The complementary Bell numbers $B^{\pm}(n)$ are defined by the alternating sum of the Stirling numbers of the second kind, $S(n,k)$: $$B^{\pm}(n)=\sum_{k=0}^n(-1)^kS(n,k),$$ and they count the difference between the set partitions of $\{1,2,\ldots,n\}$ with an even number of blocks and those with an odd number of blocks. So $B^{\pm}(2)=0$. A question attributed to H.Wilf asks if there is any $n\gt 2$ for which $B^{\pm}(n)=0$. This problem has now been almost solved, but not quite: there may be a single positive integer $n_0$ (that would be larger than $2944838$) for which $B^{\pm}(n_0)=0$. See these posts on MO:

Wilf's conjecture: complementary Bell numbers

Congruence for complementary Bell numbers

as well as the article "Wilf's conjecture", with the references there, in the June-July 2016 issue of Amer. Math. Monthly).

There is a crucial congruence that is at the heart of the method I used to study this problem, and that I will describe below. Denote by $\nu_2(n)$ the 2-adic valuation of $n$. Let $m$ be a positive integer. Let $y_m \in \{1,2,\ldots,2^m-1\}$ be such that $\nu_2(B^{\pm}(24y_m+14))=\max\{\nu_2(B^{\pm}(24i+14)): 1\leq i\leq 2^m-1\}$. In fact there is only one such element of $\{1,2,\ldots,2^m-1\}$ where the maximum is achieved. Then we have the congruence $$B^{\pm}(24\cdot 2^m+24y_m+14)\equiv B^{\pm}(24y_m+14)+2^{m+5}\pmod{2^{m+6}}.$$ In order to rule out the existence of $n_0$ above, it is enough to prove that $y_m$ is not eventually constant.

In fact, this is almost a restatement of the original problem, restricted to the subsequence $24n+14$. The above congruence does not give much information, but was enough to prove that there is at most one integer $n_0$ for which $B^{\pm}(n_0)=0$.

So I would like to know if the congruence above can be refined somehow to provide more information. For example, if it can be extended to some congruence modulo $2^n$ for $n$ larger than $m+6$.