Partial sums of primes












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.






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.
    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












$begingroup$


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







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.
    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








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.






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. 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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago







homunculus

















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














  • 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










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, $$P(k) approx frac1{p_k} sim frac1{k log k}$$



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











$endgroup$









  • 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
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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, $$P(k) approx frac1{p_k} sim frac1{k log k}$$



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











$endgroup$









  • 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, $$P(k) approx frac1{p_k} sim frac1{k log k}$$



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











$endgroup$









  • 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, $$P(k) approx frac1{p_k} sim frac1{k log k}$$



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 Mar 25 at 22:43









Peter Taylor

1536




1536










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














  • 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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Puebla de Zaragoza

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Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
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Musa

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Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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