:\  F'\subset F \text{ for some }F\in {\mathcal F}\big\}.$$ If $|{\mathcal F}| = {a_k\choose k}+\ldots +{a_s\choose s}$, then $$|\partial({\mathcal F})|\ge {a_k\choose k-1}+{a_{k-1}\choose k-1}+\ldots +{a_s\choose s-1}.$$

Next, we start preparations to state the main result of this section. Given a number $\gamma<{n-1\choose k-1}$, let us write it in the $(n-k-1)$-cascade form: $$\gamma = {n-b_1\choose n-k-1}+{n-b_{2}\choose n-k-2}+\ldots+{n-b_{s_b}\choose n-k-s_b},$$ where $1< b_1<b_2<\ldots<b_{s_b}.$[^5] Define $T_{\gamma}:=\{b_1,\ldots,b_{s_b}\}$ and put $S_{\gamma}:=T_{\gamma}\,\oplus\, [2,b_{s_b}-1],$ where $\oplus$ stands for symmetric difference. Note that $S_{\gamma}\cup T_{\gamma} = [2,b_{s_b}]$ and $S_{\gamma}\cap T_{\gamma} = \{b_{s_b}\}$. Suppose that $S_{\gamma} = \{a_1,\ldots, a_{s_a}\}$, where $1< a_1<\ldots<a_{s_a} := b_{s_b}$.

\[defresnu\] We call a nonnegative integer $\gamma$ , if either $\gamma={n-4\choose k-3}$ or the following holds:

1.  $s_a := |S_{\gamma}|\le k$ and $s_b := |T_{\gamma}|\le k-1$;

2.  $b_i>2i+2$ for each $i\in [s_b]$.

In particular, any integer $\gamma>{n-4\choose k-3}$ has ${n-4\choose k-3}$ as one of the summands in the $(n-k-1)$-cascade form, and thus is not resistant.

Let $0= \gamma_0<\gamma_1<\ldots <\gamma_{m} = {n-4\choose k-3}$ be all the resistant numbers in increasing order.

\[thmfull1\] Let $n>2k\ge6$. Consider an intersecting family ${\mathcal F}\subset {[n]\choose k}$. Suppose that $\gamma_{l-1}<\gamma ({\mathcal F})\le\gamma_l$ for $l\in [m]$ and that the representation of $\gamma_l$ in the $(n-k-1)$-cascade form is $$\gamma_l = {n-b_1\choose n-k-1}+{n-b_{2}\choose n-k-2}+\ldots +{n-b_{s_b}\choose n-k-s_b},$$ 