The structure of late-growing permutations

walks that never go underground

With Arthemy Kiselev. The permutations behind the Wronskian constant $c(p)$ (OEIS A392714) and why their signs almost perfectly cancel.

Preprint · Oct 2026

arXiv:2610.35779 ↗ arXiv:2605.11137 ↗ arXiv:2610.08636 ↗ OEIS A147681 ↗ OEIS A392714 ↗

What they are

never ahead of the average

A permutation $\sigma \in S_{2p-1}$ is late-growing if no prefix sum gets ahead of the average value $p$:

$$\sigma_1 + \cdots + \sigma_k \;\leqslant\; k\,p \qquad (1 \leqslant k \leqslant 2p-1).$$

The small entries come early and the large ones late. Write $\Phi_p$ for the set of them.

Reading $\sigma$ from the right turns it into a walk. The steps are $\tau_j = \sigma_{2p-j} - p$, which use each of $-(p-1), \dots, p-1$ once, and the heights are $T_k = \tau_1 + \cdots + \tau_k$. The walk ends at $0$, and $\sigma$ is late-growing exactly when it never goes below $0$.

For $p = 3$ the steps $\tau = (0, 1, -1, 2, -2)$ have heights $0, 1, 0, 2, 0$, so $\sigma = (1, 5, 2, 4, 3) \in \Phi_3$. There are 35 such permutations.

Why they matter: with the weights $w_j = x^{j-1}$, the height $T_k$ is the power of $x$ left after the $k$-th block of $p$ derivatives, and a term dies as soon as that power would go negative. What survives is a signed sum over $\Phi_p$ (arXiv:2610.08636, Eq. 8):

$$c(p)\cdot\prod_{k=0}^{2p-1}k! \;=\; \sum_{\sigma\in\Phi_p}(-1)^{\sigma}\prod_{k=1}^{2p-1}E_k^{\,\underline{p}}, \qquad E_k = T_k + p,$$

$$\begin{aligned}c(p)\cdot\prod_{k=0}^{2p-1}k! &= \sum_{\sigma\in\Phi_p}(-1)^{\sigma}\prod_{k=1}^{2p-1}E_k^{\,\underline{p}},\\ E_k &= T_k + p,\end{aligned}$$

where $E^{\underline{p}} = E(E-1)\cdots(E-p+1)$ is a falling factorial. Each walk is weighted by the falling factorials of its heights.

Every walk at once

pick one, or build your own

even odd

Build your own: tap the steps in order. Steps that would go below the ground are greyed out.

The explorer needs JavaScript. The walk above is one of the 35 for $p = 3$.

Each thumbnail is one walk in $\Phi_p$, coloured by the sign of its permutation. The line under the grid adds up every signed weight and divides by $\prod k!$, which gives $c(p)$.

Pinned at one step

the ornament

Fix a step $k$ and keep only the walks that are back on the ground there, $T_k = 0$. Their signed weights still add up to $c(p)$, up to a sign (arXiv:2610.35779, Proposition 6.1, verified for $p \leqslant 14$):

$$\sum_{\sigma\in\Phi_p,\ T_k = 0}(-1)^{\sigma}\prod_{i=1}^{2p-1}E_i^{\,\underline{p}} \;=\; \pm\, c(p)\cdot\prod_{j=0}^{2p-1}j!, \qquad 1 \leqslant k \leqslant 2p-2.$$

$$\begin{aligned}&\sum_{\sigma\in\Phi_p,\ T_k = 0}(-1)^{\sigma}\prod_{i=1}^{2p-1}E_i^{\,\underline{p}}\\ &\qquad= \pm\, c(p)\cdot\prod_{j=0}^{2p-1}j!\end{aligned}$$

So $c(p)$ can be read off a much smaller set of walks. The signs form what the paper calls the ornament: blue for $+c(p)$, orange for $-c(p)$. The last cell of each row, $k = 2p-1$, pins nothing and is the full sum. The pattern has period 4 in $k$ and period 2 in $p$, and the paper conjectures that it continues for every $p$ (Conjecture 6.2).

The ornament needs JavaScript; it is Table 4 of the paper.

Hover or tap a cell, or use the arrow keys. For $p \leqslant 4$ every walk in $\Phi_p$ is drawn, with the pinned ones in colour and a gold dot where they touch the ground; the sum underneath is computed on the spot.

How many

counted by parity

Counted with a dynamic programme over the set of values used so far, up to $p = 10$; the rows from $p = 11$ are Table 6 of arXiv:2610.35779. The even and odd columns never differ by more than one.

$p$	$\|\Phi_p\|$	even	odd
1	1	1	0
2	3	1	2
3	35	18	17
4	1,001	500	501
5	53,109	26,555	26,554
6	4,605,271	2,302,635	2,302,636
7	589,809,987	294,904,994	294,904,993
8	104,899,483,845	52,449,741,922	52,449,741,923
9	24,698,838,710,457	12,349,419,355,229	12,349,419,355,228
10	7,444,522,779,300,351	3,722,261,389,650,175	3,722,261,389,650,176
11	2,792,467,448,952,670,671	1,396,233,724,476,335,336	1,396,233,724,476,335,335
12	1,276,369,340,371,154,144,337	638,184,670,185,577,072,168	638,184,670,185,577,072,169
13	698,008,560,075,731,437,652,425	349,004,280,037,865,718,826,213	349,004,280,037,865,718,826,212
14	450,111,715,263,775,132,783,135,875	225,055,857,631,887,566,391,567,937	225,055,857,631,887,566,391,567,938

$|\Phi_p|$ divided by $(2p-2)!$. The ratio falls slowly, from 1.5 at $p = 2$ to 1.12 at $p = 14$; the dashed line is 1.

back to research

arXiv · Shah, K. C.with A. V. Kiselev

2605.11137 · math.CO · may 2026

Wronskian identity for iterated differential operators

$\bigl(\operatorname{const}(p)\bigr)_{p\geqslant 1}=\bigl(1,\,2,\,90,\,586656,\,\dots\bigr)$
2610.08636 · math.NT · oct 2026expanded extract

Fast calculation of OEIS A392714

$c(18)=4.881\ldots\cdot 10^{462}$
2610.35779 · math.CO · oct 2026proof of the claim

Even vs odd late-growing permutations

$\#\Phi_p^{+}-\#\Phi_p^{-}=(-1)^{p+1}$

arXiv:2605.11137 · math.CO

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

Kian C. Shah, Arthemy V. Kiselev

\begin{aligned}&\sum_{\sigma\in S_N}(-1)^{\sigma}\,\bigl(w_{\sigma(1)}\partial_x^{\,p}\bigr)\circ\cdots\circ\bigl(w_{\sigma(N)}\partial_x^{\,p}\bigr)\\&\qquad=\operatorname{const}(p)\cdot\operatorname{Wr}(w_1,\dots,w_N)\cdot\partial_x^{\,p}\end{aligned}

/now

Current state vector

Preparing for the Caltech Mathathon - super excited to be a part of it!
Back from the ASC School on Generalized Symmetries - Arnold Sommerfeld Center, LMU Munich, 5 - 9 October 2026 - school page ↗.
New preprint with A. V. Kiselev: The counts of parity-odd vs even late-growing permutations in the group S2p−1 differ by (−1)p - arXiv:2610.35779. The ornament from it is now on the site.

OEIS

A147681

Late-growing permutations: permutations of 1..n with every partial sum at most the average partial sum.

1, 1, 1, 3, 7, 35, 139, 1001, 5701, 53109, 402985, 4605271, …

Gold: the odd-indexed terms, which count the walks behind c(p):

a(2p-1) = |\Phi_p|