Partial sums of primes
$begingroup$
$2+3+5+7+11+13...$ 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?
nt.number-theory
asked 11 hours ago
homunculushomunculus
314
$endgroup$
add a comment |
$begingroup$
$2+3+5+7+11+13...$ 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?
nt.number-theory
asked 11 hours ago
homunculushomunculus
314
$endgroup$
2
$begingroup$
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.
4 hours ago
add a comment |
$begingroup$
$2+3+5+7+11+13...$ 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?
nt.number-theory
asked 11 hours ago
homunculushomunculus
314
$endgroup$
$2+3+5+7+11+13...$ 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?
nt.number-theory
nt.number-theory
asked 11 hours ago
homunculushomunculus
314
asked 11 hours ago
homunculushomunculus
314
edited 10 hours ago
homunculus
asked 11 hours ago
homunculushomunculus
314
asked 11 hours ago
homunculushomunculus
314
asked 11 hours ago
homunculushomunculus
314
314
2
$begingroup$
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.
4 hours ago
add a comment |
2
$begingroup$
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.
4 hours ago
2
2
$begingroup$
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.
4 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.
4 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
edited 4 hours ago
Peter Taylor
1536
answered 10 hours ago
Mark FischlerMark Fischler
900313
$endgroup$
$begingroup$
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
3 hours ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
edited 4 hours ago
Peter Taylor
1536
answered 10 hours ago
Mark FischlerMark Fischler
900313
$endgroup$
$begingroup$
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
3 hours ago
add a comment |
$begingroup$
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.
edited 4 hours ago
Peter Taylor
1536
answered 10 hours ago
Mark FischlerMark Fischler
900313
$endgroup$
$begingroup$
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
3 hours ago
add a comment |
$begingroup$
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.
edited 4 hours ago
Peter Taylor
1536
answered 10 hours ago
Mark FischlerMark Fischler
900313
$endgroup$
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.
edited 4 hours ago
Peter Taylor
1536
answered 10 hours ago
Mark FischlerMark Fischler
900313
edited 4 hours ago
Peter Taylor
1536
edited 4 hours ago
Peter Taylor
1536
edited 4 hours ago
Peter Taylor
1536
1536
answered 10 hours ago
Mark FischlerMark Fischler
900313
answered 10 hours ago
Mark FischlerMark Fischler
900313
answered 10 hours ago
Mark FischlerMark Fischler
900313
900313
$begingroup$
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
3 hours ago
add a comment |
$begingroup$
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
3 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
3 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
3 hours ago
add a comment |
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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.
4 hours ago