Suppose a polynomial of the form
$$\prod\_i^d \sum\_j^p x\_i^{f\_j}$$
clearly symmetric, where $f\_j\in \mathbb{N}$. There is a way to find the set of $f$ numbers such that this polynomial is Schur positive? I have been trying to build a Kashiwara crystal and impose conditions on the weight function but without success beyond the fact that $\lbrace f\_0,...,f\_p\rbrace$ must be a sequence of consecutive numbers so $f\_{i+1}= f\_{i}+1 \text{ or } f\_{i+1}= f\_{i}, i<p$.



Schur positivity of a polynomial
Given $f\_1,\dots,f\_p$ and $d\geq \max f\_i$, a necessary and sufficient condition is that all zeros of the polynomial $\sum x^{f\_j}$ are real. See *Enumerative Combinatorics*, vol. 2, Exercise 7.91. **Note.** Your necessary condition need not hold for small $d$. If $d=1$, then $\sum x\_1^{f\_j}= \sum s\_{f\_j}(x\_1)$.


