Let $$a_{n,k}=\sum_{s_i \geq 1 \atop \sum_{i=1}^{n-k} s_i \leq n} \frac{2^{n}}{(2(n-\sum_{i=1}^{n-k} s_i)+1)!\prod_{i=1}^{n-k} (2s_i)! }$$ for $0 \leq k \leq n-1$. Prove for $1 \leq k \leq n-1$ that $$b_{n,k}=\sum_{l=1}^k (-1)^{k-l} \sum_{s_i \geq 1 \atop \sum_{i=1}^l s_i =k} \prod_{i=1}^l a_{n,s_i}>0.$$ Motivation and alternative formulation can be found here
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$\begingroup$ Any chance that $b_{n,k}$ is an inclusion-exclusion summation? $\endgroup$– Brendan McKayCommented May 10, 2013 at 6:01
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$\begingroup$ Yes, it may be possible to interpret it as a inclusion-exclusion summation which would prove the positivity but I don't know how. The formula for $b_{n,k}$ comes from the condition $$\sum_{i=0}^ka_{n,i}b_{n,k-i}=0$$ $\endgroup$– MarkusCommented May 10, 2013 at 7:56
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$\begingroup$ Forgot a $(-1)^i$ in the formula $\endgroup$– MarkusCommented May 10, 2013 at 7:57
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