Does there exist a strictly increasing sequence of primes $(q_i)_{i \in \mathbb N}$ such that $\text {ds} (\prod_{k=1}^l q_k)$ is prime for every $l \in \mathbb N$?

Here $\text{ds}(n)$ denotes a digit sum of $n$.

I do not know from where should I attack this problem, main thoughts are:

a) If we start with some prime then there is probably an infinite number of primes that will not work with that starting-point prime, so that still leaves us with probably an infinite number of primes that we could choose as to be our second prime. Then those two first primes again probably determine an infinite number of primes that will not work but again probably leave us with an infinite number of primes that could serve as a third prime to choose. So, maybe some sieving could do the job, if there are some specialized sieves for this type of problems.

b) On the other hand, every prime that does not have digit-sum a composite number seems as a suitable candidate to be a starting point because there are so many possible multiplications that we can do here that it would be a sort of a miracle if some primes are better-behaved than some others as a candidates for a starting point.

c) If we choose some starting-point prime a reason why an infinite number of primes will not be suitable as a second prime is because probably it is often that digit sum of product of two primes is a composite number, and this should also hold for products of $l$ primes, for every $l \in \mathbb N \setminus \{1\}$.

d) All of this leads to a conclusion that if every prime that does not have digit-sum a composite number could be a suitable starting-point prime that it could also be that sequence associated to that prime could be unique and strictly determined by a starting point, but also it could be that it is not so.

Also, in an answer or in a comment, I would welcome if someone gives me some works (books/articles) on "digit-sum theory" because it seems to me even if there are glimpses of such a theory that such a theory is in a very poor stage of a development

  • $\begingroup$ Is that base ten digit sum? If so, it seems to me it is a rather unnatural thing to consider. Indeed, I expect there are no books on "digit-sum theory" because it has (so far) been use for very little. $\endgroup$ – Gerald Edgar Mar 23 '18 at 17:38
  • $\begingroup$ @GeraldEdgar Yes, base 10. It does not need to be "of use". But there is no reason why it shouldn´t be developed, first as a part of pure mathematics, more specially ,number theory, and an "uses" come later, if there would be any. $\endgroup$ – Shalom Mar 23 '18 at 17:48
  • $\begingroup$ Do you mean a strictly increasing sequence of primes or? $\endgroup$ – მამუკა ჯიბლაძე Mar 23 '18 at 18:50
  • $\begingroup$ @მამუკაჯიბლაძე Yes. $\endgroup$ – Shalom Mar 23 '18 at 18:55
  • 1
    $\begingroup$ Here is Mathematica code in case you would like to use it genseq[n_, m_] := genseq[n, m] = If[n == 1, If[Not[PrimeQ[Total[IntegerDigits[Prime[m]]]]], {}, {Prime[m]}], Module[{s = genseq[n - 1, m], pr, k}, pr = Times @@ s; If[Not[PrimeQ[Total[IntegerDigits[pr]]]], Return[{}]]; k = PrimePi[Last[s]] + 1; While[Not[PrimeQ[Total[IntegerDigits[pr Prime[k]]]]], k++]; Append[s, Prime[k]] ] ] $\endgroup$ – მამუკა ჯიბლაძე Mar 23 '18 at 19:44

I agree that the sum of the base ten digits seems rather peripheral. However: given a positive integer $M;$ let $S_M$ be the set of primes $p$ so that the base $10$ digit sum of $Mp$ is a prime.

Q: is $S_M$ infinite as long as the digit sum of $M$ is not a multiple of $3?$

I would guess that one could find encouraging evidence of YES (I haven’t checked) although no proof. You only care about $M$ with prime digit sum but that would follow from the more general question I gave.

Flimsy reasoning for why I suspect yes: Suppose $M$ has $t$ digits and consider primes with $s$ digits. Then there will be a very large number of products $Mp$ and the average digit sum should be roughly $5(s+t)$. There seems no reason to doubt that these digit sums behave as if they were randomly chosen (among non-multiples of $3$) in the appropriate range. Estimating the number of sums and density of primes in that range should suggest a healthy number of successes.


I just looked at $M=314159$ and the $1228$ primes $p \neq 3$ less than $10^5.$ The digit sums for $Mp$ range from $13$ to $70.$ The only missing sums from that little experiment (among non-multiples of $3$ from $13$ to $70$ are $14, 16.$ Here are the counts in the middle of that range

$[28, 34], [29, 25], [31, 39], [32, 44], [34, 54], [35, 74], [37, 67], [38, 82], [40, 75], [41, 75], $

$[43, 81], [44, 69], [46, 69], [47, 62], [49, 58], [50, 60], [52, 44], [53, 33], [55, 29], [56, 29]$

One might expect from this that in general among the $Mp$ are every possible digit sum (prime and otherwise, but not multiples of $3$) multiple times (with perhaps a small number of small exceptions.)

I'm not sure I would have predicted it but the comment by მამუკა ჯიბლაძე suggests that there are infinitely many primes such that $3p$ has digit sum $3$

There are $k-1$ odd integers $x$ with digit sum $3$ and $k$ digits. The $k-1$ values $\frac{x}{3}$ are all relatively prime to $6$ so one might expect the number of primes there to be roughly $\frac3{\ln(10^k/3)}(k-1) \sim 1.3$. Up to $k=100$ there are on average about $1.6$ in each range. I wonder if something related is true for $M$ with digit sum a multiple of $3.$

| cite | improve this answer | |
  • 2
    $\begingroup$ For $M$ not divisible by $3$, more than $1/3$ of all primes seems to be in $S_M$. But is it clear that $S_M$ (although much much sparser) is finite for multiples of $3$? For example, with $M=3$ there is certain pattern that might well continue forever: 7, 37, 67, 337, 367, 333337, 333367, 333667, 336667, 666667, 33333667, 33666667, 66666667, ...; (multiplying these by 3 gives, respectively, 21, 111, 201, 1011, 1101, 1000011, 1000101, 1001001, 1010001, 2000001, 100001001, 101000001, 200000001) $\endgroup$ – მამუკა ჯიბლაძე Mar 24 '18 at 7:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.