The Pascal descent polynomial twin and anti-twin conjecture

Let [n]={1,,n}[n]=\{1,\ldots,n\}, and let I,J[n]I,J\subseteq[n] have the same cardinality sn/2s\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+1in+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.

Sources & referencesView supporting material

Primary source

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

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