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I am interested in the number of "primal" sequences of consecutive numbers of the form $p_1, 2 p_2, 3 p_3,\ldots, k p_k$ for primes $p_1, p_2,\ldots, p_k$.

For instance, there are $56,157$ sequences of the form $p_1, 2 p_2 = p_1 + 1$ for $p_1 < 20,000,000$ where $p_1$ and $p_2$ are both prime. I have found sequences up to length 7 of which the earliest begins at $5,516,281$.

Are there any results limiting the existence or number of such primal sequences? Is there even an infinity of pairs of the form $p_1$, $2 p_2$? Is there an upper bound to the length of such sequences? If not, then clearly there are an infinitude of sequences of each particular length.

I am a beginner here and have checked many posts about twin primes but not seen something like this, apologies if it is a trivial or answered question already.

Kenta S
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fairflow
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  • So far I have searched primes up to $10^8$ without finding a primal sequence of length 8. – fairflow Aug 12 '20 at 14:22
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    I think, the generalized Bunyakovsky conjecture implies infinite many primes with every length, but already the case $n=2$ with length $2$ is unknown. Almost surely do infinite many primes $p$ exist, such that $2p-1$ is prime as well. but noone could yet prove it. – Peter Aug 12 '20 at 14:49
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    According to my calculations, $$p_1=7\ 321\ 991\ 041$$ does the job for length $8$. Please check. – Peter Aug 12 '20 at 15:05
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    And for length $9$ , the prime $$p_1=363\ 500\ 177\ 041$$ should do the job. – Peter Aug 12 '20 at 15:24
  • @Peter That's great. I assume you made use of the fact that you must have $p_1=2520k+1$. Did you use brute force or have any better techniques to reduce the search space? – Jaap Scherphuis Aug 12 '20 at 15:34
  • @JaapScherphuis You guessed it exactly, brute force apart from the necessary condition. – Peter Aug 12 '20 at 15:38
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    And finally, this should give length $10$ : $$p_1=2\ 394\ 196\ 081\ 201$$ – Peter Aug 12 '20 at 15:39
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    A093553 gives the value of $p_1$ for a given value of $k$. – Varun Vejalla Aug 13 '20 at 01:50
  • @Peter there should be as many primes $2p+1$ where $p$ is prime, no? In fact all those sequences could be generalised so that the largest prime is not the first. – fairflow Aug 13 '20 at 07:41
  • @fairflow We have $p_1=2p_2-1$ , so we need a prime $p$ (which is then $p_2$) , such that $2p-1$ (which is then $p_1$) is prime as well. – Peter Aug 13 '20 at 08:19
  • @VarunVejalla Fortunately, my calculations did not take long. Did not expect this sequence in OEIS. – Peter Aug 13 '20 at 08:21
  • @fairflow Neither case has been proven. We do not know whether infinite many primes $p$ exist such that $2p+1$ is prime as well , same for $2p-1$. – Peter Aug 13 '20 at 11:43
  • @Peter thanks, yes, that is what I understood; what I meant is that it would be bizarre if infinitely many primes $p$ exist for which $2p+1$ is prime too and yet there were only finitely many primes $p$ for which $2p - 1$ is prime, or vice versa. And yet, without a proof of either, we could not be sure: unless one condition implied the other. I suppose the generalized Bunyakovsky conjecture implies both, I still do not understand it that well. – fairflow Aug 13 '20 at 16:17
  • @fairflow Yes, the generalized Bunyakovsky conjecture implies infinite many primes $p$ such that $2p-1$ is prime and infinite many primes $p$ such that $2p+1$ is prime. In fact, if we have irreducible polynomials with no forced common factor and positive leading coefficients, the conjecture states that for infinite many $p$ , all polynomials give simultaneously a prime number. Special cases are the Sophie-Germain primes, the twin-pair primes and primes $p$ such that $2p-1$ is prime as well. – Peter Aug 14 '20 at 09:46
  • @fairflow Extremely hard to imagine, but we cannot rule out that there are finite many primes of both kinds. So, if one would imply the other, this would still not help. – Peter Sep 18 '20 at 13:39

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