Does the algorithm of the Greeks produce all prime numbers?











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Let ${cal P}$ be the set of prime numbers. Define a subset ${cal P}'={p_1,p_2,p_3,cdots}$ of ${cal P}$ by setting $p_1=2$ and defining $p_{n+1}$ to be the smallest element of ${cal P}$ dividing $1+p_1cdots p_n$. Is there any obstruction to ${cal P}'={cal P}$ ?










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  • 24




    I don't think I would call this "the algorithm of the Greeks", since Eratosthenes produced an algorithm which definitely captures all the primes.
    – Todd Trimble
    Dec 2 at 22:28






  • 2




    This and variations have been considered. There is no nice (and known to me) alternate characterization of any of these variants. You should check the OEIS for more information. Gerhard "Should Always Check The OEIS" Paseman, 2018.12.02.
    – Gerhard Paseman
    Dec 2 at 22:33






  • 8




    In the spirit of mathoverflow.net/questions/45951/sexy-vacuity, let me point out that the special case $p_1 = 2$ is unnecessary here — the single general case “$p_n$ is the smallest prime dividing $1 + p_1 cdots p_{n-1}$” suffices to define the whole sequence.
    – Peter LeFanu Lumsdaine
    Dec 3 at 0:21








  • 7




    Note that this is not an algorithm at all, just a recursively defined sequence. An algorithm also needs to specify the way to compute it; especially in this case how to compute the smallest prime divisor.
    – YCor
    Dec 3 at 2:17






  • 3




    @Acccumulation Which is exactly how empty products are treated, yes.
    – Johannes Hahn
    Dec 4 at 1:11















up vote
31
down vote

favorite
12












Let ${cal P}$ be the set of prime numbers. Define a subset ${cal P}'={p_1,p_2,p_3,cdots}$ of ${cal P}$ by setting $p_1=2$ and defining $p_{n+1}$ to be the smallest element of ${cal P}$ dividing $1+p_1cdots p_n$. Is there any obstruction to ${cal P}'={cal P}$ ?










share|cite|improve this question




















  • 24




    I don't think I would call this "the algorithm of the Greeks", since Eratosthenes produced an algorithm which definitely captures all the primes.
    – Todd Trimble
    Dec 2 at 22:28






  • 2




    This and variations have been considered. There is no nice (and known to me) alternate characterization of any of these variants. You should check the OEIS for more information. Gerhard "Should Always Check The OEIS" Paseman, 2018.12.02.
    – Gerhard Paseman
    Dec 2 at 22:33






  • 8




    In the spirit of mathoverflow.net/questions/45951/sexy-vacuity, let me point out that the special case $p_1 = 2$ is unnecessary here — the single general case “$p_n$ is the smallest prime dividing $1 + p_1 cdots p_{n-1}$” suffices to define the whole sequence.
    – Peter LeFanu Lumsdaine
    Dec 3 at 0:21








  • 7




    Note that this is not an algorithm at all, just a recursively defined sequence. An algorithm also needs to specify the way to compute it; especially in this case how to compute the smallest prime divisor.
    – YCor
    Dec 3 at 2:17






  • 3




    @Acccumulation Which is exactly how empty products are treated, yes.
    – Johannes Hahn
    Dec 4 at 1:11













up vote
31
down vote

favorite
12









up vote
31
down vote

favorite
12






12





Let ${cal P}$ be the set of prime numbers. Define a subset ${cal P}'={p_1,p_2,p_3,cdots}$ of ${cal P}$ by setting $p_1=2$ and defining $p_{n+1}$ to be the smallest element of ${cal P}$ dividing $1+p_1cdots p_n$. Is there any obstruction to ${cal P}'={cal P}$ ?










share|cite|improve this question















Let ${cal P}$ be the set of prime numbers. Define a subset ${cal P}'={p_1,p_2,p_3,cdots}$ of ${cal P}$ by setting $p_1=2$ and defining $p_{n+1}$ to be the smallest element of ${cal P}$ dividing $1+p_1cdots p_n$. Is there any obstruction to ${cal P}'={cal P}$ ?







nt.number-theory prime-numbers algorithms






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edited Dec 4 at 0:58









Konstantinos Kanakoglou

3,01021033




3,01021033










asked Dec 2 at 22:19









Zidane

23625




23625








  • 24




    I don't think I would call this "the algorithm of the Greeks", since Eratosthenes produced an algorithm which definitely captures all the primes.
    – Todd Trimble
    Dec 2 at 22:28






  • 2




    This and variations have been considered. There is no nice (and known to me) alternate characterization of any of these variants. You should check the OEIS for more information. Gerhard "Should Always Check The OEIS" Paseman, 2018.12.02.
    – Gerhard Paseman
    Dec 2 at 22:33






  • 8




    In the spirit of mathoverflow.net/questions/45951/sexy-vacuity, let me point out that the special case $p_1 = 2$ is unnecessary here — the single general case “$p_n$ is the smallest prime dividing $1 + p_1 cdots p_{n-1}$” suffices to define the whole sequence.
    – Peter LeFanu Lumsdaine
    Dec 3 at 0:21








  • 7




    Note that this is not an algorithm at all, just a recursively defined sequence. An algorithm also needs to specify the way to compute it; especially in this case how to compute the smallest prime divisor.
    – YCor
    Dec 3 at 2:17






  • 3




    @Acccumulation Which is exactly how empty products are treated, yes.
    – Johannes Hahn
    Dec 4 at 1:11














  • 24




    I don't think I would call this "the algorithm of the Greeks", since Eratosthenes produced an algorithm which definitely captures all the primes.
    – Todd Trimble
    Dec 2 at 22:28






  • 2




    This and variations have been considered. There is no nice (and known to me) alternate characterization of any of these variants. You should check the OEIS for more information. Gerhard "Should Always Check The OEIS" Paseman, 2018.12.02.
    – Gerhard Paseman
    Dec 2 at 22:33






  • 8




    In the spirit of mathoverflow.net/questions/45951/sexy-vacuity, let me point out that the special case $p_1 = 2$ is unnecessary here — the single general case “$p_n$ is the smallest prime dividing $1 + p_1 cdots p_{n-1}$” suffices to define the whole sequence.
    – Peter LeFanu Lumsdaine
    Dec 3 at 0:21








  • 7




    Note that this is not an algorithm at all, just a recursively defined sequence. An algorithm also needs to specify the way to compute it; especially in this case how to compute the smallest prime divisor.
    – YCor
    Dec 3 at 2:17






  • 3




    @Acccumulation Which is exactly how empty products are treated, yes.
    – Johannes Hahn
    Dec 4 at 1:11








24




24




I don't think I would call this "the algorithm of the Greeks", since Eratosthenes produced an algorithm which definitely captures all the primes.
– Todd Trimble
Dec 2 at 22:28




I don't think I would call this "the algorithm of the Greeks", since Eratosthenes produced an algorithm which definitely captures all the primes.
– Todd Trimble
Dec 2 at 22:28




2




2




This and variations have been considered. There is no nice (and known to me) alternate characterization of any of these variants. You should check the OEIS for more information. Gerhard "Should Always Check The OEIS" Paseman, 2018.12.02.
– Gerhard Paseman
Dec 2 at 22:33




This and variations have been considered. There is no nice (and known to me) alternate characterization of any of these variants. You should check the OEIS for more information. Gerhard "Should Always Check The OEIS" Paseman, 2018.12.02.
– Gerhard Paseman
Dec 2 at 22:33




8




8




In the spirit of mathoverflow.net/questions/45951/sexy-vacuity, let me point out that the special case $p_1 = 2$ is unnecessary here — the single general case “$p_n$ is the smallest prime dividing $1 + p_1 cdots p_{n-1}$” suffices to define the whole sequence.
– Peter LeFanu Lumsdaine
Dec 3 at 0:21






In the spirit of mathoverflow.net/questions/45951/sexy-vacuity, let me point out that the special case $p_1 = 2$ is unnecessary here — the single general case “$p_n$ is the smallest prime dividing $1 + p_1 cdots p_{n-1}$” suffices to define the whole sequence.
– Peter LeFanu Lumsdaine
Dec 3 at 0:21






7




7




Note that this is not an algorithm at all, just a recursively defined sequence. An algorithm also needs to specify the way to compute it; especially in this case how to compute the smallest prime divisor.
– YCor
Dec 3 at 2:17




Note that this is not an algorithm at all, just a recursively defined sequence. An algorithm also needs to specify the way to compute it; especially in this case how to compute the smallest prime divisor.
– YCor
Dec 3 at 2:17




3




3




@Acccumulation Which is exactly how empty products are treated, yes.
– Johannes Hahn
Dec 4 at 1:11




@Acccumulation Which is exactly how empty products are treated, yes.
– Johannes Hahn
Dec 4 at 1:11










1 Answer
1






active

oldest

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up vote
29
down vote



accepted










According to Booker - A variant of the Euclid-Mullin sequence containing every prime, as of 2016, this question remains open.




One of the central questions in this area was posed by Mullin [6] in 1963: Does the Euclid–Mullin sequence contain every prime number? Despite a compelling heuristic argument of Shanks [9] that the answer is yes, even the broader question of whether there is any Euclid sequence containing every prime number remains open.




The OEIS contains a decent amount of information. For example, the primes up to 37 do appear within the first 50 terms of the sequence.






share|cite|improve this answer























  • Any idea what the euristic expected asymptotic growth of the least missed prime $p_n$ at step $n$ would be? (Sorry for this sentence's violence to the English language.)
    – Yaakov Baruch
    Dec 3 at 14:07






  • 1




    @YaakovBaruch I don't have an answer for you, but I do see multiple references to a heuristic argument by Daniel Shanks in the Bulletin of the Institute of Combinatorics and its Applications (specifically, in 1991). Unfortunately, I can't find the argument itself.
    – user44191
    Dec 3 at 19:10








  • 2




    The OEIS A000945 b-file gives $p_{50}=31$
    – Henry
    Dec 3 at 21:32











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1 Answer
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active

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes








up vote
29
down vote



accepted










According to Booker - A variant of the Euclid-Mullin sequence containing every prime, as of 2016, this question remains open.




One of the central questions in this area was posed by Mullin [6] in 1963: Does the Euclid–Mullin sequence contain every prime number? Despite a compelling heuristic argument of Shanks [9] that the answer is yes, even the broader question of whether there is any Euclid sequence containing every prime number remains open.




The OEIS contains a decent amount of information. For example, the primes up to 37 do appear within the first 50 terms of the sequence.






share|cite|improve this answer























  • Any idea what the euristic expected asymptotic growth of the least missed prime $p_n$ at step $n$ would be? (Sorry for this sentence's violence to the English language.)
    – Yaakov Baruch
    Dec 3 at 14:07






  • 1




    @YaakovBaruch I don't have an answer for you, but I do see multiple references to a heuristic argument by Daniel Shanks in the Bulletin of the Institute of Combinatorics and its Applications (specifically, in 1991). Unfortunately, I can't find the argument itself.
    – user44191
    Dec 3 at 19:10








  • 2




    The OEIS A000945 b-file gives $p_{50}=31$
    – Henry
    Dec 3 at 21:32















up vote
29
down vote



accepted










According to Booker - A variant of the Euclid-Mullin sequence containing every prime, as of 2016, this question remains open.




One of the central questions in this area was posed by Mullin [6] in 1963: Does the Euclid–Mullin sequence contain every prime number? Despite a compelling heuristic argument of Shanks [9] that the answer is yes, even the broader question of whether there is any Euclid sequence containing every prime number remains open.




The OEIS contains a decent amount of information. For example, the primes up to 37 do appear within the first 50 terms of the sequence.






share|cite|improve this answer























  • Any idea what the euristic expected asymptotic growth of the least missed prime $p_n$ at step $n$ would be? (Sorry for this sentence's violence to the English language.)
    – Yaakov Baruch
    Dec 3 at 14:07






  • 1




    @YaakovBaruch I don't have an answer for you, but I do see multiple references to a heuristic argument by Daniel Shanks in the Bulletin of the Institute of Combinatorics and its Applications (specifically, in 1991). Unfortunately, I can't find the argument itself.
    – user44191
    Dec 3 at 19:10








  • 2




    The OEIS A000945 b-file gives $p_{50}=31$
    – Henry
    Dec 3 at 21:32













up vote
29
down vote



accepted







up vote
29
down vote



accepted






According to Booker - A variant of the Euclid-Mullin sequence containing every prime, as of 2016, this question remains open.




One of the central questions in this area was posed by Mullin [6] in 1963: Does the Euclid–Mullin sequence contain every prime number? Despite a compelling heuristic argument of Shanks [9] that the answer is yes, even the broader question of whether there is any Euclid sequence containing every prime number remains open.




The OEIS contains a decent amount of information. For example, the primes up to 37 do appear within the first 50 terms of the sequence.






share|cite|improve this answer














According to Booker - A variant of the Euclid-Mullin sequence containing every prime, as of 2016, this question remains open.




One of the central questions in this area was posed by Mullin [6] in 1963: Does the Euclid–Mullin sequence contain every prime number? Despite a compelling heuristic argument of Shanks [9] that the answer is yes, even the broader question of whether there is any Euclid sequence containing every prime number remains open.




The OEIS contains a decent amount of information. For example, the primes up to 37 do appear within the first 50 terms of the sequence.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 3 at 21:32

























answered Dec 2 at 22:26









user44191

2,5231026




2,5231026












  • Any idea what the euristic expected asymptotic growth of the least missed prime $p_n$ at step $n$ would be? (Sorry for this sentence's violence to the English language.)
    – Yaakov Baruch
    Dec 3 at 14:07






  • 1




    @YaakovBaruch I don't have an answer for you, but I do see multiple references to a heuristic argument by Daniel Shanks in the Bulletin of the Institute of Combinatorics and its Applications (specifically, in 1991). Unfortunately, I can't find the argument itself.
    – user44191
    Dec 3 at 19:10








  • 2




    The OEIS A000945 b-file gives $p_{50}=31$
    – Henry
    Dec 3 at 21:32


















  • Any idea what the euristic expected asymptotic growth of the least missed prime $p_n$ at step $n$ would be? (Sorry for this sentence's violence to the English language.)
    – Yaakov Baruch
    Dec 3 at 14:07






  • 1




    @YaakovBaruch I don't have an answer for you, but I do see multiple references to a heuristic argument by Daniel Shanks in the Bulletin of the Institute of Combinatorics and its Applications (specifically, in 1991). Unfortunately, I can't find the argument itself.
    – user44191
    Dec 3 at 19:10








  • 2




    The OEIS A000945 b-file gives $p_{50}=31$
    – Henry
    Dec 3 at 21:32
















Any idea what the euristic expected asymptotic growth of the least missed prime $p_n$ at step $n$ would be? (Sorry for this sentence's violence to the English language.)
– Yaakov Baruch
Dec 3 at 14:07




Any idea what the euristic expected asymptotic growth of the least missed prime $p_n$ at step $n$ would be? (Sorry for this sentence's violence to the English language.)
– Yaakov Baruch
Dec 3 at 14:07




1




1




@YaakovBaruch I don't have an answer for you, but I do see multiple references to a heuristic argument by Daniel Shanks in the Bulletin of the Institute of Combinatorics and its Applications (specifically, in 1991). Unfortunately, I can't find the argument itself.
– user44191
Dec 3 at 19:10






@YaakovBaruch I don't have an answer for you, but I do see multiple references to a heuristic argument by Daniel Shanks in the Bulletin of the Institute of Combinatorics and its Applications (specifically, in 1991). Unfortunately, I can't find the argument itself.
– user44191
Dec 3 at 19:10






2




2




The OEIS A000945 b-file gives $p_{50}=31$
– Henry
Dec 3 at 21:32




The OEIS A000945 b-file gives $p_{50}=31$
– Henry
Dec 3 at 21:32


















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