Estimation of the number of solutions for the equation $sigma(varphi(n))=sigma(operatorname{rad}(n))$
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For integers $ngeq 1$ in this post we denote the square-free kernel as $$operatorname{rad}(n)=prod_{substack{pmid n\ptext{ prime}}}p,$$ that is the product of distinct primes dividing an integer $n>1$ with the definition $operatorname{rad}(1)=1$ (the Wikipedia's article dedicated to this multiplicative function is Radical of an integer).
And I denote the Euler's totient function as $varphi(n)$. We consider the solutions over integers $ngeq 1$ of the equation $$sigma(varphi(n))=sigma(operatorname{rad}(n)).tag{1}$$
The first few solutions are $n=1,4,18,87,260,362$ and $732$. We denote the set of all solutions of $(1)$ as $mathcal{A}$ and for a real number $x>1$ let $mathcal{A}(x)$ defined as $mathcal{A}(x)=mathcal{A}cap[1,x]$ with cardinality denoted as $#mathcal{A}(x)$.
After I did a table using a program with my computer and since my equation is a variant of an equation from the literature [1], I wrote next conjecture (any case I believe that it can be wrong thus I am asking my Question).
Conjecture. The estimate $$#mathcal{A}(x)=Oleft(frac{x}{(log x)^3}right)tag{2}$$
is true for enough large $x>1$.
Question. Can you prove or refute previous Conjecture? Many thanks.
I'm curious to know if we can refute previous conjeture, what methods/reasonings can one use to refute it?
References:
[1] Jean-Marie De Koninck and Florian Luca, Positive Integers $n$ Such That $sigma(phi(n))=sigma(n)$, Journal of Integer Sequences, Vol. 11 (2008), Article 08.1.5.
asymptotics analytic-number-theory prime-factorization totient-function divisor-sum
edited Nov 22 at 12:57
Klangen
1,43711232
asked Jun 7 at 10:03
user243301
1
add a comment |
up vote
6
down vote
favorite
For integers $ngeq 1$ in this post we denote the square-free kernel as $$operatorname{rad}(n)=prod_{substack{pmid n\ptext{ prime}}}p,$$ that is the product of distinct primes dividing an integer $n>1$ with the definition $operatorname{rad}(1)=1$ (the Wikipedia's article dedicated to this multiplicative function is Radical of an integer).
And I denote the Euler's totient function as $varphi(n)$. We consider the solutions over integers $ngeq 1$ of the equation $$sigma(varphi(n))=sigma(operatorname{rad}(n)).tag{1}$$
The first few solutions are $n=1,4,18,87,260,362$ and $732$. We denote the set of all solutions of $(1)$ as $mathcal{A}$ and for a real number $x>1$ let $mathcal{A}(x)$ defined as $mathcal{A}(x)=mathcal{A}cap[1,x]$ with cardinality denoted as $#mathcal{A}(x)$.
After I did a table using a program with my computer and since my equation is a variant of an equation from the literature [1], I wrote next conjecture (any case I believe that it can be wrong thus I am asking my Question).
Conjecture. The estimate $$#mathcal{A}(x)=Oleft(frac{x}{(log x)^3}right)tag{2}$$
is true for enough large $x>1$.
Question. Can you prove or refute previous Conjecture? Many thanks.
I'm curious to know if we can refute previous conjeture, what methods/reasonings can one use to refute it?
References:
[1] Jean-Marie De Koninck and Florian Luca, Positive Integers $n$ Such That $sigma(phi(n))=sigma(n)$, Journal of Integer Sequences, Vol. 11 (2008), Article 08.1.5.
asymptotics analytic-number-theory prime-factorization totient-function divisor-sum
edited Nov 22 at 12:57
Klangen
1,43711232
asked Jun 7 at 10:03
user243301
1
1
Upto $10^8$ , $2 942$ solutions exist
– Peter
Jun 7 at 11:17
Many thanks for your calculation @Peter
– user243301
Jun 7 at 14:28
My suggestion is that you walk through the method in the paper you cited, and try on your equation.
– i707107
Jun 24 at 13:34
Thanks for your suggestion @i707107
– user243301
Jun 24 at 18:09
add a comment |
up vote
6
down vote
favorite
up vote
6
down vote
favorite
For integers $ngeq 1$ in this post we denote the square-free kernel as $$operatorname{rad}(n)=prod_{substack{pmid n\ptext{ prime}}}p,$$ that is the product of distinct primes dividing an integer $n>1$ with the definition $operatorname{rad}(1)=1$ (the Wikipedia's article dedicated to this multiplicative function is Radical of an integer).
And I denote the Euler's totient function as $varphi(n)$. We consider the solutions over integers $ngeq 1$ of the equation $$sigma(varphi(n))=sigma(operatorname{rad}(n)).tag{1}$$
The first few solutions are $n=1,4,18,87,260,362$ and $732$. We denote the set of all solutions of $(1)$ as $mathcal{A}$ and for a real number $x>1$ let $mathcal{A}(x)$ defined as $mathcal{A}(x)=mathcal{A}cap[1,x]$ with cardinality denoted as $#mathcal{A}(x)$.
After I did a table using a program with my computer and since my equation is a variant of an equation from the literature [1], I wrote next conjecture (any case I believe that it can be wrong thus I am asking my Question).
Conjecture. The estimate $$#mathcal{A}(x)=Oleft(frac{x}{(log x)^3}right)tag{2}$$
is true for enough large $x>1$.
Question. Can you prove or refute previous Conjecture? Many thanks.
I'm curious to know if we can refute previous conjeture, what methods/reasonings can one use to refute it?
References:
[1] Jean-Marie De Koninck and Florian Luca, Positive Integers $n$ Such That $sigma(phi(n))=sigma(n)$, Journal of Integer Sequences, Vol. 11 (2008), Article 08.1.5.
asymptotics analytic-number-theory prime-factorization totient-function divisor-sum
edited Nov 22 at 12:57
Klangen
1,43711232
asked Jun 7 at 10:03
user243301
1
For integers $ngeq 1$ in this post we denote the square-free kernel as $$operatorname{rad}(n)=prod_{substack{pmid n\ptext{ prime}}}p,$$ that is the product of distinct primes dividing an integer $n>1$ with the definition $operatorname{rad}(1)=1$ (the Wikipedia's article dedicated to this multiplicative function is Radical of an integer).
And I denote the Euler's totient function as $varphi(n)$. We consider the solutions over integers $ngeq 1$ of the equation $$sigma(varphi(n))=sigma(operatorname{rad}(n)).tag{1}$$
The first few solutions are $n=1,4,18,87,260,362$ and $732$. We denote the set of all solutions of $(1)$ as $mathcal{A}$ and for a real number $x>1$ let $mathcal{A}(x)$ defined as $mathcal{A}(x)=mathcal{A}cap[1,x]$ with cardinality denoted as $#mathcal{A}(x)$.
After I did a table using a program with my computer and since my equation is a variant of an equation from the literature [1], I wrote next conjecture (any case I believe that it can be wrong thus I am asking my Question).
Conjecture. The estimate $$#mathcal{A}(x)=Oleft(frac{x}{(log x)^3}right)tag{2}$$
is true for enough large $x>1$.
Question. Can you prove or refute previous Conjecture? Many thanks.
I'm curious to know if we can refute previous conjeture, what methods/reasonings can one use to refute it?
References:
[1] Jean-Marie De Koninck and Florian Luca, Positive Integers $n$ Such That $sigma(phi(n))=sigma(n)$, Journal of Integer Sequences, Vol. 11 (2008), Article 08.1.5.
asymptotics analytic-number-theory prime-factorization totient-function divisor-sum
asymptotics analytic-number-theory prime-factorization totient-function divisor-sum
edited Nov 22 at 12:57
Klangen
1,43711232
asked Jun 7 at 10:03
user243301
1
edited Nov 22 at 12:57
Klangen
1,43711232
asked Jun 7 at 10:03
user243301
1
edited Nov 22 at 12:57
Klangen
1,43711232
edited Nov 22 at 12:57
Klangen
1,43711232
edited Nov 22 at 12:57
Klangen
1,43711232
1,43711232
asked Jun 7 at 10:03
user243301
1
asked Jun 7 at 10:03
user243301
1
asked Jun 7 at 10:03
user243301
1
1
1
Upto $10^8$ , $2 942$ solutions exist
– Peter
Jun 7 at 11:17
Many thanks for your calculation @Peter
– user243301
Jun 7 at 14:28
My suggestion is that you walk through the method in the paper you cited, and try on your equation.
– i707107
Jun 24 at 13:34
Thanks for your suggestion @i707107
– user243301
Jun 24 at 18:09
add a comment |
1
Upto $10^8$ , $2 942$ solutions exist
– Peter
Jun 7 at 11:17
Many thanks for your calculation @Peter
– user243301
Jun 7 at 14:28
My suggestion is that you walk through the method in the paper you cited, and try on your equation.
– i707107
Jun 24 at 13:34
Thanks for your suggestion @i707107
– user243301
Jun 24 at 18:09
1
1
Upto $10^8$ , $2 942$ solutions exist
– Peter
Jun 7 at 11:17
Upto $10^8$ , $2 942$ solutions exist
– Peter
Jun 7 at 11:17
Many thanks for your calculation @Peter
– user243301
Jun 7 at 14:28
Many thanks for your calculation @Peter
– user243301
Jun 7 at 14:28
My suggestion is that you walk through the method in the paper you cited, and try on your equation.
– i707107
Jun 24 at 13:34
My suggestion is that you walk through the method in the paper you cited, and try on your equation.
– i707107
Jun 24 at 13:34
Thanks for your suggestion @i707107
– user243301
Jun 24 at 18:09
Thanks for your suggestion @i707107
– user243301
Jun 24 at 18:09
add a comment |
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1
Upto $10^8$ , $2 942$ solutions exist
– Peter
Jun 7 at 11:17
Many thanks for your calculation @Peter
– user243301
Jun 7 at 14:28
My suggestion is that you walk through the method in the paper you cited, and try on your equation.
– i707107
Jun 24 at 13:34
Thanks for your suggestion @i707107
– user243301
Jun 24 at 18:09