That turned dark quickly. Instead  of a single $5$-gram, we now need to consider all occurrences of $n\big(\lfloor b_1/2\rfloor, \ldots , \lfloor b_{k-1}/2\rfloor, b_{k-1}\big)$ and $n\big(\lfloor b_1/2\rfloor, \ldots , \lfloor b_{k-1}/2\rfloor, b_{k}\big)$ to produce an additional $O(n)$-term. In other words, there are now $k-1$ ‘classes’ of $5$-grams that require additional consideration. We might now assume that there are $m_0$ $5$-grams of the first type, $m_1$ of the second, and so forth. Thus, the entire set of $5$-grams is now encoded by $$\label{eq:first_enc} \sum_{\ell = 0}^{k-1} m_\ell n\big(\lfloor b_1/2\rfloor, \ldots , \lfloor b_{\ell-1}/2\rfloor, b_{\ell-1}\big).$$ We must then find the frequencies of each combination, say $$n\big(b_{i_1}/2, \ldots , b_{i_r}/2\big).$$ However, as the number of bits in each $b_i$ grows linearly with $i$, it is actually simpler to find the $5$-gram $n\big(b_i, b_j, \ldots, b_k\big)$ with $i$ and $j$ the smallest indices satisfying $b_i\neq b_j$ and then use the fact that $n\big(b_i, b_j, \ldots, b_k\big) = n\big(b_i, b_j, \ldots, b_k\big) n\big(b_j, b_i, \ldots, b_k\big)$ and $n\big(b_i, b_j, \ldots, b_k\big) = n\big(b_i, b_j, \ldots, b_k\big) + n\big(b_j, b_i, \ldots, b_k\big)$. As a result, we would get for each $5$-gram $$\begin{aligned} n\big(b_{i_1}, b_{i_2}, \ldots, b_{i_r}\big) &=& m_0 n\big(b_{i_1}/2, \ldots , b_{i_r}/2\big)\nonumber\\ &+& m_1 n\big(b_{i_1}/2, \ldots , b_{i_r}/2\big) + m_2 n\big(b_{i_1}/2, \ldots , b_{i_r}/2\big)\nonumber\\ &+& \cdots \nonumber\\ &+& m_{k-1} n\big(b_{i_1}/2, \ldots , b_{i_r}/2\big). \label{eq:second_enc}\end{aligned}$$ Thus, the entire problem is still encoded as a sum with polynomial complexity; the only difference is that each term now has a polynomial number of terms as well. Of course, the number of $5$-grams grows rapidly with $n$, and even more rapidly with $k$ and $m_\ell$. For example, if $b_i$ takes $1+i$ binary digits, the number of $5$-grams becomes $k \choose i$ which grows factorially. The same would be the case if we were to simply consider all possible combinations of $b_i/2$, or, in other words, the $k$-ary representation of the numbers $b_i/2$ are written in base $k$ (here $k$ is a power of $2$). Thus, a more reasonable measure of the size of the first and second encodings would be $$\begin{aligned} n\big(b_i/2\big) &=& n\big(b_i/2, b_j/2, \ldots, b_k/2\big)\nonumber\\ &+& n\big(b_j/2, b_i/2, \ldots, b_k/2\big)\nonumber\\ &+& \cdots \nonumber\\ &+& n\big(b_k/2, b_i/2, \ldots, b_j/2\big)\nonumber\\ &=& m_0 n\big(b_i/2, b_j/2, \ldots, b_k/2\big) + \cdots \nonumber\\ &+& m_{k-1} n\big(b_i/2, b_j/2, \ldots, b_k/2\big). \label{eq:third_enc}\end{aligned}$$ If $k$ is a power of $2$ and we are dealing with powers of $2$ to base $k$, as is necessary to represent integers (let us say $i_1 = \cdots = i_r = i$ such that $k^i = n$), then is the number of $5$-grams as given in the third line. We do not know a general solution to the sum of such numbers. Let us consider a simple example: $b_1=5$ and $b_2=b_3 = 3$. In , for $k=5$, this is $$\begin{aligned} n\big(b_1/2, b_2/2, b_3/2, b_4/2, b_5/2\big) &=& m_0 n\big(b_1/2, b_2/2, b_3/2, b_4/2, b_5/2\big)\nonumber\\ &+& m_1 n\big(b_1/2, b_2/2, b_3/2, b_4/2, b_5/2\big) \nonumber\\ &+& m_2 n\big(b_1/2, b_2/2, b_3/2, b_4/2, b_5/2\big)\nonumber\\ &+& m_3 n\big(b_1/2, b_2/2, b_3/2, b_4/2, b_5/2\big) \nonumber\\ &+& m_4 n\big(b_1/2, b_2/2, b_3/2, b_4/2, b_5/2\big)\nonumber\\ &+& m_5 n\big(b_1/2, b_2/2, b_3/2, b_4/2, b_5/2\big).\end{aligned}$$ If the first of the sums is $$\begin{aligned} m_0 n\big(5/2, 3/2, 5/2, 5/2, 3/2\big) &=& 10 \cdot 3^4\nonumber\\ &=& 3600,\end{aligned}$$ and the second is $$\begin{aligned} m_1 n\big(3/2, 5/2, 3/2, 5/2, 3/2\big) &=& m_2 n\big(3/2, 5/2, 3/2, 5/2, 3/2\big)\nonumber\\ &=& m_3 n\big(3/2, 5/2, 3/2, 5/2, 3/2\big)\nonumber\\ &=& m_4 n\big(3/2, 5/2, 3/2, 5/2, 3/2\big)\nonumber\\ &=& m_5 n\big(3/2, 5/2, 3/2, 5/2, 3/2\big),\end{aligned}$$ then, using $n=15$, we would obtain for $n(15,3)$ $$\begin{aligned} n(15,3) &=& 3600\nonumber\\ &+& 10 \cdot 3^4\nonumber\\ &+& m_2 \left(3600 + 10 \cdot 3^4\right)\nonumber\\ &+& m_3 \left(3600 + 10 \cdot 3^4\right)\nonumber\\ &+& m_4 \left(3600 + 10 \cdot 3^4\right)\nonumber\\ &+& m_5 \left(3600 + 10 \cdot 3^4\right)\nonumber\\ &+& m_2^2 \left(3600 + 10 \cdot 3^4\right)\nonumber\\ &+& m_3^2 \left(3600 + 10 \cdot 3^4\right)\nonumber\\ &+& m_4^2 \left(3600 + 10 \cdot 3^4\right)\nonumber\\ &+