$x_1^{c_1}\cdots x_n^{c_n}\in G(J)$ with $c_i\leq a_i$ for $i=1, \ldots, n$.

Now let $\alpha=(1,1,\ldots,1)$. Then $(I^k)_{\leq \alpha}=I^{[k]}$. Therefore it follows from [@HHZ1 Lemma 4.4] that $\beta_{i,j}(I^{[k]})\leq \beta_{i,j}(I^k)$ for all $k$. This together with Theorem \[main\] implies:

1.  $\index(I^{[k]})\geq \index(I^k)$ for all $k$;

2.  if $G$ is gap free and $I=I(G)$, then $\index(I^{[k]})>1$ for all $k$.

Here we use the convention that the index of the zero ideal is infinity.

The inequality (i) may be strict or not. Indeed, if $I$ is the monomial ideal given by Nevo and Peeva in [@NP Counterexample 1.10], then it can be shown by computer that $\index(I^k)=\index(I^{[k]})$ for $k=1,\ldots,4$. On the other hand, if $G$ is a $9$-cycle, then $\index(I)=1$, $\index(I^{[2]})=1$, $\index(I^{[3]})=2$ and $\index(I^{[k]})=\infty$ for $k> 3$, while by Theorem \[main\], $\index(I^k)=1$ for all $k$.

The converse of (ii) is not true, that is, $G$ may not be gap free but $\index(I^{[k]})>1$ for some $k$. Of course, $\index(I^{[k]})>1$ for $k>n/2$, since for such powers $I^{[k]}=0$. But even if $I^{[k]}\neq 0$ and $G$ is not gap free we may have $\index(I^{[k]})>1$. For example, if $G$ is the graph with vertex set $[4]$ and edges $\{1,2\},\{3,4\}$. Then $G$ is not gap free, but $\index(I(G)^{[2]})=\infty$, because in this case 