Partial sums of primes Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Why do primes dislike dividing the sum of all the preceding primes?Partial sums of multiplicative functionsAre all primes in a PAP-3?Uniform distribution of digits of 1/pTwin primes and D primesGeneralizations of Chen's theoremSums of primes that are themselves primeNew proofs of Euclid's theorem of the infinitude of primes?For $k>3$ does there exist an odd prime $q_k$ such that $p_k=2^kq_k+1$ is prime and $p_k$ divides $a_k=dfrac3^2^k-1+12$?Two equivalent statements about primesPermutations $piin S_n$ with $p_k+p_pi(k)+1$ prime for all $k=1,ldots,n$

Partial sums of primes



Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Why do primes dislike dividing the sum of all the preceding primes?Partial sums of multiplicative functionsAre all primes in a PAP-3?Uniform distribution of digits of 1/pTwin primes and D primesGeneralizations of Chen's theoremSums of primes that are themselves primeNew proofs of Euclid's theorem of the infinitude of primes?For $k>3$ does there exist an odd prime $q_k$ such that $p_k=2^kq_k+1$ is prime and $p_k$ divides $a_k=dfrac3^2^k-1+12$?Two equivalent statements about primesPermutations $piin S_n$ with $p_k+p_pi(k)+1$ prime for all $k=1,ldots,n$










5












$begingroup$


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










share|cite|improve this question











$endgroup$







  • 6




    $begingroup$
    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







  • 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






  • 3




    $begingroup$
    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
    Mar 28 at 18:54






  • 2




    $begingroup$
    Version 16. Please, homunc, give it a rest.
    $endgroup$
    – Gerry Myerson
    Mar 28 at 21:30















5












$begingroup$


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










share|cite|improve this question











$endgroup$







  • 6




    $begingroup$
    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







  • 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






  • 3




    $begingroup$
    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
    Mar 28 at 18:54






  • 2




    $begingroup$
    Version 16. Please, homunc, give it a rest.
    $endgroup$
    – Gerry Myerson
    Mar 28 at 21:30













5












5








5


1



$begingroup$


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










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. 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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 28 at 18:09







Enzo Creti

















asked Mar 25 at 15:56









Enzo CretiEnzo Creti

63119




63119







  • 6




    $begingroup$
    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







  • 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






  • 3




    $begingroup$
    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
    Mar 28 at 18:54






  • 2




    $begingroup$
    Version 16. Please, homunc, give it a rest.
    $endgroup$
    – Gerry Myerson
    Mar 28 at 21:30












  • 6




    $begingroup$
    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







  • 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






  • 3




    $begingroup$
    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
    Mar 28 at 18:54






  • 2




    $begingroup$
    Version 16. Please, homunc, give it a rest.
    $endgroup$
    – Gerry Myerson
    Mar 28 at 21:30







6




6




$begingroup$
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





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




3




3




$begingroup$
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
Mar 28 at 18:54




$begingroup$
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
Mar 28 at 18:54




2




2




$begingroup$
Version 16. Please, homunc, give it a rest.
$endgroup$
– Gerry Myerson
Mar 28 at 21:30




$begingroup$
Version 16. Please, homunc, give it a rest.
$endgroup$
– Gerry Myerson
Mar 28 at 21:30










1 Answer
1






active

oldest

votes


















16












$begingroup$

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, $$P(k) approx frac1p_k sim frac1k log k$$



In that case, the expected number of primes with the property you want would be something like
$$int_2^infty frac1x log x,dx$$
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











$endgroup$








  • 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











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









16












$begingroup$

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, $$P(k) approx frac1p_k sim frac1k log k$$



In that case, the expected number of primes with the property you want would be something like
$$int_2^infty frac1x log x,dx$$
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











$endgroup$








  • 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















16












$begingroup$

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, $$P(k) approx frac1p_k sim frac1k log k$$



In that case, the expected number of primes with the property you want would be something like
$$int_2^infty frac1x log x,dx$$
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











$endgroup$








  • 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













16












16








16





$begingroup$

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, $$P(k) approx frac1p_k sim frac1k log k$$



In that case, the expected number of primes with the property you want would be something like
$$int_2^infty frac1x log x,dx$$
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, $$P(k) approx frac1p_k sim frac1k log k$$



In that case, the expected number of primes with the property you want would be something like
$$int_2^infty frac1x log x,dx$$
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 Mar 25 at 22:43









Peter Taylor

1536




1536










answered Mar 25 at 16:42









Mark FischlerMark Fischler

992313




992313







  • 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




    $begingroup$
    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

















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He _____ here since 1970 . Answer needed [closed]What does “since he was so high” mean?Meaning of “catch birds for”?How do I ensure “since” takes the meaning I want?“Who cares here” meaningWhat does “right round toward” mean?the time tense (had now been detected)What does the phrase “ring around the roses” mean here?Correct usage of “visited upon”Meaning of “foiled rail sabotage bid”It was the third time I had gone to Rome or It is the third time I had been to Rome

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