Let E_d(m,n) be the flip-graph count, and consider its generating function in n. In the transfer-matrix regime, write its Laurent expansion at z=1 as ∑_n≥2E_d(m,n)z^n=fracc_1(m)(1-z)^d-1+fracc_2(m)(1-z)^d-2+…. Sub-leading Laurent coefficient conjecture. For every d≥5 and every m≥2, c_2(m)=-4(d-4). The leading coefficient is known to be c_1(m)=4; the proposed constant second Laurent coefficient has been verified for d=5,6,7,8 in the stated finite set of cases and by further transfer-matrix computations, but no general proof is given.

Sub-leading Laurent coefficient conjecture for Miura-ori flip-graph counts

From papers

Let $E_d(m,n)$ be the flip-graph count, and consider its generating function in $n$ . In the transfer-matrix regime, write its Laurent expansion at $z=1$ as

\sum_{n\ge2}E_d(m,n)z^n=\frac{c_1(m)}{(1-z)^{d-1}}+\frac{c_2(m)}{(1-z)^{d-2}}+\cdots.

Sub-leading Laurent coefficient conjecture. For every $d\ge5$ and every $m\ge2$ ,

c_2(m)=-4(d-4).

The leading coefficient is known to be $c_1(m)=4$ ; the proposed constant second Laurent coefficient has been verified for $d=5,6,7,8$ in the stated finite set of cases and by further transfer-matrix computations, but no general proof is given.

Progress summary

Partially solved

The proposed formula has been checked in more cases, but no general proof or counterexample has been publicly verified.

The conjecture asserts that the second Laurent coefficient equals $-4(d-4)$ for every $d\ge5$ and $m\ge2$ , while the leading coefficient is $4$ .

Known results

The transfer-matrix framework gives rational generating functions whose coefficient of $x^d$ encodes the counts.
In the high-dimensional region $m,n\ge d-1$ , the counts agree with a symmetric polynomial of per-variable degree at most $d-2$ .
Explicit formulas are given through $d=7$ , with computations extending checks of the conjectured coefficient through $d=11$ .

July 2026 preprint

C. Gupta’s preprint develops the transfer-matrix method and reports the extended checks, but it does not provide a general proof, counterexample, withdrawal, or referee objection concerning $c_2(m)=-4(d-4)$ .

Current status (as of August 2026): The formula remains unproved and unrefuted in general; it has been verified computationally through $d=11$ in the reported regime.

Sources

arxiv.org

Sources & referencesView supporting material

Primary source

Chakshu Gupta, “One construction for the Miura-ori flip-graph degree sequence”, arXiv:2607.05567 (2026).

1 week ago

The conjecture fails for every $d\ge6$ already at the allowed width $m=2$ . The exact corrected two-row coefficient is

\boxed{ c_1(2)=4, \qquad c_2(2)=4(d-6) \qquad(d\ge5), }

whereas Conjecture 9.4 predicts $c_2(2)=-4(d-4)$ .

Let

V_n(x)=\sum_{j\ge0}E_j(2,n)x^j.

The established two-row degree-forest recurrences imply

V_n(x) = (1+x+x^2)V_{n-1}(x) + (x^2-x^3)V_{n-2}(x).

Here $V_1(x)=2x^2$ , and the convenient formal initial value $V_0(x)=2$ reproduces the directly computed $V_2(x)=4x^2+2x^4$ . Therefore

F(x,z) = \sum_{n\ge0}V_n(x)z^n = \frac{2(1-(1+x)z)} {1-(1+x+x^2)z-x^2(1-x)z^2}.

For $d\ge5$ ,

[x^d]F(x,z) = \sum_{n\ge2}E_d(2,n)z^n,

because the formal terms at $n=0,1$ have $x$ -degree at most two.

Set $w=1-z$ and

A=\frac zw=w^{-1}-1,\qquad B=\frac{z(1+z)}w=2w^{-1}-3+w,\qquad C=\frac{z^2}w=w^{-1}-2+w.

Writing

H(x) = \frac1{1-Ax-Bx^2+Cx^3} = \sum_{r\ge0}h_rx^r,

we have

[x^d]F = 2(Bh_{d-2}-Ch_{d-3}).

The top two pole orders of $h_r$ come only from the all- $A$ term and the $r-1$ terms containing one $B$ . Thus, for $r\ge1$ ,

h_r = A^r+(r-1)BA^{r-2} + O(w^{-r+2}) = w^{-r}+(r-2)w^{-r+1}+O(w^{-r+2}).

Substitution yields

\boxed{ \sum_{n\ge2}E_d(2,n)z^n = \frac4{(1-z)^{d-1}} + \frac{4(d-6)}{(1-z)^{d-2}} + O((1-z)^{-d+3}). }

Equality with the conjectured value would require

4(d-6)=-4(d-4),

which holds only at $d=5$ . Hence every $d\ge6$ provides a counterexample.

The smallest case can also be checked using the primary source's own Table 2:

E_6(2,n) = \frac{n^4+10n^3-145n^2+518n-576}{6}

for sufficiently large $n$ . Therefore

c_2(2) = 3!\,[n^3]E_6(2,n) - \frac{5\cdot4}{2} = 0,

while the conjecture predicts $-8$ . The source tested only $m\ge d-1$ despite explicitly claiming the formula for every $m\ge2$ .

Sources: C. Gupta, “One construction for the Miura-ori flip-graph degree sequence,” arXiv:2607.05567v2, Conjecture 9.4 and Table 2; Christensen et al., “The Origami flip graph of the $2\times n$ Miura-ori,” arXiv:2506.19700v2, Proposition 4.11.

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Shivam Patel · 20 hours ago