The five primes, 131, 157, 211, 349, 739, are neither in arithmetic or geometric progression, but are instead the sum of the five corresponding terms of an arithmetic and geometric progression.

Are there arbitrarily long sequences of primes with this property, that is, that they are the sum of the corresponding terms of an arithmetic and a (non-constant) geometric progression?


  • 1
    $\begingroup$ There are arbitrarily long APs in primes, take your geometric progression to be constant and you're good. If you require they are not in arithmetic progression, it gets harder. $\endgroup$
    – Wojowu
    Oct 2, 2019 at 12:46
  • $\begingroup$ @Wojowu I do! I shall clarify the issue. $\endgroup$ Oct 2, 2019 at 12:54
  • 3
    $\begingroup$ $1+2=3$, $3+4=7$, $5+8=13$, $7+16=23$, $9+32=41$ is a smaller set of five. $\endgroup$ Oct 2, 2019 at 13:18
  • $\begingroup$ @Wojowu I wonder how much harder it gets. A non-constant geometric progression could be "almost constant", meaning that all elements are about the same size. Then maybe a modification of Green–Tao will be able to show that there are arbitrarily long arithmetic progression such that the sums with that given geometric progression are all prime. It's just my very vague idea, and I barely have any knowledge in this domain. Perhaps we can at TerryTao. $\endgroup$
    – WhatsUp
    Oct 2, 2019 at 14:34
  • 1
    $\begingroup$ A sequence of ten primes: $14+9=23,56+27=83,98+81=179,140+243=383,182+729=911,224+2187=2411,266+6561=6827,308+19683=19991,350+59049=59399,392+177147=177539$ $\endgroup$ Oct 4, 2019 at 12:44

1 Answer 1


Claim: I actually have almost zero knowledge in this domain, and this answer is totally based on the correctness of this wiki page. Moreover, it could be that I misunderstood the content of that page, so it would be kind of some expert to point out to me if I made a mistake.

Quote from that wiki page:

More precisely, given any integer-valued polynomials $P_1,\cdots, P_k$ in one unknown $m$ all with constant term $0$, there are infinitely many integers $x$, $m$ such that $x + P_1(m), \cdots, x + P_k(m)$ are simultaneously prime.

From what I understand, this implies (or means) that there exists positive integers $x$ and $m$ such that all the $x + P_i(m)$ are primes.

Now for any fixed $k$, simply take $P_i(m) = m^i + im$. We will then have: $$x + P_i(m) = (x + im) + m^i,$$ where $(x + im)_i$ and $(m^i)_i$ are arithmetic and geometric progressions, respectively.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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