For every integer $n$, we have the following recurrence:
$a_{i}= p^i(1-p)^{n-i}\binom{n}{i} -\sum_{j=i+1}^na_j\binom{j}{i}$.
Can we prove that for every $n$ and $p<1/\sqrt{n}$, it holds that $\forall i, a_i\geq 0?$
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3$\begingroup$ It seems that this is solved by $a_i = p^i (1-2p)^{n-i} \binom{n}{i}$ by a rearrangement of the binomial formula $(1-p)^{n-i} = \sum_{k=0}^{n-i} \binom{n-i}{k} p^k (1-2p)^{n-i-k}$, so yes once $p \leq 1/2$. $\endgroup$– Terry TaoCommented Jun 24, 2023 at 18:46
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$\begingroup$ Thanks for your comment @TerryTao. This perfectly solves my question. $\endgroup$– was_nCommented Jun 25, 2023 at 15:11
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