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Let $k,n \in \mathbb{N}$, where $k$ can be thought of as being fixed constant. For each $1 \leq \ell \leq k$ and $1 \leq i \leq n$ we have a proposition symbol $p_{(\ell,i)}$ (so in total we have $nk$-proposition symbols). Consider the following propositional formula $$\bigvee_{n_1 + \dots + n_k = n} \ \bigwedge_{1 \leq \ell \leq k} \ p_{(\ell,n_\ell)}$$ of size $O(kn^{k-1})$. Is there an equivalent propositional formula which is essentially shorter (say, of length $n^{(k/2)}$)?

EDIT: Even though the original question now has a perfectly valid answer, I would still like to know whether one could do better than $O(n^{\log(k)})$. For example, is $O(n^{100})$ possible? If you know how one could achieve a bound of this shape, you could send me an email about it directly (or just post it as an answer, if you prefer).

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    $\begingroup$ Would you accept a formula that uses additional symbols and is equisatisfiable to your formula? $\endgroup$
    – D.W.
    Nov 30, 2022 at 9:23
  • $\begingroup$ @D.W. Good question. I'm only looking for shorter formulas that are equivalent with my formula (motivation comes from formula size lower bounds, not SAT-encoding related things). $\endgroup$ Nov 30, 2022 at 9:41

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How about this?

$$\bigvee_{1 \le u \le n} \left( \bigvee_{n_1+\dots+n_{k/2}=u} \bigwedge_{1 \le \ell \le k/2} p_{\ell,n_\ell} \right) \land \left( \bigvee_{n_{k/2+1}+\dots+n_k=n-u} \bigwedge_{k/2+1 \le \ell \le n} p_{\ell,n_\ell} \right)$$

This has size $O(kn^{k/2+1})$. Of course, this can be generalized by applying the same transformation recursively.

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    $\begingroup$ I guess that if you apply this recursively you can get something like $O(kn^{\log(k)})$? $\endgroup$ Nov 30, 2022 at 10:01

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