Conjecture: the sequence of sums of all consecutive primes contains an infinite number of primes












20












$begingroup$


Starting from 2, the sequence of sums of all consecutive primes is:



$$begin{array}{lcl}2 &=& 2\
2+3 &=& 5 \
2+3+5 &=& 10 \
2+3+5+7 &=& 17 \
2+3+5+7+11 &=& 28 \
&vdots&
end{array}
$$
If the $n^text{th}$ prime is $P_n$, then we can write $S_n=sum_{i=1}^n P_n$.



I conjecture that the sequence $S_n$ contains an infinite number of primes.



I doubt I'm the first person ever to think this, but I cannot find reference to the idea, nor can I conceive a proof, nor a disproof. Computationally, it can be verified that
$$S_{13,932}=998,658,581=P_{50,783,012},$$
and the sequence shows no sign of slowing down.



Can you prove the sequence $S_n$ contains an infinite number of primes?






sequences-and-series number-theory prime-numbers summation







share|cite|improve this question











$endgroup$








  • 8




    $begingroup$
    very hard to believe a proof is possible, when we can not prove that there are infinitely many primes $n^2 +1$
    $endgroup$
    – Will Jagy
    Jan 13 '14 at 2:55






  • 3




    $begingroup$
    I am surprised that the sequence 2,5,17,29,59,101 does not appear in the OEIS. I would recommend adding it.
    $endgroup$
    – RghtHndSd
    Jan 13 '14 at 3:14






  • 5




    $begingroup$
    Standard heuristics would predict that the number of $nle x$ for which $S_n$ is prime is asymptotic to $x/(2log x)$ (and computation seems to bear this out). But I agree it's hopeless to expect a proof that there are infinitely many prime $S_n$.
    $endgroup$
    – Greg Martin
    Jan 13 '14 at 3:18






  • 5




    $begingroup$
    oeis.org/A013918
    $endgroup$
    – user940
    Jan 13 '14 at 3:30






  • 2




    $begingroup$
    Related: mathoverflow.net/q/153656/12357
    $endgroup$
    – Joel Reyes Noche
    Jan 13 '14 at 3:37
















20












$begingroup$


Starting from 2, the sequence of sums of all consecutive primes is:



$$begin{array}{lcl}2 &=& 2\
2+3 &=& 5 \
2+3+5 &=& 10 \
2+3+5+7 &=& 17 \
2+3+5+7+11 &=& 28 \
&vdots&
end{array}
$$
If the $n^text{th}$ prime is $P_n$, then we can write $S_n=sum_{i=1}^n P_n$.



I conjecture that the sequence $S_n$ contains an infinite number of primes.



I doubt I'm the first person ever to think this, but I cannot find reference to the idea, nor can I conceive a proof, nor a disproof. Computationally, it can be verified that
$$S_{13,932}=998,658,581=P_{50,783,012},$$
and the sequence shows no sign of slowing down.



Can you prove the sequence $S_n$ contains an infinite number of primes?






sequences-and-series number-theory prime-numbers summation







share|cite|improve this question











$endgroup$








  • 8




    $begingroup$
    very hard to believe a proof is possible, when we can not prove that there are infinitely many primes $n^2 +1$
    $endgroup$
    – Will Jagy
    Jan 13 '14 at 2:55






  • 3




    $begingroup$
    I am surprised that the sequence 2,5,17,29,59,101 does not appear in the OEIS. I would recommend adding it.
    $endgroup$
    – RghtHndSd
    Jan 13 '14 at 3:14






  • 5




    $begingroup$
    Standard heuristics would predict that the number of $nle x$ for which $S_n$ is prime is asymptotic to $x/(2log x)$ (and computation seems to bear this out). But I agree it's hopeless to expect a proof that there are infinitely many prime $S_n$.
    $endgroup$
    – Greg Martin
    Jan 13 '14 at 3:18






  • 5




    $begingroup$
    oeis.org/A013918
    $endgroup$
    – user940
    Jan 13 '14 at 3:30






  • 2




    $begingroup$
    Related: mathoverflow.net/q/153656/12357
    $endgroup$
    – Joel Reyes Noche
    Jan 13 '14 at 3:37














20












20








20


7



$begingroup$


Starting from 2, the sequence of sums of all consecutive primes is:



$$begin{array}{lcl}2 &=& 2\
2+3 &=& 5 \
2+3+5 &=& 10 \
2+3+5+7 &=& 17 \
2+3+5+7+11 &=& 28 \
&vdots&
end{array}
$$
If the $n^text{th}$ prime is $P_n$, then we can write $S_n=sum_{i=1}^n P_n$.



I conjecture that the sequence $S_n$ contains an infinite number of primes.



I doubt I'm the first person ever to think this, but I cannot find reference to the idea, nor can I conceive a proof, nor a disproof. Computationally, it can be verified that
$$S_{13,932}=998,658,581=P_{50,783,012},$$
and the sequence shows no sign of slowing down.



Can you prove the sequence $S_n$ contains an infinite number of primes?






sequences-and-series number-theory prime-numbers summation







share|cite|improve this question











$endgroup$




Starting from 2, the sequence of sums of all consecutive primes is:



$$begin{array}{lcl}2 &=& 2\
2+3 &=& 5 \
2+3+5 &=& 10 \
2+3+5+7 &=& 17 \
2+3+5+7+11 &=& 28 \
&vdots&
end{array}
$$
If the $n^text{th}$ prime is $P_n$, then we can write $S_n=sum_{i=1}^n P_n$.



I conjecture that the sequence $S_n$ contains an infinite number of primes.



I doubt I'm the first person ever to think this, but I cannot find reference to the idea, nor can I conceive a proof, nor a disproof. Computationally, it can be verified that
$$S_{13,932}=998,658,581=P_{50,783,012},$$
and the sequence shows no sign of slowing down.



Can you prove the sequence $S_n$ contains an infinite number of primes?






sequences-and-series number-theory prime-numbers summation




sequences-and-series number-theory prime-numbers summation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 13 '14 at 12:49







Jeff Snider

















asked Jan 13 '14 at 2:48









Jeff SniderJeff Snider

2,235919




2,235919








  • 8




    $begingroup$
    very hard to believe a proof is possible, when we can not prove that there are infinitely many primes $n^2 +1$
    $endgroup$
    – Will Jagy
    Jan 13 '14 at 2:55






  • 3




    $begingroup$
    I am surprised that the sequence 2,5,17,29,59,101 does not appear in the OEIS. I would recommend adding it.
    $endgroup$
    – RghtHndSd
    Jan 13 '14 at 3:14






  • 5




    $begingroup$
    Standard heuristics would predict that the number of $nle x$ for which $S_n$ is prime is asymptotic to $x/(2log x)$ (and computation seems to bear this out). But I agree it's hopeless to expect a proof that there are infinitely many prime $S_n$.
    $endgroup$
    – Greg Martin
    Jan 13 '14 at 3:18






  • 5




    $begingroup$
    oeis.org/A013918
    $endgroup$
    – user940
    Jan 13 '14 at 3:30






  • 2




    $begingroup$
    Related: mathoverflow.net/q/153656/12357
    $endgroup$
    – Joel Reyes Noche
    Jan 13 '14 at 3:37














  • 8




    $begingroup$
    very hard to believe a proof is possible, when we can not prove that there are infinitely many primes $n^2 +1$
    $endgroup$
    – Will Jagy
    Jan 13 '14 at 2:55






  • 3




    $begingroup$
    I am surprised that the sequence 2,5,17,29,59,101 does not appear in the OEIS. I would recommend adding it.
    $endgroup$
    – RghtHndSd
    Jan 13 '14 at 3:14






  • 5




    $begingroup$
    Standard heuristics would predict that the number of $nle x$ for which $S_n$ is prime is asymptotic to $x/(2log x)$ (and computation seems to bear this out). But I agree it's hopeless to expect a proof that there are infinitely many prime $S_n$.
    $endgroup$
    – Greg Martin
    Jan 13 '14 at 3:18






  • 5




    $begingroup$
    oeis.org/A013918
    $endgroup$
    – user940
    Jan 13 '14 at 3:30






  • 2




    $begingroup$
    Related: mathoverflow.net/q/153656/12357
    $endgroup$
    – Joel Reyes Noche
    Jan 13 '14 at 3:37








8




8




$begingroup$
very hard to believe a proof is possible, when we can not prove that there are infinitely many primes $n^2 +1$
$endgroup$
– Will Jagy
Jan 13 '14 at 2:55




$begingroup$
very hard to believe a proof is possible, when we can not prove that there are infinitely many primes $n^2 +1$
$endgroup$
– Will Jagy
Jan 13 '14 at 2:55




3




3




$begingroup$
I am surprised that the sequence 2,5,17,29,59,101 does not appear in the OEIS. I would recommend adding it.
$endgroup$
– RghtHndSd
Jan 13 '14 at 3:14




$begingroup$
I am surprised that the sequence 2,5,17,29,59,101 does not appear in the OEIS. I would recommend adding it.
$endgroup$
– RghtHndSd
Jan 13 '14 at 3:14




5




5




$begingroup$
Standard heuristics would predict that the number of $nle x$ for which $S_n$ is prime is asymptotic to $x/(2log x)$ (and computation seems to bear this out). But I agree it's hopeless to expect a proof that there are infinitely many prime $S_n$.
$endgroup$
– Greg Martin
Jan 13 '14 at 3:18




$begingroup$
Standard heuristics would predict that the number of $nle x$ for which $S_n$ is prime is asymptotic to $x/(2log x)$ (and computation seems to bear this out). But I agree it's hopeless to expect a proof that there are infinitely many prime $S_n$.
$endgroup$
– Greg Martin
Jan 13 '14 at 3:18




5




5




$begingroup$
oeis.org/A013918
$endgroup$
– user940
Jan 13 '14 at 3:30




$begingroup$
oeis.org/A013918
$endgroup$
– user940
Jan 13 '14 at 3:30




2




2




$begingroup$
Related: mathoverflow.net/q/153656/12357
$endgroup$
– Joel Reyes Noche
Jan 13 '14 at 3:37




$begingroup$
Related: mathoverflow.net/q/153656/12357
$endgroup$
– Joel Reyes Noche
Jan 13 '14 at 3:37










3 Answers
3






active

oldest

votes


















8












$begingroup$

As Will Jagy points out, a proof is probably out of reach, but the statement is probably true.



One can give a heuristic argument. The $n$-th prime is roughtly $nlog n$, so $S_n$ is roughly



$$S_n approx sum_{i=1}^n i log i approx frac{n^2(2log n - 1)}{4}.$$



The density of primes around $x$ is approximately $1/log x$, so the probability that $S_n$ is prime is roughly



$$1/logleft(frac{n^2(2log n - 1)}{4}right) approx frac{1}{2log n + log(2log n-1) - log 4} approx frac{1}{2log n}.$$



So the expected number of primes among $S_1, dots, S_n$ is roughly



$$sum_{i=1}^n frac{1}{2log i} approx frac{text{li}(n)}{2},$$



which goes to $infty$ as $nto infty$.



(I just saw that Greg Martin came to the same conclusion in the comments.)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The bound for $sum_{p leq n} p$ you gave is too rough, why not improve it by successive partial sums? Also, perhaps improve the probabilistic arguments by coming up with bounds instead of asymptotics? In any case, the above heuristic can be definitely improved.
    $endgroup$
    – Balarka Sen
    Jan 13 '14 at 8:50










  • $begingroup$
    @BalarkaSen No doubt it can be improved.
    $endgroup$
    – Bruno Joyal
    Jan 13 '14 at 13:56



















3












$begingroup$

The recent paper "Curious conjectures on the distribution of primes among the sums of the first $2n$ primes" by Romeo Meštrović at https://arxiv.org/abs/1804.04198 states your conjecture, along with certain details like the estimate of the number of primes, in section 3. This paper's other sections examine the distribution of primes, using heuristic, computational and analytic arguments that they are similar to the distribution of primes among the positive integers.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Nice reference!
    $endgroup$
    – Klangen
    Dec 17 '18 at 16:18



















0












$begingroup$

Looking for "prime partitions" might help to find prior information for this conjecture.



Another conjecture would be the sum of the gaps of consecutive primes, $g_n = p_{n+1} - p_n$: $$s = sum_{i=1}^{n}g_i,$$ having an infinite number of primes. Considering the fact that the next prime is $p_{n+1} = s + 2$ one might think it be easy, but it is not.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The sum you mentioned is $p_{n+1}-2$, the conjecture you mentioned is therefore equivalent to the twin prime conjecture.
    $endgroup$
    – Peter
    Sep 2 '15 at 10:15













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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









8












$begingroup$

As Will Jagy points out, a proof is probably out of reach, but the statement is probably true.



One can give a heuristic argument. The $n$-th prime is roughtly $nlog n$, so $S_n$ is roughly



$$S_n approx sum_{i=1}^n i log i approx frac{n^2(2log n - 1)}{4}.$$



The density of primes around $x$ is approximately $1/log x$, so the probability that $S_n$ is prime is roughly



$$1/logleft(frac{n^2(2log n - 1)}{4}right) approx frac{1}{2log n + log(2log n-1) - log 4} approx frac{1}{2log n}.$$



So the expected number of primes among $S_1, dots, S_n$ is roughly



$$sum_{i=1}^n frac{1}{2log i} approx frac{text{li}(n)}{2},$$



which goes to $infty$ as $nto infty$.



(I just saw that Greg Martin came to the same conclusion in the comments.)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The bound for $sum_{p leq n} p$ you gave is too rough, why not improve it by successive partial sums? Also, perhaps improve the probabilistic arguments by coming up with bounds instead of asymptotics? In any case, the above heuristic can be definitely improved.
    $endgroup$
    – Balarka Sen
    Jan 13 '14 at 8:50










  • $begingroup$
    @BalarkaSen No doubt it can be improved.
    $endgroup$
    – Bruno Joyal
    Jan 13 '14 at 13:56
















8












$begingroup$

As Will Jagy points out, a proof is probably out of reach, but the statement is probably true.



One can give a heuristic argument. The $n$-th prime is roughtly $nlog n$, so $S_n$ is roughly



$$S_n approx sum_{i=1}^n i log i approx frac{n^2(2log n - 1)}{4}.$$



The density of primes around $x$ is approximately $1/log x$, so the probability that $S_n$ is prime is roughly



$$1/logleft(frac{n^2(2log n - 1)}{4}right) approx frac{1}{2log n + log(2log n-1) - log 4} approx frac{1}{2log n}.$$



So the expected number of primes among $S_1, dots, S_n$ is roughly



$$sum_{i=1}^n frac{1}{2log i} approx frac{text{li}(n)}{2},$$



which goes to $infty$ as $nto infty$.



(I just saw that Greg Martin came to the same conclusion in the comments.)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The bound for $sum_{p leq n} p$ you gave is too rough, why not improve it by successive partial sums? Also, perhaps improve the probabilistic arguments by coming up with bounds instead of asymptotics? In any case, the above heuristic can be definitely improved.
    $endgroup$
    – Balarka Sen
    Jan 13 '14 at 8:50










  • $begingroup$
    @BalarkaSen No doubt it can be improved.
    $endgroup$
    – Bruno Joyal
    Jan 13 '14 at 13:56














8












8








8





$begingroup$

As Will Jagy points out, a proof is probably out of reach, but the statement is probably true.



One can give a heuristic argument. The $n$-th prime is roughtly $nlog n$, so $S_n$ is roughly



$$S_n approx sum_{i=1}^n i log i approx frac{n^2(2log n - 1)}{4}.$$



The density of primes around $x$ is approximately $1/log x$, so the probability that $S_n$ is prime is roughly



$$1/logleft(frac{n^2(2log n - 1)}{4}right) approx frac{1}{2log n + log(2log n-1) - log 4} approx frac{1}{2log n}.$$



So the expected number of primes among $S_1, dots, S_n$ is roughly



$$sum_{i=1}^n frac{1}{2log i} approx frac{text{li}(n)}{2},$$



which goes to $infty$ as $nto infty$.



(I just saw that Greg Martin came to the same conclusion in the comments.)






share|cite|improve this answer









$endgroup$



As Will Jagy points out, a proof is probably out of reach, but the statement is probably true.



One can give a heuristic argument. The $n$-th prime is roughtly $nlog n$, so $S_n$ is roughly



$$S_n approx sum_{i=1}^n i log i approx frac{n^2(2log n - 1)}{4}.$$



The density of primes around $x$ is approximately $1/log x$, so the probability that $S_n$ is prime is roughly



$$1/logleft(frac{n^2(2log n - 1)}{4}right) approx frac{1}{2log n + log(2log n-1) - log 4} approx frac{1}{2log n}.$$



So the expected number of primes among $S_1, dots, S_n$ is roughly



$$sum_{i=1}^n frac{1}{2log i} approx frac{text{li}(n)}{2},$$



which goes to $infty$ as $nto infty$.



(I just saw that Greg Martin came to the same conclusion in the comments.)







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 13 '14 at 3:30









Bruno JoyalBruno Joyal

42.9k695186




42.9k695186












  • $begingroup$
    The bound for $sum_{p leq n} p$ you gave is too rough, why not improve it by successive partial sums? Also, perhaps improve the probabilistic arguments by coming up with bounds instead of asymptotics? In any case, the above heuristic can be definitely improved.
    $endgroup$
    – Balarka Sen
    Jan 13 '14 at 8:50










  • $begingroup$
    @BalarkaSen No doubt it can be improved.
    $endgroup$
    – Bruno Joyal
    Jan 13 '14 at 13:56


















  • $begingroup$
    The bound for $sum_{p leq n} p$ you gave is too rough, why not improve it by successive partial sums? Also, perhaps improve the probabilistic arguments by coming up with bounds instead of asymptotics? In any case, the above heuristic can be definitely improved.
    $endgroup$
    – Balarka Sen
    Jan 13 '14 at 8:50










  • $begingroup$
    @BalarkaSen No doubt it can be improved.
    $endgroup$
    – Bruno Joyal
    Jan 13 '14 at 13:56
















$begingroup$
The bound for $sum_{p leq n} p$ you gave is too rough, why not improve it by successive partial sums? Also, perhaps improve the probabilistic arguments by coming up with bounds instead of asymptotics? In any case, the above heuristic can be definitely improved.
$endgroup$
– Balarka Sen
Jan 13 '14 at 8:50




$begingroup$
The bound for $sum_{p leq n} p$ you gave is too rough, why not improve it by successive partial sums? Also, perhaps improve the probabilistic arguments by coming up with bounds instead of asymptotics? In any case, the above heuristic can be definitely improved.
$endgroup$
– Balarka Sen
Jan 13 '14 at 8:50












$begingroup$
@BalarkaSen No doubt it can be improved.
$endgroup$
– Bruno Joyal
Jan 13 '14 at 13:56




$begingroup$
@BalarkaSen No doubt it can be improved.
$endgroup$
– Bruno Joyal
Jan 13 '14 at 13:56











3












$begingroup$

The recent paper "Curious conjectures on the distribution of primes among the sums of the first $2n$ primes" by Romeo Meštrović at https://arxiv.org/abs/1804.04198 states your conjecture, along with certain details like the estimate of the number of primes, in section 3. This paper's other sections examine the distribution of primes, using heuristic, computational and analytic arguments that they are similar to the distribution of primes among the positive integers.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Nice reference!
    $endgroup$
    – Klangen
    Dec 17 '18 at 16:18
















3












$begingroup$

The recent paper "Curious conjectures on the distribution of primes among the sums of the first $2n$ primes" by Romeo Meštrović at https://arxiv.org/abs/1804.04198 states your conjecture, along with certain details like the estimate of the number of primes, in section 3. This paper's other sections examine the distribution of primes, using heuristic, computational and analytic arguments that they are similar to the distribution of primes among the positive integers.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Nice reference!
    $endgroup$
    – Klangen
    Dec 17 '18 at 16:18














3












3








3





$begingroup$

The recent paper "Curious conjectures on the distribution of primes among the sums of the first $2n$ primes" by Romeo Meštrović at https://arxiv.org/abs/1804.04198 states your conjecture, along with certain details like the estimate of the number of primes, in section 3. This paper's other sections examine the distribution of primes, using heuristic, computational and analytic arguments that they are similar to the distribution of primes among the positive integers.






share|cite|improve this answer









$endgroup$



The recent paper "Curious conjectures on the distribution of primes among the sums of the first $2n$ primes" by Romeo Meštrović at https://arxiv.org/abs/1804.04198 states your conjecture, along with certain details like the estimate of the number of primes, in section 3. This paper's other sections examine the distribution of primes, using heuristic, computational and analytic arguments that they are similar to the distribution of primes among the positive integers.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 16 '18 at 22:37









John OmielanJohn Omielan

3,9251215




3,9251215








  • 1




    $begingroup$
    Nice reference!
    $endgroup$
    – Klangen
    Dec 17 '18 at 16:18














  • 1




    $begingroup$
    Nice reference!
    $endgroup$
    – Klangen
    Dec 17 '18 at 16:18








1




1




$begingroup$
Nice reference!
$endgroup$
– Klangen
Dec 17 '18 at 16:18




$begingroup$
Nice reference!
$endgroup$
– Klangen
Dec 17 '18 at 16:18











0












$begingroup$

Looking for "prime partitions" might help to find prior information for this conjecture.



Another conjecture would be the sum of the gaps of consecutive primes, $g_n = p_{n+1} - p_n$: $$s = sum_{i=1}^{n}g_i,$$ having an infinite number of primes. Considering the fact that the next prime is $p_{n+1} = s + 2$ one might think it be easy, but it is not.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The sum you mentioned is $p_{n+1}-2$, the conjecture you mentioned is therefore equivalent to the twin prime conjecture.
    $endgroup$
    – Peter
    Sep 2 '15 at 10:15


















0












$begingroup$

Looking for "prime partitions" might help to find prior information for this conjecture.



Another conjecture would be the sum of the gaps of consecutive primes, $g_n = p_{n+1} - p_n$: $$s = sum_{i=1}^{n}g_i,$$ having an infinite number of primes. Considering the fact that the next prime is $p_{n+1} = s + 2$ one might think it be easy, but it is not.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The sum you mentioned is $p_{n+1}-2$, the conjecture you mentioned is therefore equivalent to the twin prime conjecture.
    $endgroup$
    – Peter
    Sep 2 '15 at 10:15
















0












0








0





$begingroup$

Looking for "prime partitions" might help to find prior information for this conjecture.



Another conjecture would be the sum of the gaps of consecutive primes, $g_n = p_{n+1} - p_n$: $$s = sum_{i=1}^{n}g_i,$$ having an infinite number of primes. Considering the fact that the next prime is $p_{n+1} = s + 2$ one might think it be easy, but it is not.






share|cite|improve this answer









$endgroup$



Looking for "prime partitions" might help to find prior information for this conjecture.



Another conjecture would be the sum of the gaps of consecutive primes, $g_n = p_{n+1} - p_n$: $$s = sum_{i=1}^{n}g_i,$$ having an infinite number of primes. Considering the fact that the next prime is $p_{n+1} = s + 2$ one might think it be easy, but it is not.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Apr 16 '14 at 2:08









user160140user160140

778




778












  • $begingroup$
    The sum you mentioned is $p_{n+1}-2$, the conjecture you mentioned is therefore equivalent to the twin prime conjecture.
    $endgroup$
    – Peter
    Sep 2 '15 at 10:15




















  • $begingroup$
    The sum you mentioned is $p_{n+1}-2$, the conjecture you mentioned is therefore equivalent to the twin prime conjecture.
    $endgroup$
    – Peter
    Sep 2 '15 at 10:15


















$begingroup$
The sum you mentioned is $p_{n+1}-2$, the conjecture you mentioned is therefore equivalent to the twin prime conjecture.
$endgroup$
– Peter
Sep 2 '15 at 10:15






$begingroup$
The sum you mentioned is $p_{n+1}-2$, the conjecture you mentioned is therefore equivalent to the twin prime conjecture.
$endgroup$
– Peter
Sep 2 '15 at 10:15




















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