$2+3+5+7+11+13...$ 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.

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    $\begingroup$ 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, 2019 at 22:31
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    $\begingroup$ Seven edits in the last 12 hours. $\endgroup$ Mar 26, 2019 at 21:11
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    $\begingroup$ Now up to Version 13. $\endgroup$ Mar 27, 2019 at 21:35
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    $\begingroup$ 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. $\endgroup$ Mar 28, 2019 at 18:54
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    $\begingroup$ Version 16. Please, homunc, give it a rest. $\endgroup$ Mar 28, 2019 at 21:30

1 Answer 1


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.

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    $\begingroup$ 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, 2019 at 23:30
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    $\begingroup$ "nobody ever sees it do so." - you made my day! $\endgroup$
    – Wolfgang
    Mar 26, 2019 at 9:26

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