The alternating compositions of weighted differential operators yield the weights' Wronskian with which constant?

Abstract

The alternated composition of $N=2p$ differential operators $w_j(x)\,\partial_x^p$ of strict order $p$ on the line $\mathbb{R}\ni x$ is again a differential operator of strict order $p$; its coefficient is the constant $\mathrm{const}(p)$, depending only on the arity $N$, times the Wronskian determinant of the originally taken coefficients $w_1,\dots,w_N$. The case $p=1$ of the Lie bracket for two vector fields fixes $\mathrm{const}(1)=1$, and $\mathrm{const}(2)=2$ is found easily by hand; $\mathrm{const}(3)=90$ can still be obtained symbolically. The problem is to determine $\mathrm{const}(p\geqslant4)$. We compute $\mathrm{const}(p)$ exactly for all $p\leqslant14$ -- a 241-digit integer at $p=14$ -- and record the resulting integer sequence as OEIS A392714. We prove that $v_p(\mathrm{const}(p))\geqslant p-1$ for every prime $p$, matching the exact equality observed numerically throughout our range, and conjecture that this equality holds in general. We show that $\log\mathrm{const}(p)$ grows like $\alpha p^2\log p$, with the leading coefficient close to $2$, nearly saturating the bound $p^2\log p\,(1+O(1/\log p))\leqslant\log\mathrm{const}(p)\leqslant2p^2\log p\,(1+O(1/\log p))$ obtained by O. Zaboronski (private communication). A naturally arising reduced constant is found to decay to zero super-exponentially rather than grow, a direct consequence of this near-saturation.