The Pascal descent polynomial twin and anti-twin conjecture

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Let [n]={1,…,n}[n]=\{1,\ldots,n\}, and let I,J⊆[n]I,J\subseteq[n] have the same cardinality s≤⌊n/2⌋s\leq\lfloor n/2\rfloor. Let pn(I)p_n(I) and pn(J)p_n(J) be the Pascal descent polynomials associated with these subsets, and let τn\tau_n send each position ii to n+1−in+1-i. Assume pn(I)eq0p_n(I) eq0 and pn(J)eq0p_n(J) eq0.

Pascal descent polynomial conjecture. One has

pn(I)=pn(J)p_n(I)=p_n(J)

if and only if I=JI=J, or nn is odd and I=τn(J)I=\tau_n(J); and

pn(I)=−pn(J)p_n(I)=-p_n(J)

if and only if nn is even and I=τn(J)I=\tau_n(J).

This is the Pascal-descent-polynomial reduction of the tabloid conjecture and would classify the relevant polynomial coincidences up to reversal and sign. The source gives no resolution status.

References

Primary source

Ioannis Michos, “On the support of the free Lie algebra: the Schützenberger problems”, arXiv:0807.3519 (2008).

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