Does this infinite primes snake-product converge?












33












$begingroup$


Form an infinite product of prime ratios as follows.
Start with
$$
frac{2}{3}cdotfrac{7}{5}=frac{14}{15} approx 0.93 ;.
$$
Continue alternating a fraction $< 1$ times the next fraction $>1$,
progressively through the primes:
$$
frac{2}{3}cdotfrac{7}{5}cdotfrac{11}{13}cdotfrac{19}{17}
= frac{2926}{3315} approx 0.88 ;,
$$
$$
frac{2}{3}cdotfrac{7}{5}cdotfrac{11}{13}cdotfrac{19}{17}cdotfrac{23}{29}cdotfrac{37}{31}
=frac{2490026}{2980185} approx 0.83 ;.
$$
Continue this process to $infty$. One way to write the product is
$$
xi = prod_{1,5,9,ldots}^infty
frac{p_i}{p_{i+1}}cdotfrac{p_{i+3}}{p_{i+2}}
$$
where $p_i$ is the $i$-th prime.
I call this the primes snake-product:



         
Snake

My questions are:


Q1. Does the product converge?



Q2. If so, to what value $xi$ does it converge?




Up to the $1$-millionth prime ($15485863$),
the product is about $0.9056$:



         
Prod_1M

Update (26Jan2019): @Peter has calculated out to $p_i=10^{10}$ when the product is
$approx 0.9048$.




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







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Using the prime number theorem as an approximation, if you set $p_n=nln(n)$, does your product converge?
    $endgroup$
    – quarague
    Jan 25 at 11:55






  • 2




    $begingroup$
    $10^8$ th prime : $$0.90482881546cdots $$
    $endgroup$
    – Peter
    Jan 25 at 15:19






  • 2




    $begingroup$
    You could prove that the value of $prod_{i=1}^{infty} frac{p_{2n-1}}{p_{2n}} = P$ converges. Then, you can easily see that $P < xi < frac{1}{P}$, which would show that your product converges.
    $endgroup$
    – Haran
    Jan 25 at 16:58








  • 2




    $begingroup$
    @Haran: But perhaps $P=0$, and I'd be left with $0 < xi < infty$.
    $endgroup$
    – Joseph O'Rourke
    Jan 25 at 21:20






  • 2




    $begingroup$
    Since (as I note above) the products of $frac{p_i}{p_{i+1}}$ and of $frac{p_{i+3}}{p_{i+2}}$ do not converge absolutely, we are lead to considering cancelations between the terms. This is closely related to asymptotics of alternating sums of primes, of which we have only a limited understanding. See this MO question. I will elaborate in an answer.
    $endgroup$
    – Wojowu
    Jan 26 at 11:57


















33












$begingroup$


Form an infinite product of prime ratios as follows.
Start with
$$
frac{2}{3}cdotfrac{7}{5}=frac{14}{15} approx 0.93 ;.
$$
Continue alternating a fraction $< 1$ times the next fraction $>1$,
progressively through the primes:
$$
frac{2}{3}cdotfrac{7}{5}cdotfrac{11}{13}cdotfrac{19}{17}
= frac{2926}{3315} approx 0.88 ;,
$$
$$
frac{2}{3}cdotfrac{7}{5}cdotfrac{11}{13}cdotfrac{19}{17}cdotfrac{23}{29}cdotfrac{37}{31}
=frac{2490026}{2980185} approx 0.83 ;.
$$
Continue this process to $infty$. One way to write the product is
$$
xi = prod_{1,5,9,ldots}^infty
frac{p_i}{p_{i+1}}cdotfrac{p_{i+3}}{p_{i+2}}
$$
where $p_i$ is the $i$-th prime.
I call this the primes snake-product:



         
Snake

My questions are:


Q1. Does the product converge?



Q2. If so, to what value $xi$ does it converge?




Up to the $1$-millionth prime ($15485863$),
the product is about $0.9056$:



         
Prod_1M

Update (26Jan2019): @Peter has calculated out to $p_i=10^{10}$ when the product is
$approx 0.9048$.




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







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Using the prime number theorem as an approximation, if you set $p_n=nln(n)$, does your product converge?
    $endgroup$
    – quarague
    Jan 25 at 11:55






  • 2




    $begingroup$
    $10^8$ th prime : $$0.90482881546cdots $$
    $endgroup$
    – Peter
    Jan 25 at 15:19






  • 2




    $begingroup$
    You could prove that the value of $prod_{i=1}^{infty} frac{p_{2n-1}}{p_{2n}} = P$ converges. Then, you can easily see that $P < xi < frac{1}{P}$, which would show that your product converges.
    $endgroup$
    – Haran
    Jan 25 at 16:58








  • 2




    $begingroup$
    @Haran: But perhaps $P=0$, and I'd be left with $0 < xi < infty$.
    $endgroup$
    – Joseph O'Rourke
    Jan 25 at 21:20






  • 2




    $begingroup$
    Since (as I note above) the products of $frac{p_i}{p_{i+1}}$ and of $frac{p_{i+3}}{p_{i+2}}$ do not converge absolutely, we are lead to considering cancelations between the terms. This is closely related to asymptotics of alternating sums of primes, of which we have only a limited understanding. See this MO question. I will elaborate in an answer.
    $endgroup$
    – Wojowu
    Jan 26 at 11:57
















33












33








33


18



$begingroup$


Form an infinite product of prime ratios as follows.
Start with
$$
frac{2}{3}cdotfrac{7}{5}=frac{14}{15} approx 0.93 ;.
$$
Continue alternating a fraction $< 1$ times the next fraction $>1$,
progressively through the primes:
$$
frac{2}{3}cdotfrac{7}{5}cdotfrac{11}{13}cdotfrac{19}{17}
= frac{2926}{3315} approx 0.88 ;,
$$
$$
frac{2}{3}cdotfrac{7}{5}cdotfrac{11}{13}cdotfrac{19}{17}cdotfrac{23}{29}cdotfrac{37}{31}
=frac{2490026}{2980185} approx 0.83 ;.
$$
Continue this process to $infty$. One way to write the product is
$$
xi = prod_{1,5,9,ldots}^infty
frac{p_i}{p_{i+1}}cdotfrac{p_{i+3}}{p_{i+2}}
$$
where $p_i$ is the $i$-th prime.
I call this the primes snake-product:



         
Snake

My questions are:


Q1. Does the product converge?



Q2. If so, to what value $xi$ does it converge?




Up to the $1$-millionth prime ($15485863$),
the product is about $0.9056$:



         
Prod_1M

Update (26Jan2019): @Peter has calculated out to $p_i=10^{10}$ when the product is
$approx 0.9048$.




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







share|cite|improve this question











$endgroup$




Form an infinite product of prime ratios as follows.
Start with
$$
frac{2}{3}cdotfrac{7}{5}=frac{14}{15} approx 0.93 ;.
$$
Continue alternating a fraction $< 1$ times the next fraction $>1$,
progressively through the primes:
$$
frac{2}{3}cdotfrac{7}{5}cdotfrac{11}{13}cdotfrac{19}{17}
= frac{2926}{3315} approx 0.88 ;,
$$
$$
frac{2}{3}cdotfrac{7}{5}cdotfrac{11}{13}cdotfrac{19}{17}cdotfrac{23}{29}cdotfrac{37}{31}
=frac{2490026}{2980185} approx 0.83 ;.
$$
Continue this process to $infty$. One way to write the product is
$$
xi = prod_{1,5,9,ldots}^infty
frac{p_i}{p_{i+1}}cdotfrac{p_{i+3}}{p_{i+2}}
$$
where $p_i$ is the $i$-th prime.
I call this the primes snake-product:



         
Snake

My questions are:


Q1. Does the product converge?



Q2. If so, to what value $xi$ does it converge?




Up to the $1$-millionth prime ($15485863$),
the product is about $0.9056$:



         
Prod_1M

Update (26Jan2019): @Peter has calculated out to $p_i=10^{10}$ when the product is
$approx 0.9048$.




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




sequences-and-series number-theory elementary-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 Jan 26 at 13:26







Joseph O'Rourke

















asked Jan 25 at 11:39









Joseph O'RourkeJoseph O'Rourke

18.2k350112




18.2k350112








  • 3




    $begingroup$
    Using the prime number theorem as an approximation, if you set $p_n=nln(n)$, does your product converge?
    $endgroup$
    – quarague
    Jan 25 at 11:55






  • 2




    $begingroup$
    $10^8$ th prime : $$0.90482881546cdots $$
    $endgroup$
    – Peter
    Jan 25 at 15:19






  • 2




    $begingroup$
    You could prove that the value of $prod_{i=1}^{infty} frac{p_{2n-1}}{p_{2n}} = P$ converges. Then, you can easily see that $P < xi < frac{1}{P}$, which would show that your product converges.
    $endgroup$
    – Haran
    Jan 25 at 16:58








  • 2




    $begingroup$
    @Haran: But perhaps $P=0$, and I'd be left with $0 < xi < infty$.
    $endgroup$
    – Joseph O'Rourke
    Jan 25 at 21:20






  • 2




    $begingroup$
    Since (as I note above) the products of $frac{p_i}{p_{i+1}}$ and of $frac{p_{i+3}}{p_{i+2}}$ do not converge absolutely, we are lead to considering cancelations between the terms. This is closely related to asymptotics of alternating sums of primes, of which we have only a limited understanding. See this MO question. I will elaborate in an answer.
    $endgroup$
    – Wojowu
    Jan 26 at 11:57
















  • 3




    $begingroup$
    Using the prime number theorem as an approximation, if you set $p_n=nln(n)$, does your product converge?
    $endgroup$
    – quarague
    Jan 25 at 11:55






  • 2




    $begingroup$
    $10^8$ th prime : $$0.90482881546cdots $$
    $endgroup$
    – Peter
    Jan 25 at 15:19






  • 2




    $begingroup$
    You could prove that the value of $prod_{i=1}^{infty} frac{p_{2n-1}}{p_{2n}} = P$ converges. Then, you can easily see that $P < xi < frac{1}{P}$, which would show that your product converges.
    $endgroup$
    – Haran
    Jan 25 at 16:58








  • 2




    $begingroup$
    @Haran: But perhaps $P=0$, and I'd be left with $0 < xi < infty$.
    $endgroup$
    – Joseph O'Rourke
    Jan 25 at 21:20






  • 2




    $begingroup$
    Since (as I note above) the products of $frac{p_i}{p_{i+1}}$ and of $frac{p_{i+3}}{p_{i+2}}$ do not converge absolutely, we are lead to considering cancelations between the terms. This is closely related to asymptotics of alternating sums of primes, of which we have only a limited understanding. See this MO question. I will elaborate in an answer.
    $endgroup$
    – Wojowu
    Jan 26 at 11:57










3




3




$begingroup$
Using the prime number theorem as an approximation, if you set $p_n=nln(n)$, does your product converge?
$endgroup$
– quarague
Jan 25 at 11:55




$begingroup$
Using the prime number theorem as an approximation, if you set $p_n=nln(n)$, does your product converge?
$endgroup$
– quarague
Jan 25 at 11:55




2




2




$begingroup$
$10^8$ th prime : $$0.90482881546cdots $$
$endgroup$
– Peter
Jan 25 at 15:19




$begingroup$
$10^8$ th prime : $$0.90482881546cdots $$
$endgroup$
– Peter
Jan 25 at 15:19




2




2




$begingroup$
You could prove that the value of $prod_{i=1}^{infty} frac{p_{2n-1}}{p_{2n}} = P$ converges. Then, you can easily see that $P < xi < frac{1}{P}$, which would show that your product converges.
$endgroup$
– Haran
Jan 25 at 16:58






$begingroup$
You could prove that the value of $prod_{i=1}^{infty} frac{p_{2n-1}}{p_{2n}} = P$ converges. Then, you can easily see that $P < xi < frac{1}{P}$, which would show that your product converges.
$endgroup$
– Haran
Jan 25 at 16:58






2




2




$begingroup$
@Haran: But perhaps $P=0$, and I'd be left with $0 < xi < infty$.
$endgroup$
– Joseph O'Rourke
Jan 25 at 21:20




$begingroup$
@Haran: But perhaps $P=0$, and I'd be left with $0 < xi < infty$.
$endgroup$
– Joseph O'Rourke
Jan 25 at 21:20




2




2




$begingroup$
Since (as I note above) the products of $frac{p_i}{p_{i+1}}$ and of $frac{p_{i+3}}{p_{i+2}}$ do not converge absolutely, we are lead to considering cancelations between the terms. This is closely related to asymptotics of alternating sums of primes, of which we have only a limited understanding. See this MO question. I will elaborate in an answer.
$endgroup$
– Wojowu
Jan 26 at 11:57






$begingroup$
Since (as I note above) the products of $frac{p_i}{p_{i+1}}$ and of $frac{p_{i+3}}{p_{i+2}}$ do not converge absolutely, we are lead to considering cancelations between the terms. This is closely related to asymptotics of alternating sums of primes, of which we have only a limited understanding. See this MO question. I will elaborate in an answer.
$endgroup$
– Wojowu
Jan 26 at 11:57












1 Answer
1






active

oldest

votes


















16












$begingroup$

A very extended comment explaining why this problem is probably difficult.



Let $g_n=p_{2n}-p_{2n-1}$. The product we are looking at is then
$$prod_{n=1}^inftyleft(1-frac{g_n}{p_{2n}}right)^{(-1)^n}.$$
Taking logarithms, we are met with a sum of the form
$$sum_{n=1}^inftyleft((-1)^{n-1}frac{g_n}{p_{2n}}+Oleft(left(frac{g_n}{p_{2n}}right)^2right)right).$$
Using results due to Heath-Brown on second moments on prime gaps (see here), namely $sum_{k=1}^ng_k^2=O(x^{7/6+varepsilon})$, by summation by parts we can bound the sum of the error terms by a finite value.



Hence we are left with an alternating sum of $g_n/p_{2n}$. To deal with this, we essentially have to show the sums $sum_{n=1}^N(-1)^{n-1}g_n$ are asymptotically smaller than $p_{2N}$ (this won't guarantee convergence, but we definitely want that to hold). Using the notation of this MO answer (and the paper it cites), this sum is equal to $S(2N;1,4)-S(2N;3,4)$. What we would like to know is that this difference is $o(p_{2N})$. So you see we are quickly lead to investigating asymptotics of $S(N;a,q)$. Conjecturally, we have
$$S(N;a,q)simfrac{p_N}{q}$$
(so that the gaps are in some sense equidistributed), but available bounds are much weaker. In the cases we are interested in, we only get
$$liminffrac{S(N;a,4)}{p_N}geqfrac{1}{256}$$
unconditionally, and even conditionally on prime tuples conjecture we get $geq 1/32$, while what we would like is for the limits to exist and be equal.



Hence, as you can see, the available methods are not capable of showing that the difference $S(2N;1,4)-S(2N;3,4)$ is asymptotically small.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Thank you for your knowledgeable remarks. May I ask: Do you know if there a well-known conjecture that would imply the snake-product converges?
    $endgroup$
    – Joseph O'Rourke
    Jan 26 at 13:10








  • 1




    $begingroup$
    @JosephO'Rourke Not really, no. It should follow from suitable error term estimates for $S(N;a,q)$, but I'm not aware of such conjectures. I haven't exactly checked the references in the linked answer, you might want to look there.
    $endgroup$
    – Wojowu
    Jan 26 at 13:29






  • 3




    $begingroup$
    One of the strongest conjecture for the prime gap is that $g(n),n in [N,2N]$ looks like a random variable $G_N$ with $P(G_N > alog N) = e^{-a} log N$ and you are essentially asking $g(n)$ to be at least slightly independent in $n$ @JosephO'Rourke
    $endgroup$
    – reuns
    Jan 27 at 21:44













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1 Answer
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1 Answer
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active

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active

oldest

votes









16












$begingroup$

A very extended comment explaining why this problem is probably difficult.



Let $g_n=p_{2n}-p_{2n-1}$. The product we are looking at is then
$$prod_{n=1}^inftyleft(1-frac{g_n}{p_{2n}}right)^{(-1)^n}.$$
Taking logarithms, we are met with a sum of the form
$$sum_{n=1}^inftyleft((-1)^{n-1}frac{g_n}{p_{2n}}+Oleft(left(frac{g_n}{p_{2n}}right)^2right)right).$$
Using results due to Heath-Brown on second moments on prime gaps (see here), namely $sum_{k=1}^ng_k^2=O(x^{7/6+varepsilon})$, by summation by parts we can bound the sum of the error terms by a finite value.



Hence we are left with an alternating sum of $g_n/p_{2n}$. To deal with this, we essentially have to show the sums $sum_{n=1}^N(-1)^{n-1}g_n$ are asymptotically smaller than $p_{2N}$ (this won't guarantee convergence, but we definitely want that to hold). Using the notation of this MO answer (and the paper it cites), this sum is equal to $S(2N;1,4)-S(2N;3,4)$. What we would like to know is that this difference is $o(p_{2N})$. So you see we are quickly lead to investigating asymptotics of $S(N;a,q)$. Conjecturally, we have
$$S(N;a,q)simfrac{p_N}{q}$$
(so that the gaps are in some sense equidistributed), but available bounds are much weaker. In the cases we are interested in, we only get
$$liminffrac{S(N;a,4)}{p_N}geqfrac{1}{256}$$
unconditionally, and even conditionally on prime tuples conjecture we get $geq 1/32$, while what we would like is for the limits to exist and be equal.



Hence, as you can see, the available methods are not capable of showing that the difference $S(2N;1,4)-S(2N;3,4)$ is asymptotically small.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Thank you for your knowledgeable remarks. May I ask: Do you know if there a well-known conjecture that would imply the snake-product converges?
    $endgroup$
    – Joseph O'Rourke
    Jan 26 at 13:10








  • 1




    $begingroup$
    @JosephO'Rourke Not really, no. It should follow from suitable error term estimates for $S(N;a,q)$, but I'm not aware of such conjectures. I haven't exactly checked the references in the linked answer, you might want to look there.
    $endgroup$
    – Wojowu
    Jan 26 at 13:29






  • 3




    $begingroup$
    One of the strongest conjecture for the prime gap is that $g(n),n in [N,2N]$ looks like a random variable $G_N$ with $P(G_N > alog N) = e^{-a} log N$ and you are essentially asking $g(n)$ to be at least slightly independent in $n$ @JosephO'Rourke
    $endgroup$
    – reuns
    Jan 27 at 21:44


















16












$begingroup$

A very extended comment explaining why this problem is probably difficult.



Let $g_n=p_{2n}-p_{2n-1}$. The product we are looking at is then
$$prod_{n=1}^inftyleft(1-frac{g_n}{p_{2n}}right)^{(-1)^n}.$$
Taking logarithms, we are met with a sum of the form
$$sum_{n=1}^inftyleft((-1)^{n-1}frac{g_n}{p_{2n}}+Oleft(left(frac{g_n}{p_{2n}}right)^2right)right).$$
Using results due to Heath-Brown on second moments on prime gaps (see here), namely $sum_{k=1}^ng_k^2=O(x^{7/6+varepsilon})$, by summation by parts we can bound the sum of the error terms by a finite value.



Hence we are left with an alternating sum of $g_n/p_{2n}$. To deal with this, we essentially have to show the sums $sum_{n=1}^N(-1)^{n-1}g_n$ are asymptotically smaller than $p_{2N}$ (this won't guarantee convergence, but we definitely want that to hold). Using the notation of this MO answer (and the paper it cites), this sum is equal to $S(2N;1,4)-S(2N;3,4)$. What we would like to know is that this difference is $o(p_{2N})$. So you see we are quickly lead to investigating asymptotics of $S(N;a,q)$. Conjecturally, we have
$$S(N;a,q)simfrac{p_N}{q}$$
(so that the gaps are in some sense equidistributed), but available bounds are much weaker. In the cases we are interested in, we only get
$$liminffrac{S(N;a,4)}{p_N}geqfrac{1}{256}$$
unconditionally, and even conditionally on prime tuples conjecture we get $geq 1/32$, while what we would like is for the limits to exist and be equal.



Hence, as you can see, the available methods are not capable of showing that the difference $S(2N;1,4)-S(2N;3,4)$ is asymptotically small.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Thank you for your knowledgeable remarks. May I ask: Do you know if there a well-known conjecture that would imply the snake-product converges?
    $endgroup$
    – Joseph O'Rourke
    Jan 26 at 13:10








  • 1




    $begingroup$
    @JosephO'Rourke Not really, no. It should follow from suitable error term estimates for $S(N;a,q)$, but I'm not aware of such conjectures. I haven't exactly checked the references in the linked answer, you might want to look there.
    $endgroup$
    – Wojowu
    Jan 26 at 13:29






  • 3




    $begingroup$
    One of the strongest conjecture for the prime gap is that $g(n),n in [N,2N]$ looks like a random variable $G_N$ with $P(G_N > alog N) = e^{-a} log N$ and you are essentially asking $g(n)$ to be at least slightly independent in $n$ @JosephO'Rourke
    $endgroup$
    – reuns
    Jan 27 at 21:44
















16












16








16





$begingroup$

A very extended comment explaining why this problem is probably difficult.



Let $g_n=p_{2n}-p_{2n-1}$. The product we are looking at is then
$$prod_{n=1}^inftyleft(1-frac{g_n}{p_{2n}}right)^{(-1)^n}.$$
Taking logarithms, we are met with a sum of the form
$$sum_{n=1}^inftyleft((-1)^{n-1}frac{g_n}{p_{2n}}+Oleft(left(frac{g_n}{p_{2n}}right)^2right)right).$$
Using results due to Heath-Brown on second moments on prime gaps (see here), namely $sum_{k=1}^ng_k^2=O(x^{7/6+varepsilon})$, by summation by parts we can bound the sum of the error terms by a finite value.



Hence we are left with an alternating sum of $g_n/p_{2n}$. To deal with this, we essentially have to show the sums $sum_{n=1}^N(-1)^{n-1}g_n$ are asymptotically smaller than $p_{2N}$ (this won't guarantee convergence, but we definitely want that to hold). Using the notation of this MO answer (and the paper it cites), this sum is equal to $S(2N;1,4)-S(2N;3,4)$. What we would like to know is that this difference is $o(p_{2N})$. So you see we are quickly lead to investigating asymptotics of $S(N;a,q)$. Conjecturally, we have
$$S(N;a,q)simfrac{p_N}{q}$$
(so that the gaps are in some sense equidistributed), but available bounds are much weaker. In the cases we are interested in, we only get
$$liminffrac{S(N;a,4)}{p_N}geqfrac{1}{256}$$
unconditionally, and even conditionally on prime tuples conjecture we get $geq 1/32$, while what we would like is for the limits to exist and be equal.



Hence, as you can see, the available methods are not capable of showing that the difference $S(2N;1,4)-S(2N;3,4)$ is asymptotically small.






share|cite|improve this answer











$endgroup$



A very extended comment explaining why this problem is probably difficult.



Let $g_n=p_{2n}-p_{2n-1}$. The product we are looking at is then
$$prod_{n=1}^inftyleft(1-frac{g_n}{p_{2n}}right)^{(-1)^n}.$$
Taking logarithms, we are met with a sum of the form
$$sum_{n=1}^inftyleft((-1)^{n-1}frac{g_n}{p_{2n}}+Oleft(left(frac{g_n}{p_{2n}}right)^2right)right).$$
Using results due to Heath-Brown on second moments on prime gaps (see here), namely $sum_{k=1}^ng_k^2=O(x^{7/6+varepsilon})$, by summation by parts we can bound the sum of the error terms by a finite value.



Hence we are left with an alternating sum of $g_n/p_{2n}$. To deal with this, we essentially have to show the sums $sum_{n=1}^N(-1)^{n-1}g_n$ are asymptotically smaller than $p_{2N}$ (this won't guarantee convergence, but we definitely want that to hold). Using the notation of this MO answer (and the paper it cites), this sum is equal to $S(2N;1,4)-S(2N;3,4)$. What we would like to know is that this difference is $o(p_{2N})$. So you see we are quickly lead to investigating asymptotics of $S(N;a,q)$. Conjecturally, we have
$$S(N;a,q)simfrac{p_N}{q}$$
(so that the gaps are in some sense equidistributed), but available bounds are much weaker. In the cases we are interested in, we only get
$$liminffrac{S(N;a,4)}{p_N}geqfrac{1}{256}$$
unconditionally, and even conditionally on prime tuples conjecture we get $geq 1/32$, while what we would like is for the limits to exist and be equal.



Hence, as you can see, the available methods are not capable of showing that the difference $S(2N;1,4)-S(2N;3,4)$ is asymptotically small.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 26 at 16:15

























answered Jan 26 at 12:46









WojowuWojowu

18.9k23173




18.9k23173








  • 1




    $begingroup$
    Thank you for your knowledgeable remarks. May I ask: Do you know if there a well-known conjecture that would imply the snake-product converges?
    $endgroup$
    – Joseph O'Rourke
    Jan 26 at 13:10








  • 1




    $begingroup$
    @JosephO'Rourke Not really, no. It should follow from suitable error term estimates for $S(N;a,q)$, but I'm not aware of such conjectures. I haven't exactly checked the references in the linked answer, you might want to look there.
    $endgroup$
    – Wojowu
    Jan 26 at 13:29






  • 3




    $begingroup$
    One of the strongest conjecture for the prime gap is that $g(n),n in [N,2N]$ looks like a random variable $G_N$ with $P(G_N > alog N) = e^{-a} log N$ and you are essentially asking $g(n)$ to be at least slightly independent in $n$ @JosephO'Rourke
    $endgroup$
    – reuns
    Jan 27 at 21:44
















  • 1




    $begingroup$
    Thank you for your knowledgeable remarks. May I ask: Do you know if there a well-known conjecture that would imply the snake-product converges?
    $endgroup$
    – Joseph O'Rourke
    Jan 26 at 13:10








  • 1




    $begingroup$
    @JosephO'Rourke Not really, no. It should follow from suitable error term estimates for $S(N;a,q)$, but I'm not aware of such conjectures. I haven't exactly checked the references in the linked answer, you might want to look there.
    $endgroup$
    – Wojowu
    Jan 26 at 13:29






  • 3




    $begingroup$
    One of the strongest conjecture for the prime gap is that $g(n),n in [N,2N]$ looks like a random variable $G_N$ with $P(G_N > alog N) = e^{-a} log N$ and you are essentially asking $g(n)$ to be at least slightly independent in $n$ @JosephO'Rourke
    $endgroup$
    – reuns
    Jan 27 at 21:44










1




1




$begingroup$
Thank you for your knowledgeable remarks. May I ask: Do you know if there a well-known conjecture that would imply the snake-product converges?
$endgroup$
– Joseph O'Rourke
Jan 26 at 13:10






$begingroup$
Thank you for your knowledgeable remarks. May I ask: Do you know if there a well-known conjecture that would imply the snake-product converges?
$endgroup$
– Joseph O'Rourke
Jan 26 at 13:10






1




1




$begingroup$
@JosephO'Rourke Not really, no. It should follow from suitable error term estimates for $S(N;a,q)$, but I'm not aware of such conjectures. I haven't exactly checked the references in the linked answer, you might want to look there.
$endgroup$
– Wojowu
Jan 26 at 13:29




$begingroup$
@JosephO'Rourke Not really, no. It should follow from suitable error term estimates for $S(N;a,q)$, but I'm not aware of such conjectures. I haven't exactly checked the references in the linked answer, you might want to look there.
$endgroup$
– Wojowu
Jan 26 at 13:29




3




3




$begingroup$
One of the strongest conjecture for the prime gap is that $g(n),n in [N,2N]$ looks like a random variable $G_N$ with $P(G_N > alog N) = e^{-a} log N$ and you are essentially asking $g(n)$ to be at least slightly independent in $n$ @JosephO'Rourke
$endgroup$
– reuns
Jan 27 at 21:44






$begingroup$
One of the strongest conjecture for the prime gap is that $g(n),n in [N,2N]$ looks like a random variable $G_N$ with $P(G_N > alog N) = e^{-a} log N$ and you are essentially asking $g(n)$ to be at least slightly independent in $n$ @JosephO'Rourke
$endgroup$
– reuns
Jan 27 at 21:44




















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