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

Less than 1 year old · traced to 2026

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.

References

Primary source

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

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A reader-posted calculation claims the conjecture is false from six dimensions onward, but that counterexample has not been independently checked.

Chakshu Gupta’s 2026 preprint formulates the conjecture that the second Laurent coefficient is $c_2(m)=-4(d-4)$ for every $d\ge5$ and $m\ge2$ , with leading coefficient $c_1(m)=4$ . It supplies transfer-matrix computations but no general proof.

Known results

The relevant generating functions are rational in the transfer variable (Gupta, 2026).
Closed-form degree polynomials are computed through $d=10$ , with the underlying degree bound verified through $d=7$ (Gupta, 2026).
Transfer-matrix computations reportedly agree with the conjectured coefficient through $d=11$ in the tested regime (Gupta, 2026).

Posted attempt

A posted calculation claims an exact two-row recurrence gives $c_2(2)=4(d-6)$ , contradicting the conjecture for every $d\ge6$ ; it also claims $c_2(2)=0$ at $d=6$ . The claimed counterexample is not independently verified.

Current status (as of August 2026): The preprint’s computational evidence and general proof remain incomplete, while the posted two-row counterexample claim would refute the conjecture for $d\ge6$ if verified.

Sources

arxiv.org

CounterexampleThis solution needs a summarySee full solutionHide full solution

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.