absence of further assumptions, can this bound be improved?

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    I see a seemingly useless geometric series, if it happens to be of use to someone. – 2017-01-20
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    @SimpleArt Can you expand on that? – 2017-01-20
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    The "expanding" is that$$2^{a_k}-1=\sum_{n=0}^{a_k-1}2^n$$ – 2017-01-20
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    Indeed. Unclear (to me at least) what to do with it, however. (But thanks!) – 2017-01-20
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    On general principles, I suspect (but do not know for sure) that the maximum value of $\Delta$ for fixed $a, n$ will occur when the $a_k$ are equal.. In that case $\Delta = \frac an(2^{\frac an} - 1)$. To examine it, you can see what happens when you allow the $a_k$ to be any real number, not just positive integers. Then you can bring all those handy calculus tools into play. – 2017-01-20

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