The reduced -simplex recurrence conjecture
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.
Progress summary
No publicly verified discussion or progress on this recurrence conjecture was found.
No public discussion or published progress specific to the reduced -simplex recurrence conjecture was found.
Current status (as of August 2026): The conjecture appears open, with no recorded proof, counterexample, or verified progress.
Sources & referencesView supporting material
Primary source
Matthew Crawford, Pavan Kartik and Reese Lance, “Integrals of stable envelopes for cotangent bundles to Grassmannians”, arXiv:2510.21573 (2026).
Solutions 1
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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.