# Complementary Bell numbers $B^{\pm}(24n+14)$

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 $$n_0 \gt 2$$ 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 $$m+6$$.