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Conjectural formula for the next-to-leading coefficient of path distance matrices

About 12 years old · traced to

Let PnP_n be a path on n>5n>5 vertices. Define its characteristic polynomial by

CharPolyDPn(x)=∑i=02n−3aixi.\mathrm{CharPoly}_{\mathfrak{D}_{P_n}}(x)=\sum_{i=0}^{2n-3}a_i x^i.

Coefficient formula conjecture. The coefficient of x2n−5x^{2n-5} is

a2n−5=−16(n−1)(n−2)(2n2+6n−15).a_{2n-5}=-\frac{1}{6}(n-1)(n-2)(2n^2+6n-15).

The formula is proposed from computational data and is intended to estimate the coefficient needed for a lower-bound analysis of the peak location. No proof or resolution is given in the source.

References

Primary source

Rakesh Jana, Iswar Mahato and Sivaramakrishnan Sivasubramanian, “Unimodality and peak location of the characteristic polynomials of two distance matrices of trees”, arXiv:2407.03309 (2024).

Additional references

6 papers in this index state this conjecture (2014–2024). The statement above is taken from the most recent of them; the others are arXiv:2312.10910, arXiv:2210.01059, arXiv:1703.06550, arXiv:1703.00271, arXiv:1403.1235.

characteristic-polynomialsdistance-matricestrees
Posted

Progress summary

Refreshed
Claimed progress

A newly submitted proof claims to establish the formula, but no independent source has verified it.

In July 2024, a paper recorded this coefficient formula as Conjecture 30, based on computations for 5<n<155<n<15 and intended for a lower-bound analysis of the coefficient peak. The paper gave no proof or resolution.

Community submission (unverified), September 5, 2026

A submitted proof argues that a restricted distance matrix has trace 3n−53n-5 and computes its squared-entry sum, then applies Newton's identity to obtain a2n−5=−16(n−1)(n−2)(2n2+6n−15)a_{2n-5}=-\frac{1}{6}(n-1)(n-2)(2n^2+6n-15). The argument therefore claims to prove the conjecture, but it has no independent verification in the retrieved material.

Current status (as of September 2026): The formula remains unverified; the only purported proof is the September 5, 2026 community submission.

Sources

Solutions 1

ProofAn AI-assisted proof of Conjecture 30: computing the trace and squared trace of the restricted 2-Steiner path matrix and applying Newton’s identity yields the conjectured coefficient formula.See full solutionHide full solution

Proof. For n>5n>5, use the matrix D=D2(Pn)[B,B]D=\mathfrak D_2(P_n)[B,B] from Remark 23 of the source paper, where

B={{i,i+1}:1≤i<n}∪{{i,i+2}:1≤i<n−1},DA,C=max⁡(A∪C)−min⁡(A∪C).B=\{\{i,i+1\}:1\le i<n\}\cup\{\{i,i+2\}:1\le i<n-1\}, \qquad D_{A,C}=\max(A\cup C)-\min(A\cup C).

It has order 2n−32n-3 and trace 3n−53n-5. Let E,F,HE,F,H be the squared-entry sums of its adjacent-pair, distance-two-pair, and one mixed block. Their entries are ∣i−j∣+1|i-j|+1, ∣i−j∣+2|i-j|+2, and max⁡(i+1,j+2)−min⁡(i,j)\max(i+1,j+2)-\min(i,j), respectively. Counting by separation (the two mixed-block halves are equal) gives

E=(n−1)+2∑h=1n−2(n−1−h)(h+1)2=(n−1)(n3+n2−6)6,F=4(n−2)+2∑h=1n−3(n−2−h)(h+2)2=(n−2)(n3+2n2+3n−30)6,H=2∑h=0n−3(n−2−h)(h+2)2=(n−1)(n−2)(n2+3n+6)6.\begin{aligned} E&=(n-1)+2\sum_{h=1}^{n-2}(n-1-h)(h+1)^2 =\frac{(n-1)(n^3+n^2-6)}6,\\ F&=4(n-2)+2\sum_{h=1}^{n-3}(n-2-h)(h+2)^2 =\frac{(n-2)(n^3+2n^2+3n-30)}6,\\ H&=2\sum_{h=0}^{n-3}(n-2-h)(h+2)^2 =\frac{(n-1)(n-2)(n^2+3n+6)}6. \end{aligned}

Since DD is symmetric,

tr⁡(D2)=E+F+2H=2n4−2n2−33n+453.\operatorname{tr}(D^2)=E+F+2H=\frac{2n^4-2n^2-33n+45}{3}.

Newton's identity now yields

a2n−5=(tr⁡D)2−tr⁡(D2)2=−(n−1)(n−2)(2n2+6n−15)6,a_{2n-5} =\frac{(\operatorname{tr}D)^2-\operatorname{tr}(D^2)}2 =-\frac{(n-1)(n-2)(2n^2+6n-15)}6,

proving Conjecture 30. □\square

Developed with OpenAI Codex and reviewed by Claude Opus 5; the trace identities were also checked by exact symbolic summation.

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Augustinas Malinauskas Posted