Equal third and unequal second principal specializations for a caterpillar family

Aram Bingham, Kyle Celano, Patricia Commins, Rosa Orellana, Jianping Pan, and Mary Claire Simone, Principal specializations of the t-chromatic symmetric function, arXiv:2610.31785v1, Remark 3.11, first question:

Do (5,j,j,4,3) and (4,j,j,5,3) exhibit this phenomenon for all j?

The single target is the conjunction, for every positive integer j, that the actual caterpillars of Definition 2.1 associated with these two compositions have unequal ordinary second principal specializations and equal ordinary third principal specializations. Equality and inequality are those of formal integer polynomials in X. Equation (2.1) uses colors 0 through k−1 and sums over all proper colorings; every color need not occur. The coefficient inclusion into the source's complex polynomial ring is injective. The t refinement is outside this question.

The source reports verification through j = 7. The family theorem replaces that finite range by every positive j, including j = 1, for the source graphs themselves. The two specialization comparisons form one question and one resolution.

The source's finite verification does not establish an unbounded family. The missing uniform step is supplied by an identity in arbitrary repeated middle weights, after expressing the actual coloring sum as a sum over spine colors. The separate uniqueness question in Remark 3.11 is not part of this target.

Restrict a proper coloring to the spine and its leaves. This gives a bijection with a proper spine coloring and independent leaf color choices different from the incident spine color. For a positive part m and spine color c, write

$$B_m(c)=X^c\left(\sum_{\substack{0\le d<k\d\ne c}}X^d\right)^{m-1}.$$

The coloring polynomial is the sum, over consecutive unequal spine colors, of the product of these block weights. A part equal to one has no leaves and contributes the zeroth power one.

For k = 3, let F(A,B,C,D,E) be the sum of A(c₀)B(c₁)C(c₂)D(c₃)E(c₄) over c₀,…,c₄ in {0,1,2} with consecutive colors unequal. The identity diagonal_identity states

$$F(B_5,w,w,B_4,B_3)=F(B_4,w,w,B_5,B_3)$$

for every function w from {0,1,2} to Z[X]. Substituting w = B_j proves the third-specialization equality without enumerating j. For k = 2, the two alternating spine colorings give exactly

$$\operatorname{ps}_2(5,j,j,4,3)=X^{j+5}+X^{j+7},\qquad \operatorname{ps}_2(4,j,j,5,3)=X^{j+6}+X^{j+6}.$$

Their coefficients at degree j+5 are respectively one and zero.

A positive j for which the ordinary third coloring polynomials differ, or for which the second coloring polynomials agree, would contradict the target. Numerical agreement at a chosen X does not establish formal polynomial equality.

D5/S3/Combinatorics/Graph/PrincipalSpecialization/BinghamCaterpillarPs3Family.result establishes the displayed conjunction for every positive j. The dependent vertex carrier, symmetrized edge relation and proper-coloring sum represent the source graphs and polynomials. The coloring factorization includes empty leaf sets. The repeated-weight identity keeps the outside blocks 5 and 4 and the final block 3 fixed; it does not assert arbitrary endpoint replacement. The two-color formulas separate the coefficients at degree j+5.

theorem; this dossier concerns the first question of Remark 3.11.

The separate BinghamCaterpillarUniquenessRefutation.result refutes the exhaustive-pairs assertion using an eighteen-vertex witness. The family theorem supplies no classification. The source's four-color questions and analogous t-refined statements remain unresolved here.

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