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. 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.
nt.number-theory prime-numbers
asked Mar 25 at 15:56
homunculushomunculus
304
$endgroup$
|
show 6 more comments
$begingroup$
$2+3+5+7+11+13...$ 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. 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.
nt.number-theory prime-numbers
asked Mar 25 at 15:56
homunculushomunculus
304
$endgroup$
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.
Mar 25 at 22:31
1
$begingroup$
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
Mar 26 at 9:22
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
2
$begingroup$
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
2 days ago
|
show 6 more comments
$begingroup$
$2+3+5+7+11+13...$ 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. 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.
nt.number-theory prime-numbers
asked Mar 25 at 15:56
homunculushomunculus
304
$endgroup$
$2+3+5+7+11+13...$ 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. 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.
nt.number-theory prime-numbers
nt.number-theory prime-numbers
asked Mar 25 at 15:56
homunculushomunculus
304
asked Mar 25 at 15:56
homunculushomunculus
304
edited 2 days ago
homunculus
asked Mar 25 at 15:56
homunculushomunculus
304
asked Mar 25 at 15:56
homunculushomunculus
304
asked Mar 25 at 15:56
homunculushomunculus
304
304
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.
Mar 25 at 22:31
1
$begingroup$
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
Mar 26 at 9:22
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
2
$begingroup$
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
2 days ago
|
show 6 more comments
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.
Mar 25 at 22:31
1
$begingroup$
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
Mar 26 at 9:22
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
2
$begingroup$
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
2 days 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.
Mar 25 at 22:31
$begingroup$
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
1
1
$begingroup$
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
Mar 26 at 9:22
$begingroup$
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
Mar 26 at 9:22
3
3
$begingroup$
Seven edits in the last 12 hours.
$endgroup$
– Gerry Myerson
Mar 26 at 21:11
$begingroup$
Seven edits in the last 12 hours.
$endgroup$
– Gerry Myerson
Mar 26 at 21:11
2
2
$begingroup$
Now up to Version 13.
$endgroup$
– Gerry Myerson
Mar 27 at 21:35
$begingroup$
Now up to Version 13.
$endgroup$
– Gerry Myerson
Mar 27 at 21:35
2
2
$begingroup$
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
2 days ago
$begingroup$
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
2 days ago
|
show 6 more comments
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 Mar 25 at 22:43
Peter Taylor
1536
answered Mar 25 at 16:42
Mark FischlerMark Fischler
955313
$endgroup$
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
add a comment |
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1 Answer
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$begingroup$
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 Mar 25 at 22:43
Peter Taylor
1536
answered Mar 25 at 16:42
Mark FischlerMark Fischler
955313
$endgroup$
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
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 Mar 25 at 22:43
Peter Taylor
1536
answered Mar 25 at 16:42
Mark FischlerMark Fischler
955313
$endgroup$
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
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 Mar 25 at 22:43
Peter Taylor
1536
answered Mar 25 at 16:42
Mark FischlerMark Fischler
955313
$endgroup$
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 Mar 25 at 22:43
Peter Taylor
1536
answered Mar 25 at 16:42
Mark FischlerMark Fischler
955313
edited Mar 25 at 22:43
Peter Taylor
1536
edited Mar 25 at 22:43
Peter Taylor
1536
edited Mar 25 at 22:43
Peter Taylor
1536
1536
answered Mar 25 at 16:42
Mark FischlerMark Fischler
955313
answered Mar 25 at 16:42
Mark FischlerMark Fischler
955313
answered Mar 25 at 16:42
Mark FischlerMark Fischler
955313
955313
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
add a comment |
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
2
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
$begingroup$
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
1
$begingroup$
"nobody ever sees it do so." - you made my day!
$endgroup$
– Wolfgang
Mar 26 at 9:26
$begingroup$
"nobody ever sees it do so." - you made my day!
$endgroup$
– Wolfgang
Mar 26 at 9:26
add a comment |
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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.
Mar 25 at 22:31
1
$begingroup$
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
Mar 26 at 9:22
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
2
$begingroup$
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
2 days ago