$2+3+5+7+11+13...$ is clearly the sum of the primes.

Now I consider partial sums such:

$2+3+5+7+11=28$ which is divisible by $7$

My question is:

are there infinitely many partial sums such that:

$p_1+p_2+p_3+...+p_{k}+p_{k+1}=m*p_{k}?$ with $m$ some positive integer? With Pari/gp apparently up to 10^10 there are only two examples $7$=$p_k$ and $8263=p_k$. Heuristically do you think that infinitely many such partial sums should exist? Note: 7 and 8263 are both primes belonging to primes on the left side of the triangle formed by listing successively the prime numbers in a triangular grid. See https://oeis.org/A078721 Note in both cases $2+3+5+7=17$ is prime and $2+3+5+...+p_{1036}=3974497$ is prime. I note that $17$ and $3974497$ are primes of the form $4s+1$, whereas $p_4=7$ and $p_{1036}=8263$ are primes of the form $6s+1$. $7$ and $8263$ are primes such that starting from the right, the odd positioned digits are prime and the even positioned digits are composite. But also $5$ and $8243$ which are the previous primes have this property. No other prime of this type found below $10^{12}$ I noticed that 7! has 4 digits where 4 is a palindrome. 8263! has 28782 digits where 28782 is a palindrome.

  • 6
    $\begingroup$ Strongly related: mathoverflow.net/questions/120511/…. Also crossposted on MSE: math.stackexchange.com/questions/3161810/23571113 (please don't do this anymore). $\endgroup$ – Alex M. Mar 25 at 22:31
  • 3
    $\begingroup$ Seven edits in the last 12 hours. $\endgroup$ – Gerry Myerson Mar 26 at 21:11
  • 2
    $\begingroup$ Now up to Version 13. $\endgroup$ – Gerry Myerson Mar 27 at 21:35
  • 3
    $\begingroup$ I find these frequent edits go against the purpose of this forum. If you want to record frequent observations on a daily basis (whether they are significant or not), start a blog. You have asked a main question and gotten a reasonable answer; now move on. The numerology associated with the problem does not belong here. Next week, if you find a third prime satisfying the relations, you can report that here. Gerhard "Know When To Fold 'Em" Paseman, 2019.03.28. $\endgroup$ – Gerhard Paseman Mar 28 at 18:54
  • 2
    $\begingroup$ Version 16. Please, homunc, give it a rest. $\endgroup$ – Gerry Myerson Mar 28 at 21:30

You asked for a heuristic answer.

There is an heuristic argument that infinitely many such partial sums should exist. Consider $P(k)$, an heuristic estimate of the probability that the partial sum of the first $k+1$ primes would be divisible by $p_k$. Now $$p_k \sim k \log k$$ and if only random chance were involved, $$P(k) \approx \frac1{p_k} \sim \frac1{k \log k}$$

In that case, the expected number of primes with the property you want would be something like $$\int_2^\infty \frac1{x \log x}\,dx$$ and that integral diverges to infinity.

The reason it seems so rare is that the rate of divergence is like $\log(\log x)$ and while that function goes to infinity, "nobody ever sees it do so."

On the other hand, proving that there an infinite number of such values of $k$ (in the same sense that Euclid's argument proves there is no last prime) is probably quite difficult. And if the conjecture that there are an infinite number of such values of $k$ turned out to be false, proving that some particular $k$ is the last one with this property would seem to be even harder.

  • 2
    $\begingroup$ Note this answer is essentially the same as David Speyer's in the question linked to in the comment by @Alex M. above. $\endgroup$ – Kimball Mar 25 at 23:30
  • 1
    $\begingroup$ "nobody ever sees it do so." - you made my day! $\endgroup$ – Wolfgang Mar 26 at 9:26

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.