Does the average primeness of natural numbers tend to zero?Pascal Triangle and Prime NumbersWhat might the (normalized) pair correlation function of prime numbers look like?On prime numbersErdos Kac for imaginary class numberAKS Algorithm PseudoprimesThe conjecture of Montgomery and Soundararajan on primes in short intervals: Empirical inconsistencies?On the number of consecutive divisors of an integerOn the conjectured nonexistence of even almost perfect numbers (other than powers of two) and odd perfect numbersDoes this bound on an average over character sums have a more direct proof?Riemann sum formula for definite integral using prime numbers

Does the average primeness of natural numbers tend to zero?


Pascal Triangle and Prime NumbersWhat might the (normalized) pair correlation function of prime numbers look like?On prime numbersErdos Kac for imaginary class numberAKS Algorithm PseudoprimesThe conjecture of Montgomery and Soundararajan on primes in short intervals: Empirical inconsistencies?On the number of consecutive divisors of an integerOn the conjectured nonexistence of even almost perfect numbers (other than powers of two) and odd perfect numbersDoes this bound on an average over character sums have a more direct proof?Riemann sum formula for definite integral using prime numbers













14












$begingroup$


This question was posted in MSE. It got many upvotes but no answer hence posting it in MO.




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 the 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 reason for using standard deviation was that I was already studying the distribution of the divisors of a number.



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 $2 le i le n$, computed data shows that the minimum value of $f(i)$ occurs at the largest highly composite number $le n$.

Note: 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 $(0,1)$.










share|cite|improve this question











$endgroup$











  • $begingroup$
    From the linked question it seems that $s_n$ grows faster than $n$ so that $f(n)$ doesn't go to zero.
    $endgroup$
    – lcv
    16 hours ago






  • 1




    $begingroup$
    @lcv No $s_n$ doesn't grow faster than $n$. What are you looking at?
    $endgroup$
    – Nilos
    16 hours ago






  • 1




    $begingroup$
    "...I wanted to have a continuous function...". In what topology is $f$ continuous? If you put discrete topology on natural numbers, then any function is continuous so you probably have something else in mind.
    $endgroup$
    – Aknazar Kazhymurat
    12 hours ago










  • $begingroup$
    I have verified that $f$ is injective over composites less than 10,000,000.
    $endgroup$
    – Matt F.
    12 hours ago










  • $begingroup$
    @AknazarKazhymurat I have reworded that line. Hope it is clearer now?
    $endgroup$
    – Nilos
    12 hours ago















14












$begingroup$


This question was posted in MSE. It got many upvotes but no answer hence posting it in MO.




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 the 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 reason for using standard deviation was that I was already studying the distribution of the divisors of a number.



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 $2 le i le n$, computed data shows that the minimum value of $f(i)$ occurs at the largest highly composite number $le n$.

Note: 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 $(0,1)$.










share|cite|improve this question











$endgroup$











  • $begingroup$
    From the linked question it seems that $s_n$ grows faster than $n$ so that $f(n)$ doesn't go to zero.
    $endgroup$
    – lcv
    16 hours ago






  • 1




    $begingroup$
    @lcv No $s_n$ doesn't grow faster than $n$. What are you looking at?
    $endgroup$
    – Nilos
    16 hours ago






  • 1




    $begingroup$
    "...I wanted to have a continuous function...". In what topology is $f$ continuous? If you put discrete topology on natural numbers, then any function is continuous so you probably have something else in mind.
    $endgroup$
    – Aknazar Kazhymurat
    12 hours ago










  • $begingroup$
    I have verified that $f$ is injective over composites less than 10,000,000.
    $endgroup$
    – Matt F.
    12 hours ago










  • $begingroup$
    @AknazarKazhymurat I have reworded that line. Hope it is clearer now?
    $endgroup$
    – Nilos
    12 hours ago













14












14








14


6



$begingroup$


This question was posted in MSE. It got many upvotes but no answer hence posting it in MO.




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 the 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 reason for using standard deviation was that I was already studying the distribution of the divisors of a number.



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 $2 le i le n$, computed data shows that the minimum value of $f(i)$ occurs at the largest highly composite number $le n$.

Note: 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 $(0,1)$.










share|cite|improve this question











$endgroup$




This question was posted in MSE. It got many upvotes but no answer hence posting it in MO.




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 the 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 reason for using standard deviation was that I was already studying the distribution of the divisors of a number.



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 $2 le i le n$, computed data shows that the minimum value of $f(i)$ occurs at the largest highly composite number $le n$.

Note: 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 $(0,1)$.







nt.number-theory real-analysis analytic-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 12 hours ago







Nilos

















asked 16 hours ago









NilosNilos

1,4411834




1,4411834











  • $begingroup$
    From the linked question it seems that $s_n$ grows faster than $n$ so that $f(n)$ doesn't go to zero.
    $endgroup$
    – lcv
    16 hours ago






  • 1




    $begingroup$
    @lcv No $s_n$ doesn't grow faster than $n$. What are you looking at?
    $endgroup$
    – Nilos
    16 hours ago






  • 1




    $begingroup$
    "...I wanted to have a continuous function...". In what topology is $f$ continuous? If you put discrete topology on natural numbers, then any function is continuous so you probably have something else in mind.
    $endgroup$
    – Aknazar Kazhymurat
    12 hours ago










  • $begingroup$
    I have verified that $f$ is injective over composites less than 10,000,000.
    $endgroup$
    – Matt F.
    12 hours ago










  • $begingroup$
    @AknazarKazhymurat I have reworded that line. Hope it is clearer now?
    $endgroup$
    – Nilos
    12 hours ago
















  • $begingroup$
    From the linked question it seems that $s_n$ grows faster than $n$ so that $f(n)$ doesn't go to zero.
    $endgroup$
    – lcv
    16 hours ago






  • 1




    $begingroup$
    @lcv No $s_n$ doesn't grow faster than $n$. What are you looking at?
    $endgroup$
    – Nilos
    16 hours ago






  • 1




    $begingroup$
    "...I wanted to have a continuous function...". In what topology is $f$ continuous? If you put discrete topology on natural numbers, then any function is continuous so you probably have something else in mind.
    $endgroup$
    – Aknazar Kazhymurat
    12 hours ago










  • $begingroup$
    I have verified that $f$ is injective over composites less than 10,000,000.
    $endgroup$
    – Matt F.
    12 hours ago










  • $begingroup$
    @AknazarKazhymurat I have reworded that line. Hope it is clearer now?
    $endgroup$
    – Nilos
    12 hours ago















$begingroup$
From the linked question it seems that $s_n$ grows faster than $n$ so that $f(n)$ doesn't go to zero.
$endgroup$
– lcv
16 hours ago




$begingroup$
From the linked question it seems that $s_n$ grows faster than $n$ so that $f(n)$ doesn't go to zero.
$endgroup$
– lcv
16 hours ago




1




1




$begingroup$
@lcv No $s_n$ doesn't grow faster than $n$. What are you looking at?
$endgroup$
– Nilos
16 hours ago




$begingroup$
@lcv No $s_n$ doesn't grow faster than $n$. What are you looking at?
$endgroup$
– Nilos
16 hours ago




1




1




$begingroup$
"...I wanted to have a continuous function...". In what topology is $f$ continuous? If you put discrete topology on natural numbers, then any function is continuous so you probably have something else in mind.
$endgroup$
– Aknazar Kazhymurat
12 hours ago




$begingroup$
"...I wanted to have a continuous function...". In what topology is $f$ continuous? If you put discrete topology on natural numbers, then any function is continuous so you probably have something else in mind.
$endgroup$
– Aknazar Kazhymurat
12 hours ago












$begingroup$
I have verified that $f$ is injective over composites less than 10,000,000.
$endgroup$
– Matt F.
12 hours ago




$begingroup$
I have verified that $f$ is injective over composites less than 10,000,000.
$endgroup$
– Matt F.
12 hours ago












$begingroup$
@AknazarKazhymurat I have reworded that line. Hope it is clearer now?
$endgroup$
– Nilos
12 hours ago




$begingroup$
@AknazarKazhymurat I have reworded that line. Hope it is clearer now?
$endgroup$
– Nilos
12 hours ago










1 Answer
1






active

oldest

votes


















20












$begingroup$

The answer to Question 1 is "yes".
To see this, notice that $s_n$ is at most square root of the average square of divisor, i.e.
$$
s_nleq sqrtfracsum_dmid nd^2sum_dmid n 1=sqrtfracsigma_2(n)sigma_0(n),
$$



where $sigma_k(n)$ is the sum of $k$-th powers of divisors of $n$. Now,



$$
sigma_2(n)=n^2sigma_-2(n),
$$



so



$$
sigma_2(n)<fracpi^26n^2
$$



for all $n$. Therefore we have



$$
f(n)leq frac2n-1 sqrtfracpi^26n^2/sigma_0(n)leq frac5.14sqrtsigma_0(n)
$$



for all $n$. Now, almost all $nleq N$ have at least $0.5lnln N$ distinct prime factors. In particular, for almost all $nleq N$ we have $sigma_0(n)geq 0.5lnln N$. Therefore, our bound for $f(n)$ together with the trivial observation that $0leq f(n)leq 1$ gives



$$
sum_nleq N f(n)leq sum_nleq N, sigma_0(n)geq 0.5lnln N frac5.14sqrtsigma_0(n)+sum_nleq N, sigma_0(n)<0.5lnln N 1= o(N),
$$



as needed.



Using contour integration method one can even prove something like



$$
sum_nleq N f(n)=O(N(ln N)^1/sqrt2-1)
$$






share|cite|improve this answer











$endgroup$








  • 4




    $begingroup$
    That last expression reminds me an XKCD alt-text: "If you ever find yourself raising $log(textanything)^1/sqrt2$, set down the marker and back away from the whiteboard; something has gone horribly wrong."
    $endgroup$
    – Michael Seifert
    8 hours ago







  • 1




    $begingroup$
    @MichaelSeifert Bah, it was $log(anything)^e$ not $frac1sqrt2$. Log to the power of one over the sqrt of 2 is mundane; log of something to the power e is a sign of insanity.
    $endgroup$
    – Yakk
    7 hours ago







  • 1




    $begingroup$
    @Yakk Yeah, but Randall also says that taking $pi$-th root of anything is insane. However, the paper "Mean values of multiplicative functions" by Montgomery and Vaughan, Theorem 5, contains $(log x)^1/pi-1$ and is totally fine! (P.S. Is inequality $n_p<p^frac14sqrte+o(1)$ ok?..)
    $endgroup$
    – Asymptotiac K
    6 hours ago











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









20












$begingroup$

The answer to Question 1 is "yes".
To see this, notice that $s_n$ is at most square root of the average square of divisor, i.e.
$$
s_nleq sqrtfracsum_dmid nd^2sum_dmid n 1=sqrtfracsigma_2(n)sigma_0(n),
$$



where $sigma_k(n)$ is the sum of $k$-th powers of divisors of $n$. Now,



$$
sigma_2(n)=n^2sigma_-2(n),
$$



so



$$
sigma_2(n)<fracpi^26n^2
$$



for all $n$. Therefore we have



$$
f(n)leq frac2n-1 sqrtfracpi^26n^2/sigma_0(n)leq frac5.14sqrtsigma_0(n)
$$



for all $n$. Now, almost all $nleq N$ have at least $0.5lnln N$ distinct prime factors. In particular, for almost all $nleq N$ we have $sigma_0(n)geq 0.5lnln N$. Therefore, our bound for $f(n)$ together with the trivial observation that $0leq f(n)leq 1$ gives



$$
sum_nleq N f(n)leq sum_nleq N, sigma_0(n)geq 0.5lnln N frac5.14sqrtsigma_0(n)+sum_nleq N, sigma_0(n)<0.5lnln N 1= o(N),
$$



as needed.



Using contour integration method one can even prove something like



$$
sum_nleq N f(n)=O(N(ln N)^1/sqrt2-1)
$$






share|cite|improve this answer











$endgroup$








  • 4




    $begingroup$
    That last expression reminds me an XKCD alt-text: "If you ever find yourself raising $log(textanything)^1/sqrt2$, set down the marker and back away from the whiteboard; something has gone horribly wrong."
    $endgroup$
    – Michael Seifert
    8 hours ago







  • 1




    $begingroup$
    @MichaelSeifert Bah, it was $log(anything)^e$ not $frac1sqrt2$. Log to the power of one over the sqrt of 2 is mundane; log of something to the power e is a sign of insanity.
    $endgroup$
    – Yakk
    7 hours ago







  • 1




    $begingroup$
    @Yakk Yeah, but Randall also says that taking $pi$-th root of anything is insane. However, the paper "Mean values of multiplicative functions" by Montgomery and Vaughan, Theorem 5, contains $(log x)^1/pi-1$ and is totally fine! (P.S. Is inequality $n_p<p^frac14sqrte+o(1)$ ok?..)
    $endgroup$
    – Asymptotiac K
    6 hours ago















20












$begingroup$

The answer to Question 1 is "yes".
To see this, notice that $s_n$ is at most square root of the average square of divisor, i.e.
$$
s_nleq sqrtfracsum_dmid nd^2sum_dmid n 1=sqrtfracsigma_2(n)sigma_0(n),
$$



where $sigma_k(n)$ is the sum of $k$-th powers of divisors of $n$. Now,



$$
sigma_2(n)=n^2sigma_-2(n),
$$



so



$$
sigma_2(n)<fracpi^26n^2
$$



for all $n$. Therefore we have



$$
f(n)leq frac2n-1 sqrtfracpi^26n^2/sigma_0(n)leq frac5.14sqrtsigma_0(n)
$$



for all $n$. Now, almost all $nleq N$ have at least $0.5lnln N$ distinct prime factors. In particular, for almost all $nleq N$ we have $sigma_0(n)geq 0.5lnln N$. Therefore, our bound for $f(n)$ together with the trivial observation that $0leq f(n)leq 1$ gives



$$
sum_nleq N f(n)leq sum_nleq N, sigma_0(n)geq 0.5lnln N frac5.14sqrtsigma_0(n)+sum_nleq N, sigma_0(n)<0.5lnln N 1= o(N),
$$



as needed.



Using contour integration method one can even prove something like



$$
sum_nleq N f(n)=O(N(ln N)^1/sqrt2-1)
$$






share|cite|improve this answer











$endgroup$








  • 4




    $begingroup$
    That last expression reminds me an XKCD alt-text: "If you ever find yourself raising $log(textanything)^1/sqrt2$, set down the marker and back away from the whiteboard; something has gone horribly wrong."
    $endgroup$
    – Michael Seifert
    8 hours ago







  • 1




    $begingroup$
    @MichaelSeifert Bah, it was $log(anything)^e$ not $frac1sqrt2$. Log to the power of one over the sqrt of 2 is mundane; log of something to the power e is a sign of insanity.
    $endgroup$
    – Yakk
    7 hours ago







  • 1




    $begingroup$
    @Yakk Yeah, but Randall also says that taking $pi$-th root of anything is insane. However, the paper "Mean values of multiplicative functions" by Montgomery and Vaughan, Theorem 5, contains $(log x)^1/pi-1$ and is totally fine! (P.S. Is inequality $n_p<p^frac14sqrte+o(1)$ ok?..)
    $endgroup$
    – Asymptotiac K
    6 hours ago













20












20








20





$begingroup$

The answer to Question 1 is "yes".
To see this, notice that $s_n$ is at most square root of the average square of divisor, i.e.
$$
s_nleq sqrtfracsum_dmid nd^2sum_dmid n 1=sqrtfracsigma_2(n)sigma_0(n),
$$



where $sigma_k(n)$ is the sum of $k$-th powers of divisors of $n$. Now,



$$
sigma_2(n)=n^2sigma_-2(n),
$$



so



$$
sigma_2(n)<fracpi^26n^2
$$



for all $n$. Therefore we have



$$
f(n)leq frac2n-1 sqrtfracpi^26n^2/sigma_0(n)leq frac5.14sqrtsigma_0(n)
$$



for all $n$. Now, almost all $nleq N$ have at least $0.5lnln N$ distinct prime factors. In particular, for almost all $nleq N$ we have $sigma_0(n)geq 0.5lnln N$. Therefore, our bound for $f(n)$ together with the trivial observation that $0leq f(n)leq 1$ gives



$$
sum_nleq N f(n)leq sum_nleq N, sigma_0(n)geq 0.5lnln N frac5.14sqrtsigma_0(n)+sum_nleq N, sigma_0(n)<0.5lnln N 1= o(N),
$$



as needed.



Using contour integration method one can even prove something like



$$
sum_nleq N f(n)=O(N(ln N)^1/sqrt2-1)
$$






share|cite|improve this answer











$endgroup$



The answer to Question 1 is "yes".
To see this, notice that $s_n$ is at most square root of the average square of divisor, i.e.
$$
s_nleq sqrtfracsum_dmid nd^2sum_dmid n 1=sqrtfracsigma_2(n)sigma_0(n),
$$



where $sigma_k(n)$ is the sum of $k$-th powers of divisors of $n$. Now,



$$
sigma_2(n)=n^2sigma_-2(n),
$$



so



$$
sigma_2(n)<fracpi^26n^2
$$



for all $n$. Therefore we have



$$
f(n)leq frac2n-1 sqrtfracpi^26n^2/sigma_0(n)leq frac5.14sqrtsigma_0(n)
$$



for all $n$. Now, almost all $nleq N$ have at least $0.5lnln N$ distinct prime factors. In particular, for almost all $nleq N$ we have $sigma_0(n)geq 0.5lnln N$. Therefore, our bound for $f(n)$ together with the trivial observation that $0leq f(n)leq 1$ gives



$$
sum_nleq N f(n)leq sum_nleq N, sigma_0(n)geq 0.5lnln N frac5.14sqrtsigma_0(n)+sum_nleq N, sigma_0(n)<0.5lnln N 1= o(N),
$$



as needed.



Using contour integration method one can even prove something like



$$
sum_nleq N f(n)=O(N(ln N)^1/sqrt2-1)
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 12 hours ago

























answered 12 hours ago









Asymptotiac KAsymptotiac K

1,4841314




1,4841314







  • 4




    $begingroup$
    That last expression reminds me an XKCD alt-text: "If you ever find yourself raising $log(textanything)^1/sqrt2$, set down the marker and back away from the whiteboard; something has gone horribly wrong."
    $endgroup$
    – Michael Seifert
    8 hours ago







  • 1




    $begingroup$
    @MichaelSeifert Bah, it was $log(anything)^e$ not $frac1sqrt2$. Log to the power of one over the sqrt of 2 is mundane; log of something to the power e is a sign of insanity.
    $endgroup$
    – Yakk
    7 hours ago







  • 1




    $begingroup$
    @Yakk Yeah, but Randall also says that taking $pi$-th root of anything is insane. However, the paper "Mean values of multiplicative functions" by Montgomery and Vaughan, Theorem 5, contains $(log x)^1/pi-1$ and is totally fine! (P.S. Is inequality $n_p<p^frac14sqrte+o(1)$ ok?..)
    $endgroup$
    – Asymptotiac K
    6 hours ago












  • 4




    $begingroup$
    That last expression reminds me an XKCD alt-text: "If you ever find yourself raising $log(textanything)^1/sqrt2$, set down the marker and back away from the whiteboard; something has gone horribly wrong."
    $endgroup$
    – Michael Seifert
    8 hours ago







  • 1




    $begingroup$
    @MichaelSeifert Bah, it was $log(anything)^e$ not $frac1sqrt2$. Log to the power of one over the sqrt of 2 is mundane; log of something to the power e is a sign of insanity.
    $endgroup$
    – Yakk
    7 hours ago







  • 1




    $begingroup$
    @Yakk Yeah, but Randall also says that taking $pi$-th root of anything is insane. However, the paper "Mean values of multiplicative functions" by Montgomery and Vaughan, Theorem 5, contains $(log x)^1/pi-1$ and is totally fine! (P.S. Is inequality $n_p<p^frac14sqrte+o(1)$ ok?..)
    $endgroup$
    – Asymptotiac K
    6 hours ago







4




4




$begingroup$
That last expression reminds me an XKCD alt-text: "If you ever find yourself raising $log(textanything)^1/sqrt2$, set down the marker and back away from the whiteboard; something has gone horribly wrong."
$endgroup$
– Michael Seifert
8 hours ago





$begingroup$
That last expression reminds me an XKCD alt-text: "If you ever find yourself raising $log(textanything)^1/sqrt2$, set down the marker and back away from the whiteboard; something has gone horribly wrong."
$endgroup$
– Michael Seifert
8 hours ago





1




1




$begingroup$
@MichaelSeifert Bah, it was $log(anything)^e$ not $frac1sqrt2$. Log to the power of one over the sqrt of 2 is mundane; log of something to the power e is a sign of insanity.
$endgroup$
– Yakk
7 hours ago





$begingroup$
@MichaelSeifert Bah, it was $log(anything)^e$ not $frac1sqrt2$. Log to the power of one over the sqrt of 2 is mundane; log of something to the power e is a sign of insanity.
$endgroup$
– Yakk
7 hours ago





1




1




$begingroup$
@Yakk Yeah, but Randall also says that taking $pi$-th root of anything is insane. However, the paper "Mean values of multiplicative functions" by Montgomery and Vaughan, Theorem 5, contains $(log x)^1/pi-1$ and is totally fine! (P.S. Is inequality $n_p<p^frac14sqrte+o(1)$ ok?..)
$endgroup$
– Asymptotiac K
6 hours ago




$begingroup$
@Yakk Yeah, but Randall also says that taking $pi$-th root of anything is insane. However, the paper "Mean values of multiplicative functions" by Montgomery and Vaughan, Theorem 5, contains $(log x)^1/pi-1$ and is totally fine! (P.S. Is inequality $n_p<p^frac14sqrte+o(1)$ ok?..)
$endgroup$
– Asymptotiac K
6 hours ago

















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