Partial sums of primes












4












$begingroup$


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






nt.number-theory prime-numbers







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$endgroup$








  • 5




    $begingroup$
    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.
    20 hours ago












  • $begingroup$
    @Alex M.@Mark Fischler the heuristic is the same for primes p(n) dividing the sum of primes up to p(n+1) and for primes p(n) dividing the sum of primes up to p(n). But it seems that in the first case primes are rarer. Why?
    $endgroup$
    – homunculus
    11 hours ago












  • $begingroup$
    The second solution ($8263$) has some amazing properties : The sum of its digits, the sum of the squares of its digits and the sum of the fifth powers of its digits are prime as well as $$8^8+2^2+6^6+3^3$$
    $endgroup$
    – Peter
    9 hours ago










  • $begingroup$
    @Peter $8+2+6+3=19$. $8263+19-1=91^2$ where 91 is 19 reversed
    $endgroup$
    – homunculus
    6 hours ago










  • $begingroup$
    Possible duplicate of Why do primes dislike dividing the sum of all the preceding primes?
    $endgroup$
    – Alex M.
    3 hours ago
















4












$begingroup$


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






nt.number-theory prime-numbers







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    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.
    20 hours ago












  • $begingroup$
    @Alex M.@Mark Fischler the heuristic is the same for primes p(n) dividing the sum of primes up to p(n+1) and for primes p(n) dividing the sum of primes up to p(n). But it seems that in the first case primes are rarer. Why?
    $endgroup$
    – homunculus
    11 hours ago












  • $begingroup$
    The second solution ($8263$) has some amazing properties : The sum of its digits, the sum of the squares of its digits and the sum of the fifth powers of its digits are prime as well as $$8^8+2^2+6^6+3^3$$
    $endgroup$
    – Peter
    9 hours ago










  • $begingroup$
    @Peter $8+2+6+3=19$. $8263+19-1=91^2$ where 91 is 19 reversed
    $endgroup$
    – homunculus
    6 hours ago










  • $begingroup$
    Possible duplicate of Why do primes dislike dividing the sum of all the preceding primes?
    $endgroup$
    – Alex M.
    3 hours ago














4












4








4





$begingroup$


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






nt.number-theory prime-numbers







share|cite|improve this question











$endgroup$




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






nt.number-theory prime-numbers




nt.number-theory prime-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 17 mins ago







homunculus

















asked yesterday









homunculushomunculus

314




314








  • 5




    $begingroup$
    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.
    20 hours ago












  • $begingroup$
    @Alex M.@Mark Fischler the heuristic is the same for primes p(n) dividing the sum of primes up to p(n+1) and for primes p(n) dividing the sum of primes up to p(n). But it seems that in the first case primes are rarer. Why?
    $endgroup$
    – homunculus
    11 hours ago












  • $begingroup$
    The second solution ($8263$) has some amazing properties : The sum of its digits, the sum of the squares of its digits and the sum of the fifth powers of its digits are prime as well as $$8^8+2^2+6^6+3^3$$
    $endgroup$
    – Peter
    9 hours ago










  • $begingroup$
    @Peter $8+2+6+3=19$. $8263+19-1=91^2$ where 91 is 19 reversed
    $endgroup$
    – homunculus
    6 hours ago










  • $begingroup$
    Possible duplicate of Why do primes dislike dividing the sum of all the preceding primes?
    $endgroup$
    – Alex M.
    3 hours ago














  • 5




    $begingroup$
    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.
    20 hours ago












  • $begingroup$
    @Alex M.@Mark Fischler the heuristic is the same for primes p(n) dividing the sum of primes up to p(n+1) and for primes p(n) dividing the sum of primes up to p(n). But it seems that in the first case primes are rarer. Why?
    $endgroup$
    – homunculus
    11 hours ago












  • $begingroup$
    The second solution ($8263$) has some amazing properties : The sum of its digits, the sum of the squares of its digits and the sum of the fifth powers of its digits are prime as well as $$8^8+2^2+6^6+3^3$$
    $endgroup$
    – Peter
    9 hours ago










  • $begingroup$
    @Peter $8+2+6+3=19$. $8263+19-1=91^2$ where 91 is 19 reversed
    $endgroup$
    – homunculus
    6 hours ago










  • $begingroup$
    Possible duplicate of Why do primes dislike dividing the sum of all the preceding primes?
    $endgroup$
    – Alex M.
    3 hours ago








5




5




$begingroup$
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.
20 hours ago






$begingroup$
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.
20 hours ago














$begingroup$
@Alex M.@Mark Fischler the heuristic is the same for primes p(n) dividing the sum of primes up to p(n+1) and for primes p(n) dividing the sum of primes up to p(n). But it seems that in the first case primes are rarer. Why?
$endgroup$
– homunculus
11 hours ago






$begingroup$
@Alex M.@Mark Fischler the heuristic is the same for primes p(n) dividing the sum of primes up to p(n+1) and for primes p(n) dividing the sum of primes up to p(n). But it seems that in the first case primes are rarer. Why?
$endgroup$
– homunculus
11 hours ago














$begingroup$
The second solution ($8263$) has some amazing properties : The sum of its digits, the sum of the squares of its digits and the sum of the fifth powers of its digits are prime as well as $$8^8+2^2+6^6+3^3$$
$endgroup$
– Peter
9 hours ago




$begingroup$
The second solution ($8263$) has some amazing properties : The sum of its digits, the sum of the squares of its digits and the sum of the fifth powers of its digits are prime as well as $$8^8+2^2+6^6+3^3$$
$endgroup$
– Peter
9 hours ago












$begingroup$
@Peter $8+2+6+3=19$. $8263+19-1=91^2$ where 91 is 19 reversed
$endgroup$
– homunculus
6 hours ago




$begingroup$
@Peter $8+2+6+3=19$. $8263+19-1=91^2$ where 91 is 19 reversed
$endgroup$
– homunculus
6 hours ago












$begingroup$
Possible duplicate of Why do primes dislike dividing the sum of all the preceding primes?
$endgroup$
– Alex M.
3 hours ago




$begingroup$
Possible duplicate of Why do primes dislike dividing the sum of all the preceding primes?
$endgroup$
– Alex M.
3 hours ago










1 Answer
1






active

oldest

votes


















14












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    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
    19 hours ago






  • 1




    $begingroup$
    "nobody ever sees it do so." - you made my day!
    $endgroup$
    – Wolfgang
    9 hours ago











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1 Answer
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1 Answer
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oldest

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active

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active

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14












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    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
    19 hours ago






  • 1




    $begingroup$
    "nobody ever sees it do so." - you made my day!
    $endgroup$
    – Wolfgang
    9 hours ago
















14












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    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
    19 hours ago






  • 1




    $begingroup$
    "nobody ever sees it do so." - you made my day!
    $endgroup$
    – Wolfgang
    9 hours ago














14












14








14





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 19 hours ago









Peter Taylor

1536




1536










answered yesterday









Mark FischlerMark Fischler

920313




920313








  • 1




    $begingroup$
    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
    19 hours ago






  • 1




    $begingroup$
    "nobody ever sees it do so." - you made my day!
    $endgroup$
    – Wolfgang
    9 hours ago














  • 1




    $begingroup$
    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
    19 hours ago






  • 1




    $begingroup$
    "nobody ever sees it do so." - you made my day!
    $endgroup$
    – Wolfgang
    9 hours ago








1




1




$begingroup$
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
19 hours ago




$begingroup$
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
19 hours ago




1




1




$begingroup$
"nobody ever sees it do so." - you made my day!
$endgroup$
– Wolfgang
9 hours ago




$begingroup$
"nobody ever sees it do so." - you made my day!
$endgroup$
– Wolfgang
9 hours ago


















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