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? 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
asked yesterday
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? 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
asked yesterday
homunculushomunculus
314
$endgroup$
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
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? 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
asked yesterday
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? 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
asked yesterday
homunculushomunculus
314
asked yesterday
homunculushomunculus
314
edited 17 mins ago
homunculus
asked yesterday
homunculushomunculus
314
asked yesterday
homunculushomunculus
314
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
add a comment |
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
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 19 hours ago
Peter Taylor
1536
answered yesterday
Mark FischlerMark Fischler
920313
$endgroup$
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
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "504"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f326315%2fpartial-sums-of-primes%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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 19 hours ago
Peter Taylor
1536
answered yesterday
Mark FischlerMark Fischler
920313
$endgroup$
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
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 19 hours ago
Peter Taylor
1536
answered yesterday
Mark FischlerMark Fischler
920313
$endgroup$
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
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 19 hours ago
Peter Taylor
1536
answered yesterday
Mark FischlerMark Fischler
920313
$endgroup$
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 19 hours ago
Peter Taylor
1536
answered yesterday
Mark FischlerMark Fischler
920313
edited 19 hours ago
Peter Taylor
1536
edited 19 hours ago
Peter Taylor
1536
edited 19 hours ago
Peter Taylor
1536
1536
answered yesterday
Mark FischlerMark Fischler
920313
answered yesterday
Mark FischlerMark Fischler
920313
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
add a comment |
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
add a comment |
Thanks for contributing an answer to MathOverflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f326315%2fpartial-sums-of-primes%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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