Longest sequence of primes where each term is obtained by appending a new digit to the previous term

Question

What is the longest known sequence of primes where each new term is obtained by appending a new decimal digit to the previous term?

Examples:

$$(2,23,233,2333,23333)$$

There are no more members in this sequence, because $233331,233333,233337,233339$ are all composite numbes.

$$(3,37,373,3733,37337,373379,3733799,37337999)$$

This one is longer, but still only $8$ terms.

An so on. Because prime gaps can become arbitrary long for large numbers, there is no way any such sequence will be infinite. On the other hand, they can be very long, I think.

What is the longest known sequence of this kind?

Can we prove that:

There is no infinite sequence of this kind. Or at least that the sum of reciprocals of any such sequence is finite.

Of course, we can study such sequences in binary or other bases.

Edit

The sequence can start with any prime. This link is very relevant. See also the comments.

The argument that an infinite sequence like this is unlikely goes like this: For each new term we have only four possibilities:

$$a_{n+1}=10 a_n+b,~~~~b=(1,3,7,9)$$

All of these numbers are contained in a very small interval, about $10$ in size. But the density of primes becomes smaller with growing number of digits, and the gaps become larger. In my opinion, the probability of finding the next term for any such sequence becomes smaller with growing $n$.

It doesn't mean that an infinite sequence is impossible, but I conjecture that it's impossible to prove that any such infinite sequence exists.

This question seems to be very closely related to an older question and a more recent question

You are *not* requiring the sequence to start with a one-digit prime, correct? — bof, Jul 13 '16 at 10:43
I guess the sum of reciprocals will be finite whether the numbers are prime or not, since each term of the sequence is more than ten times as big as the previous one. — bof, Jul 13 '16 at 10:51
If the sequence starts with a single-digit prime, then the longest possible sequences have 8 terms, according to this article on truncatable primes. — bof, Jul 13 '16 at 10:53
How come you state flatly "there is no way any such sequence will be infinite" and then ask as a question "can we prove that there is no infinite sequence of this kind"? — bof, Jul 13 '16 at 10:55
The sum of reciprocals of any such sequence is finite since it's bounded by $1+\frac1{10}+\frac1{100}+\cdots=\frac{10}9$. — joriki, Jul 13 '16 at 10:59
How do you prove that such a sequence cannot be infinite? I don't see how the fact that there are arbitrarily long gaps in the primes proves anything, seeing as your sequence also has arbitrarily long gaps, and is much sparser than the sequence of primes. — bof, Jul 13 '16 at 11:15
@bof, to all of your questions: I'm not sure of anything, I was just guessing. Now I'm at least sure that the sum of reciprocals is bounded. — Yuriy S, Jul 13 '16 at 11:32

Mastrem · Answer 1 · 2016-07-13T12:41:23.637

It appears such a sequence cannot be infinite. Basically, what you're doing is multiplying by $10$ and adding either $1$, $3$, $7$ or $9$. I wrote some python code to test whether sequences of a certain length existed.

I started with a list of all the integers from $0$ up until the highest prime below $10000$ minus one. These are all the possible values of $n\ mod\ p$ with $0<p<10000$. Then I made a new list, which contained all the the values in the old one, multiplied by $10$ and with $1$, $3$, $7$ or $9$ added to them. I removed all the multiples of primes (not the primes themselves) and repeated this proces. As it turns out, the list is empty after repeating this proces $15$ times

Code:

>>> L = []
>>> for x in range(max(primes)):
    L.append(x)


>>> for x in range(15):
    L2 = []
    for l in L:
        L2.append(10*l+1)
        L2.append(10*l+3)
        L2.append(10*l+7)
        L2.append(10*l+9)
    L3 = []
    for l in L2:
        zero = False
        for p in primes:
            if l%p==0 and l!=p:
                zero = True
                break
        if not zero:
            L3.append(l)
    L = copy(L3)


>>> len(L)
0
>>>

This means that no sequences of length $19$(the initial $4$-digit number can also be a prime with this very special property) exist. As a matter of fact, there don't even exist sequences of length $18$.

You are mistaken about $14$. See this link. It contains a prime $13302806296379339933399333$ We can truncate this prime exactly $14$ times, to the final prime $133028062963$ — Yuriy S, Jul 13 '16 at 12:39
Oh, you're right. the initial number can also be a prime of at most 4 digits, so the limit is $17$, I'll correct it. — Mastrem, Jul 13 '16 at 12:40
My understanding is that the starting number can be any prime, so we cannot make an assumption on the number of digits in $a_1$. This, quite possibly, kills any hope of resolving this via using a computer program. Thus the question still stands: is there an infinite sequence of primes where every next one is obtained from the previous number by adding a digit on the right? It was posted in Western Number Theory Problems 1996-12-17 by Gerry Myerson, then cited by R.Guy in his book "Unsolved Problems in Number Theory" (2004). It was also offered at a Russian math competition in 1972. — JimT, Jun 25 '23 at 05:46

Longest sequence of primes where each term is obtained by appending a new digit to the previous term

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