Does the average primeness of natural numbers tend to zero? The 2019 Stack Overflow Developer Survey Results Are InSum of the standard deviation of the divisors of a numberMultiplying two averages, what to do with the deviations?Making fermat's little theorem for composite numbers the ultimate test.The definition of the sample standard deviationDistribution of the RSA numbersStandard Deviation Around an Arbitrary Meanconstructing primes without primality testDerivating natural numbers.Huge extremely rough provably-composite numbers with no known factorsRunning time of the probabilistic primality testWhy does $ operatornameVar(X) = E[X^2] - (E[X])^2 $
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Does the average primeness of natural numbers tend to zero?
The 2019 Stack Overflow Developer Survey Results Are InSum of the standard deviation of the divisors of a numberMultiplying two averages, what to do with the deviations?Making fermat's little theorem for composite numbers the ultimate test.The definition of the sample standard deviationDistribution of the RSA numbersStandard Deviation Around an Arbitrary Meanconstructing primes without primality testDerivating natural numbers.Huge extremely rough provably-composite numbers with no known factorsRunning time of the probabilistic primality testWhy does $ operatornameVar(X) = E[X^2] - (E[X])^2 $
$begingroup$
Note 1: This questions requires some new definitions, namely "continuous primeness" which I have made. Everyone is welcome to improve the definition without altering the spirit of the question. Click here for a somewhat related question.
A number is either prime or composite, hence primality is a binary concept. Instead I wanted to put a value of primality to every number using some function $f$ such that $f(n) = 1$ iff $n$ is a prime otherwise, $0 < f(n) < 1$ and as the number divisors of $n$ increases, $f(n)$ decreases on average. Thus $f(n)$ is a measure of the degree of primeness of $n$ where 1 is a perfect prime and 0 is a hypothetical perfect composite. Hence $frac1Nsum_r le N f(r)$ can be interpreted as a measure of average primeness of the first $N$ integers.
After trying several definitions and going through the ones in literature, I came up with:
Define $f(n) = dfrac2s_nn-1$ for $n ge 2$, where $s_n$ is the
standard deviation of the divisors of $n$.
One advantage of using standard deviation is that even if two numbers have the same number of divisor their value of $f$ appears to be different hence their measure of primeness will be different.
Question 1: Does the average primeness tend to zero? i.e. does the following hold?
$$
lim_N to infty frac1Nsum_r = 2^N f(r) = 0
$$
Question 2: Is $f(n)$ injective over composites? i.e., do there exist composites $3 < m < n$ such that $f(m) = f(n)$?
My progress
$f(4.35times 10^8) approx 0.5919$ and decreasing so the limit if it exists must be between 0 and 0.5919.- For $2 le i le n$, the minimum value of $f(i)$ occurs at the largest highly composite number $le n$.
Note 2: Here standard deviation of $x_1, x_2, ldots , x_n$ is defined as $sqrt fracsum_i=1^n (x-x_i)^2n$. Also notice that even if we define standard deviation as $sqrt fracsum_i=1^n (x-x_i)^2n-1$ our questions remain unaffected because in this case in the definition of $f$, we will be multiplying with $sqrt 2$ instead of $2$ to normalize $f$ in the interval $(0,1)$.
Note 3: Posted this question in MO since it is unanswered here despite many upvotes which suggests that the the question is more difficult
number-theory limits statistics prime-numbers natural-numbers
asked Apr 5 at 17:59
Nilotpal Kanti SinhaNilotpal Kanti Sinha
4,63121641
$endgroup$
|
show 12 more comments
$begingroup$
Note 1: This questions requires some new definitions, namely "continuous primeness" which I have made. Everyone is welcome to improve the definition without altering the spirit of the question. Click here for a somewhat related question.
A number is either prime or composite, hence primality is a binary concept. Instead I wanted to put a value of primality to every number using some function $f$ such that $f(n) = 1$ iff $n$ is a prime otherwise, $0 < f(n) < 1$ and as the number divisors of $n$ increases, $f(n)$ decreases on average. Thus $f(n)$ is a measure of the degree of primeness of $n$ where 1 is a perfect prime and 0 is a hypothetical perfect composite. Hence $frac1Nsum_r le N f(r)$ can be interpreted as a measure of average primeness of the first $N$ integers.
After trying several definitions and going through the ones in literature, I came up with:
Define $f(n) = dfrac2s_nn-1$ for $n ge 2$, where $s_n$ is the
standard deviation of the divisors of $n$.
One advantage of using standard deviation is that even if two numbers have the same number of divisor their value of $f$ appears to be different hence their measure of primeness will be different.
Question 1: Does the average primeness tend to zero? i.e. does the following hold?
$$
lim_N to infty frac1Nsum_r = 2^N f(r) = 0
$$
Question 2: Is $f(n)$ injective over composites? i.e., do there exist composites $3 < m < n$ such that $f(m) = f(n)$?
My progress
$f(4.35times 10^8) approx 0.5919$ and decreasing so the limit if it exists must be between 0 and 0.5919.- For $2 le i le n$, the minimum value of $f(i)$ occurs at the largest highly composite number $le n$.
Note 2: Here standard deviation of $x_1, x_2, ldots , x_n$ is defined as $sqrt fracsum_i=1^n (x-x_i)^2n$. Also notice that even if we define standard deviation as $sqrt fracsum_i=1^n (x-x_i)^2n-1$ our questions remain unaffected because in this case in the definition of $f$, we will be multiplying with $sqrt 2$ instead of $2$ to normalize $f$ in the interval $(0,1)$.
Note 3: Posted this question in MO since it is unanswered here despite many upvotes which suggests that the the question is more difficult
number-theory limits statistics prime-numbers natural-numbers
asked Apr 5 at 17:59
Nilotpal Kanti SinhaNilotpal Kanti Sinha
4,63121641
$endgroup$
$begingroup$
Is your second statement about the minimum of $f$ obvious? I'm unable to come up with an argument.
$endgroup$
– user113102
Apr 5 at 21:30
4
$begingroup$
Why involve the standard deviation? Why not something simpler, like $2/d(n)$, where $d(n)$ is the number of divisors of $n$?
$endgroup$
– Gerry Myerson
Apr 6 at 4:06
3
$begingroup$
@GerryMyerson: Here is a more technical answer why standard deviation. If two numbers have the same number of divisors then the value $2/d(n)$ is same for both but the values of $f(n)$ is different. So under my definition, I will consider the number with smaller value of $f(n)$ to have a greater primness because we are not just measuring how many divisors a number has but also how scattered these divisors are. At the moment, I don't know if $f(n)$ is unique. I will add this to the question.
$endgroup$
– Nilotpal Kanti Sinha
Apr 6 at 7:27
1
$begingroup$
I have written some code for this in R:stanfun <- function(n)sd(divisors(n))/(n-1);funstan <- function(m)sum(sapply(2:m, function(i)stanfun(i)))/m, so $frac110000sumlimits_r=1^10000f(r)$ is given byfunstan(10000)which outputs $0.404801$.
$endgroup$
– TheSimpliFire
Apr 6 at 19:23
$begingroup$
@TheSimpliFire I have added my Sagemath code above
$endgroup$
– Nilotpal Kanti Sinha
Apr 7 at 3:23
|
show 12 more comments
$begingroup$
Note 1: This questions requires some new definitions, namely "continuous primeness" which I have made. Everyone is welcome to improve the definition without altering the spirit of the question. Click here for a somewhat related question.
A number is either prime or composite, hence primality is a binary concept. Instead I wanted to put a value of primality to every number using some function $f$ such that $f(n) = 1$ iff $n$ is a prime otherwise, $0 < f(n) < 1$ and as the number divisors of $n$ increases, $f(n)$ decreases on average. Thus $f(n)$ is a measure of the degree of primeness of $n$ where 1 is a perfect prime and 0 is a hypothetical perfect composite. Hence $frac1Nsum_r le N f(r)$ can be interpreted as a measure of average primeness of the first $N$ integers.
After trying several definitions and going through the ones in literature, I came up with:
Define $f(n) = dfrac2s_nn-1$ for $n ge 2$, where $s_n$ is the
standard deviation of the divisors of $n$.
One advantage of using standard deviation is that even if two numbers have the same number of divisor their value of $f$ appears to be different hence their measure of primeness will be different.
Question 1: Does the average primeness tend to zero? i.e. does the following hold?
$$
lim_N to infty frac1Nsum_r = 2^N f(r) = 0
$$
Question 2: Is $f(n)$ injective over composites? i.e., do there exist composites $3 < m < n$ such that $f(m) = f(n)$?
My progress
$f(4.35times 10^8) approx 0.5919$ and decreasing so the limit if it exists must be between 0 and 0.5919.- For $2 le i le n$, the minimum value of $f(i)$ occurs at the largest highly composite number $le n$.
Note 2: Here standard deviation of $x_1, x_2, ldots , x_n$ is defined as $sqrt fracsum_i=1^n (x-x_i)^2n$. Also notice that even if we define standard deviation as $sqrt fracsum_i=1^n (x-x_i)^2n-1$ our questions remain unaffected because in this case in the definition of $f$, we will be multiplying with $sqrt 2$ instead of $2$ to normalize $f$ in the interval $(0,1)$.
Note 3: Posted this question in MO since it is unanswered here despite many upvotes which suggests that the the question is more difficult
number-theory limits statistics prime-numbers natural-numbers
asked Apr 5 at 17:59
Nilotpal Kanti SinhaNilotpal Kanti Sinha
4,63121641
$endgroup$
Note 1: This questions requires some new definitions, namely "continuous primeness" which I have made. Everyone is welcome to improve the definition without altering the spirit of the question. Click here for a somewhat related question.
A number is either prime or composite, hence primality is a binary concept. Instead I wanted to put a value of primality to every number using some function $f$ such that $f(n) = 1$ iff $n$ is a prime otherwise, $0 < f(n) < 1$ and as the number divisors of $n$ increases, $f(n)$ decreases on average. Thus $f(n)$ is a measure of the degree of primeness of $n$ where 1 is a perfect prime and 0 is a hypothetical perfect composite. Hence $frac1Nsum_r le N f(r)$ can be interpreted as a measure of average primeness of the first $N$ integers.
After trying several definitions and going through the ones in literature, I came up with:
Define $f(n) = dfrac2s_nn-1$ for $n ge 2$, where $s_n$ is the
standard deviation of the divisors of $n$.
One advantage of using standard deviation is that even if two numbers have the same number of divisor their value of $f$ appears to be different hence their measure of primeness will be different.
Question 1: Does the average primeness tend to zero? i.e. does the following hold?
$$
lim_N to infty frac1Nsum_r = 2^N f(r) = 0
$$
Question 2: Is $f(n)$ injective over composites? i.e., do there exist composites $3 < m < n$ such that $f(m) = f(n)$?
My progress
$f(4.35times 10^8) approx 0.5919$ and decreasing so the limit if it exists must be between 0 and 0.5919.- For $2 le i le n$, the minimum value of $f(i)$ occurs at the largest highly composite number $le n$.
Note 2: Here standard deviation of $x_1, x_2, ldots , x_n$ is defined as $sqrt fracsum_i=1^n (x-x_i)^2n$. Also notice that even if we define standard deviation as $sqrt fracsum_i=1^n (x-x_i)^2n-1$ our questions remain unaffected because in this case in the definition of $f$, we will be multiplying with $sqrt 2$ instead of $2$ to normalize $f$ in the interval $(0,1)$.
Note 3: Posted this question in MO since it is unanswered here despite many upvotes which suggests that the the question is more difficult
number-theory limits statistics prime-numbers natural-numbers
number-theory limits statistics prime-numbers natural-numbers
asked Apr 5 at 17:59
Nilotpal Kanti SinhaNilotpal Kanti Sinha
4,63121641
asked Apr 5 at 17:59
Nilotpal Kanti SinhaNilotpal Kanti Sinha
4,63121641
edited yesterday
Nilotpal Kanti Sinha
asked Apr 5 at 17:59
Nilotpal Kanti SinhaNilotpal Kanti Sinha
4,63121641
asked Apr 5 at 17:59
Nilotpal Kanti SinhaNilotpal Kanti Sinha
4,63121641
asked Apr 5 at 17:59
Nilotpal Kanti SinhaNilotpal Kanti Sinha
4,63121641
4,63121641
$begingroup$
Is your second statement about the minimum of $f$ obvious? I'm unable to come up with an argument.
$endgroup$
– user113102
Apr 5 at 21:30
4
$begingroup$
Why involve the standard deviation? Why not something simpler, like $2/d(n)$, where $d(n)$ is the number of divisors of $n$?
$endgroup$
– Gerry Myerson
Apr 6 at 4:06
3
$begingroup$
@GerryMyerson: Here is a more technical answer why standard deviation. If two numbers have the same number of divisors then the value $2/d(n)$ is same for both but the values of $f(n)$ is different. So under my definition, I will consider the number with smaller value of $f(n)$ to have a greater primness because we are not just measuring how many divisors a number has but also how scattered these divisors are. At the moment, I don't know if $f(n)$ is unique. I will add this to the question.
$endgroup$
– Nilotpal Kanti Sinha
Apr 6 at 7:27
1
$begingroup$
I have written some code for this in R:stanfun <- function(n)sd(divisors(n))/(n-1);funstan <- function(m)sum(sapply(2:m, function(i)stanfun(i)))/m, so $frac110000sumlimits_r=1^10000f(r)$ is given byfunstan(10000)which outputs $0.404801$.
$endgroup$
– TheSimpliFire
Apr 6 at 19:23
$begingroup$
@TheSimpliFire I have added my Sagemath code above
$endgroup$
– Nilotpal Kanti Sinha
Apr 7 at 3:23
|
show 12 more comments
$begingroup$
Is your second statement about the minimum of $f$ obvious? I'm unable to come up with an argument.
$endgroup$
– user113102
Apr 5 at 21:30
4
$begingroup$
Why involve the standard deviation? Why not something simpler, like $2/d(n)$, where $d(n)$ is the number of divisors of $n$?
$endgroup$
– Gerry Myerson
Apr 6 at 4:06
3
$begingroup$
@GerryMyerson: Here is a more technical answer why standard deviation. If two numbers have the same number of divisors then the value $2/d(n)$ is same for both but the values of $f(n)$ is different. So under my definition, I will consider the number with smaller value of $f(n)$ to have a greater primness because we are not just measuring how many divisors a number has but also how scattered these divisors are. At the moment, I don't know if $f(n)$ is unique. I will add this to the question.
$endgroup$
– Nilotpal Kanti Sinha
Apr 6 at 7:27
1
$begingroup$
I have written some code for this in R:stanfun <- function(n)sd(divisors(n))/(n-1);funstan <- function(m)sum(sapply(2:m, function(i)stanfun(i)))/m, so $frac110000sumlimits_r=1^10000f(r)$ is given byfunstan(10000)which outputs $0.404801$.
$endgroup$
– TheSimpliFire
Apr 6 at 19:23
$begingroup$
@TheSimpliFire I have added my Sagemath code above
$endgroup$
– Nilotpal Kanti Sinha
Apr 7 at 3:23
$begingroup$
Is your second statement about the minimum of $f$ obvious? I'm unable to come up with an argument.
$endgroup$
– user113102
Apr 5 at 21:30
$begingroup$
Is your second statement about the minimum of $f$ obvious? I'm unable to come up with an argument.
$endgroup$
– user113102
Apr 5 at 21:30
4
4
$begingroup$
Why involve the standard deviation? Why not something simpler, like $2/d(n)$, where $d(n)$ is the number of divisors of $n$?
$endgroup$
– Gerry Myerson
Apr 6 at 4:06
$begingroup$
Why involve the standard deviation? Why not something simpler, like $2/d(n)$, where $d(n)$ is the number of divisors of $n$?
$endgroup$
– Gerry Myerson
Apr 6 at 4:06
3
3
$begingroup$
@GerryMyerson: Here is a more technical answer why standard deviation. If two numbers have the same number of divisors then the value $2/d(n)$ is same for both but the values of $f(n)$ is different. So under my definition, I will consider the number with smaller value of $f(n)$ to have a greater primness because we are not just measuring how many divisors a number has but also how scattered these divisors are. At the moment, I don't know if $f(n)$ is unique. I will add this to the question.
$endgroup$
– Nilotpal Kanti Sinha
Apr 6 at 7:27
$begingroup$
@GerryMyerson: Here is a more technical answer why standard deviation. If two numbers have the same number of divisors then the value $2/d(n)$ is same for both but the values of $f(n)$ is different. So under my definition, I will consider the number with smaller value of $f(n)$ to have a greater primness because we are not just measuring how many divisors a number has but also how scattered these divisors are. At the moment, I don't know if $f(n)$ is unique. I will add this to the question.
$endgroup$
– Nilotpal Kanti Sinha
Apr 6 at 7:27
1
1
$begingroup$
I have written some code for this in R:
stanfun <- function(n)sd(divisors(n))/(n-1);funstan <- function(m)sum(sapply(2:m, function(i)stanfun(i)))/m, so $frac110000sumlimits_r=1^10000f(r)$ is given by funstan(10000) which outputs $0.404801$.$endgroup$
– TheSimpliFire
Apr 6 at 19:23
$begingroup$
I have written some code for this in R:
stanfun <- function(n)sd(divisors(n))/(n-1);funstan <- function(m)sum(sapply(2:m, function(i)stanfun(i)))/m, so $frac110000sumlimits_r=1^10000f(r)$ is given by funstan(10000) which outputs $0.404801$.$endgroup$
– TheSimpliFire
Apr 6 at 19:23
$begingroup$
@TheSimpliFire I have added my Sagemath code above
$endgroup$
– Nilotpal Kanti Sinha
Apr 7 at 3:23
$begingroup$
@TheSimpliFire I have added my Sagemath code above
$endgroup$
– Nilotpal Kanti Sinha
Apr 7 at 3:23
|
show 12 more comments
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$begingroup$
Is your second statement about the minimum of $f$ obvious? I'm unable to come up with an argument.
$endgroup$
– user113102
Apr 5 at 21:30
4
$begingroup$
Why involve the standard deviation? Why not something simpler, like $2/d(n)$, where $d(n)$ is the number of divisors of $n$?
$endgroup$
– Gerry Myerson
Apr 6 at 4:06
3
$begingroup$
@GerryMyerson: Here is a more technical answer why standard deviation. If two numbers have the same number of divisors then the value $2/d(n)$ is same for both but the values of $f(n)$ is different. So under my definition, I will consider the number with smaller value of $f(n)$ to have a greater primness because we are not just measuring how many divisors a number has but also how scattered these divisors are. At the moment, I don't know if $f(n)$ is unique. I will add this to the question.
$endgroup$
– Nilotpal Kanti Sinha
Apr 6 at 7:27
1
$begingroup$
I have written some code for this in R:
stanfun <- function(n)sd(divisors(n))/(n-1);funstan <- function(m)sum(sapply(2:m, function(i)stanfun(i)))/m, so $frac110000sumlimits_r=1^10000f(r)$ is given byfunstan(10000)which outputs $0.404801$.$endgroup$
– TheSimpliFire
Apr 6 at 19:23
$begingroup$
@TheSimpliFire I have added my Sagemath code above
$endgroup$
– Nilotpal Kanti Sinha
Apr 7 at 3:23