Are there Paterson Prime chains of arbitrary length?

Question

Inspired by Paterson Primes (with 3Blue1Brown) - Numberphile,

Consider a function $F_b(p)$ that takes a prime $p$ and reads its base $b\lt 10$ digits as decimal.

For example, $F_4(5)=11$. That is, $5=1\cdot 4^1+1\cdot 4^0\xrightarrow{F}1\cdot 10^1+1\cdot 10^0=11$.

Consider applying the function repeatedly until a composite number is reached.

For example, if $b=4$ and $p=5$, we have $5 \to 11 \to 23 \to 113 \to 1301 \to 110111$.

That is, above example generated a chain of $5$ distinct consecutive primes.

Denote this as $f_4(5)=5$.

Checking all $p\lt 10^9$, maximal chain lengths I found (per given base $b$) are:

$$ \begin{array}{lcc} f_2(3893257)&=&4\\ f_3(2119939)&=&5\\ f_4(101495533)&=&6\\ f_5(10834643)&=&6\\ f_6(80915941)&=&6\\ f_7(13587367)&=&6\\ f_8(51425431)&=&7\\ f_9(137118229)&=&6\\ \end{array} $$

We can generalize to $F^{B}_b(p)$ that takes a prime $p$ and reads its base $b\lt B$ digits as base $B$.

Then consider finding longest consecutive chain of distinct primes, $f^B_b(p)$.

Checking $p\lt 10^5, B\lt 100$, I found:

$$ \begin{array}{lcc} f^{12}_{8}(24001)&=&8\\ f^{22}_{16}(881)&=&9\\ f^{52}_{46}(907)&=&10\\ f^{62}_{54}(313)&=&12\\ f^{66}_{60}(577)&=&13\\ f^{88}_{58}(337)&=&14\\ \end{array} $$

Checking $p\lt 10^3, B\lt 1000$, there is no new best.

Checking $p\lt 10^4, B\lt 1000$, I found: (edit #$1$)

$$ \begin{array}{lcc} f^{520}_{382}(1031)&=&15\\ f^{688}_{598}(8573)&=&16\\ f^{988}_{784}(1933)&=&17\\ \end{array} $$

Can we find very long chains, much longer than current best $17$?

Do we expect arbitrarily long chains? Perhaps only as $B$ keeps growing?

Is there a fixed $(b,B)$ where we can prove a maximum value $f^B_b(p)$ is achieved?

Answer 1

I think I found an example with length $18$ :

$$[b,B,p]=[1446, 1956, 3037]$$

I used brute force , but increasing the prime $p$ and using $b<B,b<p$. The routine is in progress.

Routine and output until the current record :

gp > forprime(merk=10^3,10^5,for(a=2,merk-1,for(b=a+1,merk-1,p=merk;t=0;while(ispseudoprime(p)==1,t=t+1;p=fromdigits(digits(p,a),b));if(t>15,print([a,b,merk,t])))))
[494, 734, 1523, 16]
[784, 988, 1933, 17]
[785, 989, 1933, 16]
[786, 990, 1933, 16]
[784, 988, 2341, 16]
[1442, 1952, 3037, 16]
[1443, 1953, 3037, 16]
[1444, 1954, 3037, 16]
[1445, 1955, 3037, 18]
[1446, 1956, 3037, 18]

As one can see, there are two chains with length $18$ where the bases $b$ and $B$ both differ by $1$. This might indicate a useful pattern.