Primes written in bases $2,...,10$ and sequence that counts number of primes in those representations (a question about values of that sequence)

Question

If we choose some prime number $p$ written in base $10$ then we can write that number in all the bases $2,...,10$ and investigate primeness of that representations when viewed in base $10$.

As an example I will choose prime number $11$.

In base $2$ number $11$ is $1011$.

In base $3$ number $11$ is $102$.

In base $4$ number $11$ is $23$.

In base $5$ number $11$ is $21$.

In base $6$ number $11$ is $15$.

In base $7$ number $11$ is $14$.

In base $8$ number $11$ is $13$.

In base $9$ number $11$ is $12$.

In base $10$ number $11$ is $11$.

Of the $9$ numbers $1011,102,23,21,15,14,13,12,11$ there are $3$ primes.

The sequence that counts how many times is some prime number $p$ prime in bases $2,3,...,10$ can be defined as $$s(p)=\dfrac{\text{nop}(p;2,3,4,5,6,7,8,9,10)}{9}$$ where $\text{nop}(p;2,3,4,5,6,7,8,9,10)$ means the number of times prime number $p$ is prime in bases $2,3,4,5,6,7,8,9,10$.

Since for any prime number $p$ the only values that $\text{nop}(p;2,3,4,5,6,7,8,9,10)$ can have are $1,2,3,4,5,6,7,8,9$ and there is an infinite number of primes the sequence $s(p)$ has the property that at least one of the values $\frac{1}{9}, \frac{2}{9}, \frac {3}{9}, \frac {4}{9}, \frac {5}{9}, \frac {6}{9}, \frac {7}{9}, \frac {8}{9}, \frac {9}{9}$ occurs as a value of that sequence an infinite number of times.

What values of the values $\frac{1}{9}, \frac{2}{9}, \frac {3}{9}, \frac {4}{9}, \frac {5}{9}, \frac {6}{9}, \frac {7}{9}, \frac {8}{9}, \frac {9}{9}$ occur as the value of the sequence $s$ an infinite number of times?

Answer 1

Computational data for first $10^6,10^7,10^8$ primes.

---

Among first $10^6$ primes, here are the total counts $f(k)$ of $(k=1,\dots,9)$ examples:

$$\begin{array}{} k&1&2&3&4&5&6&7&8&9\\ f(k)&507662& 356319& 111659& 21301& 2800& 238& 18& 3& 0 \end{array}$$

I do not see an obvious obstruction that would prove some of these sets are finite. Similarly, actually showing some non-trivial set of prime numbers is infinite, is usually very hard to show.

Here are the prime sets: $[9]$, $[8]$, $[7]$, $[6]$, $[5]$, $[4]$, $[3]$, $[2]$, $[1]$.

---

Among first $10^7$ primes, there are $5$ of $(k=8)$ examples, and $62$ of $(k=7)$ examples, where I haven't computed additional $(k=6,\dots,1)$ examples.

(k=8) 2, 3, 379081, 59771671, 146752831. 
(k=7) 2, 3, 5, 9241, 17791, 330289, 379081, 391231, 1005481, 1210483, 2378143, 2469241, 2779939, 2840041, 6817501, 8320831, 9865711, 10871407, 11087191, 12259603, 13645393, 15665833, 16707883, 17694463, 25751863, 27794287, 31488481, 31652611, 32701987, 33150841, 34571503, 35243881, 41124511, 44463427, 45649003, 45831691, 47639563, 48404791, 49030633, 52255897, 52704331, 54065743, 54644173, 59771671, 64398097, 78078283, 79622161, 79745221, 80564851, 94426273, 98503063, 101593033, 107004913, 107909623, 118918537, 124310113, 133564003, 137523697, 146752831, 166064851, 167979547, 170977843.

---

Among first $2\cdot10^8$ primes, there are no $(k=9)$ examples.

Thus, if such an example exits, it is larger than $2\cdot10^8$th prime $4222234741\approx 4.2\cdot 10^9$.

---