${\mathcal F}$ with $\gamma({\mathcal F})>\gamma_l$ has strictly smaller size (quantified in Corollary \[corhm\]). Moreover, even if $\gamma_{l-1}<\gamma({\mathcal F})\le \gamma_l$, we are still getting the same upper bound.

Returning to the proof of the observation, assume now that $u$ is a real number. The function in the right hand side of is monotone decreasing as $\gamma({\mathcal F})$ increases (or, equivalently, as $u$ decreases). Therefore, to show that the bound is stronger than , it is sufficient to verify it for all values of $\gamma_l$, $l=0,\ldots, m$. But for each of these values the bound is sharp, so can be only weaker than .

Here, we prove Corollary \[corhm\] for (integer) $u\ge 4$. Choose $l$ such that $\gamma_l={n-u-1\choose n-k-1}$. Assume first that $k\ge 5$. Then $\gamma_{l+1}={n-u-1\choose n-k-1}+{n-k-2\choose n-k-2}={n-u-1\choose n-k-1}+1=:\gamma$. To see this, we just have to check that $\gamma$ is a resistant number. In terms of Definition \[defresnu\], we have $T_{\gamma}=\{u+1,k+2\}$ and $S_{\gamma}=\{2,3,\ldots, u,u+2\ldots, k+2\}.$ We have $s_b=2\le k-1$ and $s_a=k$, thus, condition (1) of Definition \[defresnu\] is fulfilled. Moreover, $b_1=u+1>2\cdot 1+2$ and $b_2=k+2>2\cdot 2+2$, fulfilling condition (2) of the definition. Thus, implies that

$$\begin{aligned}
|{\mathcal F}|\le& {n-2\choose n-k}+\ldots+{n-u\choose n-k-u+2}+{n-u-2\choose n-k-u+1}+\ldots+{n-k-2\choose n-2k+1}+{n-u-1\choose n-k-1}+1\\
=&{n-1\choose n-k}-{n-u-1\choose n-u-k}+{n-u-1\choose n-k-1}+1-\epsilon,\end{aligned}$$

where

$$\epsilon={n-u-1\choose n-k-u+1}-\Big({n-u-2\choose n-k-u+1}+\ldots+{n-k-2\choose n-2k+1}\Big)={n-k-2\choose 