Assuming that $i>0$, let $p(i)$ denote an $i$-th prime, i.e. $p(1)=2, p(2)=3, p(3) = 5$ etc.
Let $$f(x, y)=\operatorname{floor}(x/2) \bmod 2^y,$$ i.e. $f(123, 4)=13, f(1234567, 8)=67, f(9876543210, 12)=2933$ etc.
Assuming that $a$ and $b$ are greater than $0$, a tuple $T(a,b)$ is defined as follows:
$$T(a,b)=(f(p(1), b), f(p(2), b), \ldots, f(p(a-1), b), f(p(a), b)).$$
Let $g(n, t)$ denote the number of occurrences of a number $n$ in a tuple $t$, i.e. $g(0, (0,1,2,0,3)) = 2, g(1, (0,1,2,0,3)) = 1$ etc.
Let $m(a,b)$ denote the smallest number in the tuple
$$(g(0, T(a,b)), g(1, T(a,b)), \ldots, g(2^b-2, T(a,b)), g(2^b-1, T(a,b))).$$
Similarly, let $M(a,b)$ denote the largest number in the tuple
$$(g(0, T(a,b)), g(1, T(a,b)), \ldots, g(2^b-2, T(a,b)), g(2^b-1, T(a,b))).$$
Then let $h(b)$ denote the smallest value of $a$ such that $m(a,b) = 1$.
For example, I took the first two million primes (from $2$ to $32452843$) and found that $$\begin{array}{l} m(658297,16) = 1,\\ M(658297,16) = 20,\\ m(1000000,16) = 4,\\ M(1000000,16) = 30,\\ m(2000000,16) = 12,\\ M(2000000,16) = 48. \end{array}$$.
Here $h(16) = 658297$ because $p(658297)=9897469$ is the smallest prime $p$ such that $\operatorname{floor}(p/2) \bmod 2^{16} = 33534$. But if we will test all primes less than $9897469$, we will find exactly $20$ values of $i$ such that $\text{floor}(p(i)/2) \bmod 2^{16} = 46175$.
We can see that $$\frac{{48}}{{12}} < \frac{{30}}{4} < \frac{{20}}{1},$$ but how close to $1$ we can get as we measure the corresponding ratio $\frac{{M(a,b)}}{{m(a,b)}}$ (incrementing $a$, but keeping $b$ constant)?
Question: given an arbitrary natural number $b>1$, does $$\lim_{k \to \infty} \frac{{M(h(b) + k,b)}}{{m(h(b) + k,b)}}$$
converge to some minimum? If no, why? If yes, how close to $1$ is this minimum? Can it be approximated by a formula?
