TITLE: Arithmetic problem for bicolored graphs
QUESTION [13 upvotes]: A bicolored graph is a graph each vertex of which has been assigned one of two colors such that each edge connects vertices of different colors. A bipartite graph is a graph $G$ which admits such a coloring. Given $j$ white and $i$ black vertices, there are $2^{ji}$ ways to join vertices of different colors. Thus the number of (labeled) bicolored graphs on n vertices is
$$b_n=\sum_{i+j=n}\binom{n}i2^{ij}.$$

Question. Is it true that $b_n$ is never divisible by $5$? If so, any proof?

Update. One combinatorial proof and two number-theoretic/algebraic proofs are supplied.

REPLY [12 votes]: Here is an illustration of Ira Gessel's comment about a number-theoretic proof.
Lemma. For $n\geq1$ and an odd prime $p$, the sequence $b_n$ mod $p$ is periodic of period $p-1$.
Proof. Let $n=m(p-1)+r$, where $1\leq r\leq p-1$ and $m\geq0$. Proceed with a successive application of Fermat's, Lucas' and Chu-Vandermonde's theorems so that
\begin{align*}
b_n&=\sum_{j=1}^{p-1}\sum_{k=0}^m\binom{m(p-1)+r}{k(p-1)+j}2^{[k(p-1)+j][m(p-1)+r-k(p-1)-j]} \\
&\equiv_p\sum_{j=1}^{p-1}\sum_{k=0}^m\binom{m(p-1)+r}{k(p-1)+j}2^{j(r-j)} \\
&\equiv_p\sum_{j=1}^{p-1}2^{j(r-j)}\sum_{k\geq0}\binom{m}k\binom{r-m}{j-k}\\
&\equiv_p\sum_{j=1}^{p-1}2^{j(r-j)}\binom{r}j\\
&=\sum_{j=1}^r\binom{r}j2^{j(r-j)}=b_r.
\end{align*}
The proof is complete. $\square$
Example. The first 4 values of $b_n$ mod 5 are $[2,2,1,1]$, as Wojowu mentioned. By the Lemma, we conclude that $\nu_5(b_n)=0$, that is, $b_n$ is never divisible by $5$.