The reduced -simplex recurrence conjecture
Let denote the integer in the reduced -simplex in layer , row , and NE-SW diagonal , and set when there is no entry in that position. Reduced -simplex recurrence conjecture.
The recurrence is asserted to hold without the exceptions needed for the unreduced -simplex, providing a uniform combinatorial rule for the reduced array.
References
Primary source
Matthew Crawford, Pavan Kartik and Reese Lance, “Integrals of stable envelopes for cotangent bundles to Grassmannians”, arXiv:2510.21573 (2026).
Progress summary
A reader-written argument claims a complete proof of the conjecture, but no independent verification has been found.
The conjecture asserts that every entry of the reduced -simplex$ satisfies the four-neighbor recurrence without boundary exceptions. The associated paper studies the relevant Grassmannian combinatorics and formulates conjectures about these integers.
Posted attempt
A reader-written argument claims a complete proof: it gives an explicit factorial formula, identifies the reduced layers with coefficients of a polynomial family, and derives the recurrence from a weighted-walk decomposition. It also claims positivity, integrality, and a companion divisibility result. The argument has not been independently verified.
Current status (as of August 2026): A complete proof has been claimed in an unverified discussion, while no published or independently checked proof, counterexample, or corroborating exposition is recorded.
Sources
Solutions 1
ProofThis solution needs a summarySee full solution
The four-neighbor recurrence holds for every reduced Gr₂-simplex entry, with no boundary exceptions. In fact, its entries admit the following explicit positive integral formula:
ξ_{ℓ,r,j} =(ℓ−1)!ℓ!(ℓ−r+1)/ [(r−j)!(j−1)!(ℓ−j+1)!(ℓ−r+j)!],
for 1≤j≤r≤ℓ, and ξ_{ℓ,r,j}=0 outside this range. Here ℓ=1 denotes the reduced singleton layer, as dictated by the actual reduced generating function.
For nonnegative A,B,C set d=A+B+C and
q(A,B,C)=d!(d+1)!(B+1)/ [A!C!(A+B+1)!(B+C+1)!].
Let q vanish if any argument is negative and define
Q_d(a,b,c)=∑_{A+B+C=d}q(A,B,C)a^A b^B c^C.
First, the source's explicit stable-envelope formula in Example 3.2 gives, with A=i−1,B=j−i−1,C=n−j,
F_n(i,j)=q(A−1,B,C)+2q(A,B−1,C)+q(A,B,C−1).
Indeed its displayed factorial bracket expands exactly as
(B+1)A(A+B+1) +2B(A+B+1)(B+C+1) +(B+1)C(B+C+1),
which proves the coefficient identity including all boundary cases. Consequently the original layer polynomial factors as
H_n(a,b,c)=(a+2b+c)Q_{n−3}(a,b,c).
Thus Q_d is precisely the reduced layer.
For a direct recurrence proof, interpret Q_d as the weight enumerator of nonnegative length-d walks starting at height zero with steps
U: height +1, weight b; D: height −1, weight ac/b; L_a: height 0, weight a; L_c: height 0, weight c.
The exponent B of b is the terminal height. For fixed A,B,C and s down-steps, the ballot principle gives
d!(B+1)/[(A−s)!(C−s)!s!(B+s+1)!]
walks. Summing over 0≤s≤min(A,C) and applying Chu–Vandermonde yields q(A,B,C), proving that these are exactly the reduced coefficients and are positive integers.
Appending the last step now gives the polynomial recurrence
Q_{d+1} =(a+b+c)Q_d+(ac/b)[Q_d−Q_d(a,0,c)].
The subtraction excludes precisely the height-zero walks for which a down-step is impossible. Identify the reduced position (ℓ,r,j) with the monomial
a^{r−j}b^{ℓ−r}c^{j−1}
in Q_{ℓ−1}. Extracting its coefficient from the four terms of the polynomial recurrence proves
ξ_{ℓ,r,j} =ξ_{ℓ−1,r,j} +ξ_{ℓ−1,r−1,j} +ξ_{ℓ−1,r−1,j−1} +ξ_{ℓ−1,r−2,j−1}
for every ℓ,r,j, using zero outside the triangular array. This is exactly the conjectured exception-free recurrence and simultaneously establishes the closed coefficient formula, integrality, positivity, and the companion divisibility conjecture.