Do the primes contain an infinite almost arithmetic progression?Constructing arithmetic progressionsProve that $sqrt[3]p$, $sqrt[3]q$ and $sqrt[3]r$ cannot be in the same arithmetic progressionA question on Primes in Arithmetic ProgressionWhich progressions and sequences are guaranteed to contain infinitely many primes?Primes and arithmetic progressionsWeak form of Dirichlet's theorem.Prime Factors of the Composit Terms of Arithmetic ProgressionsA question about arithmetic progressions and prime numbersArithmetic Progressions of PrimesA Question about the Green-Tao Theorem on Arithmetic Progressions in Primes
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Do the primes contain an infinite almost arithmetic progression?
Constructing arithmetic progressionsProve that $sqrt[3]p$, $sqrt[3]q$ and $sqrt[3]r$ cannot be in the same arithmetic progressionA question on Primes in Arithmetic ProgressionWhich progressions and sequences are guaranteed to contain infinitely many primes?Primes and arithmetic progressionsWeak form of Dirichlet's theorem.Prime Factors of the Composit Terms of Arithmetic ProgressionsA question about arithmetic progressions and prime numbersArithmetic Progressions of PrimesA Question about the Green-Tao Theorem on Arithmetic Progressions in Primes
$begingroup$
The primes contain finite arithmetic progressions of arbitrary length, but not an infinite arithmetic progression. Say we define an almost arithmetic progression to be a sequence $a_k$, $k geq 0$, such that there exist $a,d$ such that $a_k = a+kd + O(sqrtk)$. Do the primes contain an infinite almost arithmetic progression?
(The definition is ad hoc and just made up out of curiosity. I wrote the “error term” $O(sqrtk)$ in analogy to the expectation of a random walk. An obvious generalization of the question is to replace this with some other “small” error term like $O(ln k)$ or whatever.)
number-theory prime-numbers
asked 2 hours ago
Zach TeitlerZach Teitler
2,261419
$endgroup$
add a comment |
$begingroup$
The primes contain finite arithmetic progressions of arbitrary length, but not an infinite arithmetic progression. Say we define an almost arithmetic progression to be a sequence $a_k$, $k geq 0$, such that there exist $a,d$ such that $a_k = a+kd + O(sqrtk)$. Do the primes contain an infinite almost arithmetic progression?
(The definition is ad hoc and just made up out of curiosity. I wrote the “error term” $O(sqrtk)$ in analogy to the expectation of a random walk. An obvious generalization of the question is to replace this with some other “small” error term like $O(ln k)$ or whatever.)
number-theory prime-numbers
asked 2 hours ago
Zach TeitlerZach Teitler
2,261419
$endgroup$
1
$begingroup$
Do you require the terms to be distinct? Also, since the gaps are allowed to grow unboundedly large, in what sense is this kind of sequence similar to an arithmetical progression? If the gaps are allowed to grow, then as Wojowu writes you are really just asking for the asymptotic growth rate of the primes, in a sense.
$endgroup$
– Carl Mummert
1 hour ago
add a comment |
$begingroup$
The primes contain finite arithmetic progressions of arbitrary length, but not an infinite arithmetic progression. Say we define an almost arithmetic progression to be a sequence $a_k$, $k geq 0$, such that there exist $a,d$ such that $a_k = a+kd + O(sqrtk)$. Do the primes contain an infinite almost arithmetic progression?
(The definition is ad hoc and just made up out of curiosity. I wrote the “error term” $O(sqrtk)$ in analogy to the expectation of a random walk. An obvious generalization of the question is to replace this with some other “small” error term like $O(ln k)$ or whatever.)
number-theory prime-numbers
asked 2 hours ago
Zach TeitlerZach Teitler
2,261419
$endgroup$
The primes contain finite arithmetic progressions of arbitrary length, but not an infinite arithmetic progression. Say we define an almost arithmetic progression to be a sequence $a_k$, $k geq 0$, such that there exist $a,d$ such that $a_k = a+kd + O(sqrtk)$. Do the primes contain an infinite almost arithmetic progression?
(The definition is ad hoc and just made up out of curiosity. I wrote the “error term” $O(sqrtk)$ in analogy to the expectation of a random walk. An obvious generalization of the question is to replace this with some other “small” error term like $O(ln k)$ or whatever.)
number-theory prime-numbers
number-theory prime-numbers
asked 2 hours ago
Zach TeitlerZach Teitler
2,261419
asked 2 hours ago
Zach TeitlerZach Teitler
2,261419
asked 2 hours ago
Zach TeitlerZach Teitler
2,261419
asked 2 hours ago
Zach TeitlerZach Teitler
2,261419
asked 2 hours ago
Zach TeitlerZach Teitler
2,261419
2,261419
1
$begingroup$
Do you require the terms to be distinct? Also, since the gaps are allowed to grow unboundedly large, in what sense is this kind of sequence similar to an arithmetical progression? If the gaps are allowed to grow, then as Wojowu writes you are really just asking for the asymptotic growth rate of the primes, in a sense.
$endgroup$
– Carl Mummert
1 hour ago
add a comment |
1
$begingroup$
Do you require the terms to be distinct? Also, since the gaps are allowed to grow unboundedly large, in what sense is this kind of sequence similar to an arithmetical progression? If the gaps are allowed to grow, then as Wojowu writes you are really just asking for the asymptotic growth rate of the primes, in a sense.
$endgroup$
– Carl Mummert
1 hour ago
1
1
$begingroup$
Do you require the terms to be distinct? Also, since the gaps are allowed to grow unboundedly large, in what sense is this kind of sequence similar to an arithmetical progression? If the gaps are allowed to grow, then as Wojowu writes you are really just asking for the asymptotic growth rate of the primes, in a sense.
$endgroup$
– Carl Mummert
1 hour ago
$begingroup$
Do you require the terms to be distinct? Also, since the gaps are allowed to grow unboundedly large, in what sense is this kind of sequence similar to an arithmetical progression? If the gaps are allowed to grow, then as Wojowu writes you are really just asking for the asymptotic growth rate of the primes, in a sense.
$endgroup$
– Carl Mummert
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note: I allow for $a_k$ to repeat. If you wish to keep them distinct, the answer by N. S. gives a negative answer.
The answer is: we don't know, but probably. Indeed, your question is equivalent to the question of whether the prime gaps satisfy $p_n+1-p_n=O(sqrtp_n)$, as I explain below. This bound is widely believed to hold - indeed, the running conjecture (Cramér's conjecture) is that we even have $p_n+1-p_n=O((log p_n)^2)$, but we are quite far from proving it. The best unconditional bound we have is $p_n+1-p_n=O(p_n^0.525)$, and assuming the Riemann hypothesis we can push this to $O(p_n^1/2+varepsilon)$, but not to $O(sqrtp_n)$. (see Wikipedia)
To see the equivalence, take some infinite almost arithmetic progression. Suppose the $O(sqrtk)$ term is bounded by $Msqrtk$. For any prime $p_n$, let $k$ be the least such that $a+dk>p_n+Msqrtk$. Then $p_n+1leq a_kleq a+dk+Msqrtkleq d+p_n+Msqrtk+Msqrtk=p_n+O(sqrtk)=p_n+O(sqrtp_n)$.
Conversely, suppose gaps between primes are $O(sqrtp_n)$. Take $a=0,d=1$ and $a_k$ the smallest prime greater than $k$. The assumption trivially implies $a_k=k+O(sqrtk)$.
answered 1 hour ago
WojowuWojowu
19k23173
$endgroup$
$begingroup$
Thanks. So conjecturally $p_n = a + nd + O((log n)^2)$, but the "$a+nd$" part is boring, it can just be $a=0,d=1$ (and not $O(log n)$, that is too strong I take it).
$endgroup$
– Zach Teitler
1 hour ago
add a comment |
$begingroup$
The answer is NO.
Note that any set $S$ containing an infinite almost arithmetic progression satisfies
$$
underlinemboxdens(S):= liminf_n frac(mboxcard( S cap [0,n]))n geq frac1d$$
But the primes have density $0$.
answered 2 hours ago
N. S.N. S.
105k7114209
$endgroup$
$begingroup$
I don't think the question requires $a_k$ to be distinct.
$endgroup$
– Wojowu
2 hours ago
$begingroup$
@CarlMummert $a+kd$ has gaps of size $d$. For any $k>d^2$, it is easily possible for two terms to coincide.
$endgroup$
– Wojowu
1 hour ago
$begingroup$
Thanks; I was reading it as $O(sqrtd)$. This makes the question more confusing overall, because if the resulting sequence might not be syndetic, it's not really very similar to an arithmetical progression.
$endgroup$
– Carl Mummert
1 hour ago
$begingroup$
@CarlMummert True! I didn't really think it through.
$endgroup$
– Zach Teitler
1 hour ago
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Note: I allow for $a_k$ to repeat. If you wish to keep them distinct, the answer by N. S. gives a negative answer.
The answer is: we don't know, but probably. Indeed, your question is equivalent to the question of whether the prime gaps satisfy $p_n+1-p_n=O(sqrtp_n)$, as I explain below. This bound is widely believed to hold - indeed, the running conjecture (Cramér's conjecture) is that we even have $p_n+1-p_n=O((log p_n)^2)$, but we are quite far from proving it. The best unconditional bound we have is $p_n+1-p_n=O(p_n^0.525)$, and assuming the Riemann hypothesis we can push this to $O(p_n^1/2+varepsilon)$, but not to $O(sqrtp_n)$. (see Wikipedia)
To see the equivalence, take some infinite almost arithmetic progression. Suppose the $O(sqrtk)$ term is bounded by $Msqrtk$. For any prime $p_n$, let $k$ be the least such that $a+dk>p_n+Msqrtk$. Then $p_n+1leq a_kleq a+dk+Msqrtkleq d+p_n+Msqrtk+Msqrtk=p_n+O(sqrtk)=p_n+O(sqrtp_n)$.
Conversely, suppose gaps between primes are $O(sqrtp_n)$. Take $a=0,d=1$ and $a_k$ the smallest prime greater than $k$. The assumption trivially implies $a_k=k+O(sqrtk)$.
answered 1 hour ago
WojowuWojowu
19k23173
$endgroup$
$begingroup$
Thanks. So conjecturally $p_n = a + nd + O((log n)^2)$, but the "$a+nd$" part is boring, it can just be $a=0,d=1$ (and not $O(log n)$, that is too strong I take it).
$endgroup$
– Zach Teitler
1 hour ago
add a comment |
$begingroup$
Note: I allow for $a_k$ to repeat. If you wish to keep them distinct, the answer by N. S. gives a negative answer.
The answer is: we don't know, but probably. Indeed, your question is equivalent to the question of whether the prime gaps satisfy $p_n+1-p_n=O(sqrtp_n)$, as I explain below. This bound is widely believed to hold - indeed, the running conjecture (Cramér's conjecture) is that we even have $p_n+1-p_n=O((log p_n)^2)$, but we are quite far from proving it. The best unconditional bound we have is $p_n+1-p_n=O(p_n^0.525)$, and assuming the Riemann hypothesis we can push this to $O(p_n^1/2+varepsilon)$, but not to $O(sqrtp_n)$. (see Wikipedia)
To see the equivalence, take some infinite almost arithmetic progression. Suppose the $O(sqrtk)$ term is bounded by $Msqrtk$. For any prime $p_n$, let $k$ be the least such that $a+dk>p_n+Msqrtk$. Then $p_n+1leq a_kleq a+dk+Msqrtkleq d+p_n+Msqrtk+Msqrtk=p_n+O(sqrtk)=p_n+O(sqrtp_n)$.
Conversely, suppose gaps between primes are $O(sqrtp_n)$. Take $a=0,d=1$ and $a_k$ the smallest prime greater than $k$. The assumption trivially implies $a_k=k+O(sqrtk)$.
answered 1 hour ago
WojowuWojowu
19k23173
$endgroup$
$begingroup$
Thanks. So conjecturally $p_n = a + nd + O((log n)^2)$, but the "$a+nd$" part is boring, it can just be $a=0,d=1$ (and not $O(log n)$, that is too strong I take it).
$endgroup$
– Zach Teitler
1 hour ago
add a comment |
$begingroup$
Note: I allow for $a_k$ to repeat. If you wish to keep them distinct, the answer by N. S. gives a negative answer.
The answer is: we don't know, but probably. Indeed, your question is equivalent to the question of whether the prime gaps satisfy $p_n+1-p_n=O(sqrtp_n)$, as I explain below. This bound is widely believed to hold - indeed, the running conjecture (Cramér's conjecture) is that we even have $p_n+1-p_n=O((log p_n)^2)$, but we are quite far from proving it. The best unconditional bound we have is $p_n+1-p_n=O(p_n^0.525)$, and assuming the Riemann hypothesis we can push this to $O(p_n^1/2+varepsilon)$, but not to $O(sqrtp_n)$. (see Wikipedia)
To see the equivalence, take some infinite almost arithmetic progression. Suppose the $O(sqrtk)$ term is bounded by $Msqrtk$. For any prime $p_n$, let $k$ be the least such that $a+dk>p_n+Msqrtk$. Then $p_n+1leq a_kleq a+dk+Msqrtkleq d+p_n+Msqrtk+Msqrtk=p_n+O(sqrtk)=p_n+O(sqrtp_n)$.
Conversely, suppose gaps between primes are $O(sqrtp_n)$. Take $a=0,d=1$ and $a_k$ the smallest prime greater than $k$. The assumption trivially implies $a_k=k+O(sqrtk)$.
answered 1 hour ago
WojowuWojowu
19k23173
$endgroup$
Note: I allow for $a_k$ to repeat. If you wish to keep them distinct, the answer by N. S. gives a negative answer.
The answer is: we don't know, but probably. Indeed, your question is equivalent to the question of whether the prime gaps satisfy $p_n+1-p_n=O(sqrtp_n)$, as I explain below. This bound is widely believed to hold - indeed, the running conjecture (Cramér's conjecture) is that we even have $p_n+1-p_n=O((log p_n)^2)$, but we are quite far from proving it. The best unconditional bound we have is $p_n+1-p_n=O(p_n^0.525)$, and assuming the Riemann hypothesis we can push this to $O(p_n^1/2+varepsilon)$, but not to $O(sqrtp_n)$. (see Wikipedia)
To see the equivalence, take some infinite almost arithmetic progression. Suppose the $O(sqrtk)$ term is bounded by $Msqrtk$. For any prime $p_n$, let $k$ be the least such that $a+dk>p_n+Msqrtk$. Then $p_n+1leq a_kleq a+dk+Msqrtkleq d+p_n+Msqrtk+Msqrtk=p_n+O(sqrtk)=p_n+O(sqrtp_n)$.
Conversely, suppose gaps between primes are $O(sqrtp_n)$. Take $a=0,d=1$ and $a_k$ the smallest prime greater than $k$. The assumption trivially implies $a_k=k+O(sqrtk)$.
answered 1 hour ago
WojowuWojowu
19k23173
answered 1 hour ago
WojowuWojowu
19k23173
answered 1 hour ago
WojowuWojowu
19k23173
answered 1 hour ago
WojowuWojowu
19k23173
19k23173
$begingroup$
Thanks. So conjecturally $p_n = a + nd + O((log n)^2)$, but the "$a+nd$" part is boring, it can just be $a=0,d=1$ (and not $O(log n)$, that is too strong I take it).
$endgroup$
– Zach Teitler
1 hour ago
add a comment |
$begingroup$
Thanks. So conjecturally $p_n = a + nd + O((log n)^2)$, but the "$a+nd$" part is boring, it can just be $a=0,d=1$ (and not $O(log n)$, that is too strong I take it).
$endgroup$
– Zach Teitler
1 hour ago
$begingroup$
Thanks. So conjecturally $p_n = a + nd + O((log n)^2)$, but the "$a+nd$" part is boring, it can just be $a=0,d=1$ (and not $O(log n)$, that is too strong I take it).
$endgroup$
– Zach Teitler
1 hour ago
$begingroup$
Thanks. So conjecturally $p_n = a + nd + O((log n)^2)$, but the "$a+nd$" part is boring, it can just be $a=0,d=1$ (and not $O(log n)$, that is too strong I take it).
$endgroup$
– Zach Teitler
1 hour ago
add a comment |
$begingroup$
The answer is NO.
Note that any set $S$ containing an infinite almost arithmetic progression satisfies
$$
underlinemboxdens(S):= liminf_n frac(mboxcard( S cap [0,n]))n geq frac1d$$
But the primes have density $0$.
answered 2 hours ago
N. S.N. S.
105k7114209
$endgroup$
$begingroup$
I don't think the question requires $a_k$ to be distinct.
$endgroup$
– Wojowu
2 hours ago
$begingroup$
@CarlMummert $a+kd$ has gaps of size $d$. For any $k>d^2$, it is easily possible for two terms to coincide.
$endgroup$
– Wojowu
1 hour ago
$begingroup$
Thanks; I was reading it as $O(sqrtd)$. This makes the question more confusing overall, because if the resulting sequence might not be syndetic, it's not really very similar to an arithmetical progression.
$endgroup$
– Carl Mummert
1 hour ago
$begingroup$
@CarlMummert True! I didn't really think it through.
$endgroup$
– Zach Teitler
1 hour ago
add a comment |
$begingroup$
The answer is NO.
Note that any set $S$ containing an infinite almost arithmetic progression satisfies
$$
underlinemboxdens(S):= liminf_n frac(mboxcard( S cap [0,n]))n geq frac1d$$
But the primes have density $0$.
answered 2 hours ago
N. S.N. S.
105k7114209
$endgroup$
$begingroup$
I don't think the question requires $a_k$ to be distinct.
$endgroup$
– Wojowu
2 hours ago
$begingroup$
@CarlMummert $a+kd$ has gaps of size $d$. For any $k>d^2$, it is easily possible for two terms to coincide.
$endgroup$
– Wojowu
1 hour ago
$begingroup$
Thanks; I was reading it as $O(sqrtd)$. This makes the question more confusing overall, because if the resulting sequence might not be syndetic, it's not really very similar to an arithmetical progression.
$endgroup$
– Carl Mummert
1 hour ago
$begingroup$
@CarlMummert True! I didn't really think it through.
$endgroup$
– Zach Teitler
1 hour ago
add a comment |
$begingroup$
The answer is NO.
Note that any set $S$ containing an infinite almost arithmetic progression satisfies
$$
underlinemboxdens(S):= liminf_n frac(mboxcard( S cap [0,n]))n geq frac1d$$
But the primes have density $0$.
answered 2 hours ago
N. S.N. S.
105k7114209
$endgroup$
The answer is NO.
Note that any set $S$ containing an infinite almost arithmetic progression satisfies
$$
underlinemboxdens(S):= liminf_n frac(mboxcard( S cap [0,n]))n geq frac1d$$
But the primes have density $0$.
answered 2 hours ago
N. S.N. S.
105k7114209
edited 2 hours ago
answered 2 hours ago
N. S.N. S.
105k7114209
answered 2 hours ago
N. S.N. S.
105k7114209
answered 2 hours ago
N. S.N. S.
105k7114209
105k7114209
$begingroup$
I don't think the question requires $a_k$ to be distinct.
$endgroup$
– Wojowu
2 hours ago
$begingroup$
@CarlMummert $a+kd$ has gaps of size $d$. For any $k>d^2$, it is easily possible for two terms to coincide.
$endgroup$
– Wojowu
1 hour ago
$begingroup$
Thanks; I was reading it as $O(sqrtd)$. This makes the question more confusing overall, because if the resulting sequence might not be syndetic, it's not really very similar to an arithmetical progression.
$endgroup$
– Carl Mummert
1 hour ago
$begingroup$
@CarlMummert True! I didn't really think it through.
$endgroup$
– Zach Teitler
1 hour ago
add a comment |
$begingroup$
I don't think the question requires $a_k$ to be distinct.
$endgroup$
– Wojowu
2 hours ago
$begingroup$
@CarlMummert $a+kd$ has gaps of size $d$. For any $k>d^2$, it is easily possible for two terms to coincide.
$endgroup$
– Wojowu
1 hour ago
$begingroup$
Thanks; I was reading it as $O(sqrtd)$. This makes the question more confusing overall, because if the resulting sequence might not be syndetic, it's not really very similar to an arithmetical progression.
$endgroup$
– Carl Mummert
1 hour ago
$begingroup$
@CarlMummert True! I didn't really think it through.
$endgroup$
– Zach Teitler
1 hour ago
$begingroup$
I don't think the question requires $a_k$ to be distinct.
$endgroup$
– Wojowu
2 hours ago
$begingroup$
I don't think the question requires $a_k$ to be distinct.
$endgroup$
– Wojowu
2 hours ago
$begingroup$
@CarlMummert $a+kd$ has gaps of size $d$. For any $k>d^2$, it is easily possible for two terms to coincide.
$endgroup$
– Wojowu
1 hour ago
$begingroup$
@CarlMummert $a+kd$ has gaps of size $d$. For any $k>d^2$, it is easily possible for two terms to coincide.
$endgroup$
– Wojowu
1 hour ago
$begingroup$
Thanks; I was reading it as $O(sqrtd)$. This makes the question more confusing overall, because if the resulting sequence might not be syndetic, it's not really very similar to an arithmetical progression.
$endgroup$
– Carl Mummert
1 hour ago
$begingroup$
Thanks; I was reading it as $O(sqrtd)$. This makes the question more confusing overall, because if the resulting sequence might not be syndetic, it's not really very similar to an arithmetical progression.
$endgroup$
– Carl Mummert
1 hour ago
$begingroup$
@CarlMummert True! I didn't really think it through.
$endgroup$
– Zach Teitler
1 hour ago
$begingroup$
@CarlMummert True! I didn't really think it through.
$endgroup$
– Zach Teitler
1 hour ago
add a comment |
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1
$begingroup$
Do you require the terms to be distinct? Also, since the gaps are allowed to grow unboundedly large, in what sense is this kind of sequence similar to an arithmetical progression? If the gaps are allowed to grow, then as Wojowu writes you are really just asking for the asymptotic growth rate of the primes, in a sense.
$endgroup$
– Carl Mummert
1 hour ago