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Is there any other number that has similar propertie as 21?

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Is there any other number that has similar propertie as 21?

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8

1

$begingroup$

It's my observation.

Let
$$
n=p_1 times p_2 times p_3 times ldots times p_r
$$
where $p_i$ are prime factors and
$f$ and $g$ are the functions
$$
f(n)=1+2+ ldots +n
$$
And
$$
g(n)=p_1+p_2+ ldots +p_r
$$
If we put $n=21$
with
$$
g(f(21))=g(231)=21.
$$
I checked it upto $n=10~000$, I did not find another number with this property $g(f(n))=n$.
Can we prove that such other numbers do not exist?

elementary-number-theory prime-factorization

share|cite|improve this question

edited 9 mins ago

Especially Lime

22.3k22858

asked 1 hour ago

Pruthviraj HajariPruthviraj Hajari

513

New contributor

Pruthviraj Hajari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To clarify, is $g(12)=5$ or $=7$?
  $endgroup$
  – Hagen von Eitzen
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  12=3*2*2 then g(12)=2+2+3=7
  $endgroup$
  – Pruthviraj Hajari
  53 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @PruthvirajHajari Did you come up with this problem by yourself? If so, please state that in your question and include the program code used to run it.
  $endgroup$
  – TheSimpliFire
  41 mins ago
  

  
  
  
  

add a comment | 

8

1

$begingroup$

It's my observation.

Let
$$
n=p_1 times p_2 times p_3 times ldots times p_r
$$
where $p_i$ are prime factors and
$f$ and $g$ are the functions
$$
f(n)=1+2+ ldots +n
$$
And
$$
g(n)=p_1+p_2+ ldots +p_r
$$
If we put $n=21$
with
$$
g(f(21))=g(231)=21.
$$
I checked it upto $n=10~000$, I did not find another number with this property $g(f(n))=n$.
Can we prove that such other numbers do not exist?

elementary-number-theory prime-factorization

share|cite|improve this question

edited 9 mins ago

Especially Lime

22.3k22858

asked 1 hour ago

Pruthviraj HajariPruthviraj Hajari

513

New contributor

Pruthviraj Hajari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To clarify, is $g(12)=5$ or $=7$?
  $endgroup$
  – Hagen von Eitzen
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  12=3*2*2 then g(12)=2+2+3=7
  $endgroup$
  – Pruthviraj Hajari
  53 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @PruthvirajHajari Did you come up with this problem by yourself? If so, please state that in your question and include the program code used to run it.
  $endgroup$
  – TheSimpliFire
  41 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

It's my observation.

Let
$$
n=p_1 times p_2 times p_3 times ldots times p_r
$$
where $p_i$ are prime factors and
$f$ and $g$ are the functions
$$
f(n)=1+2+ ldots +n
$$
And
$$
g(n)=p_1+p_2+ ldots +p_r
$$
If we put $n=21$
with
$$
g(f(21))=g(231)=21.
$$
I checked it upto $n=10~000$, I did not find another number with this property $g(f(n))=n$.
Can we prove that such other numbers do not exist?

elementary-number-theory prime-factorization

share|cite|improve this question

edited 9 mins ago

Especially Lime

22.3k22858

asked 1 hour ago

Pruthviraj HajariPruthviraj Hajari

513

New contributor

Pruthviraj Hajari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 9 mins ago

Especially Lime

22.3k22858

asked 1 hour ago

Pruthviraj HajariPruthviraj Hajari

513

New contributor

Pruthviraj Hajari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 9 mins ago

Especially Lime

22.3k22858

asked 1 hour ago

Pruthviraj HajariPruthviraj Hajari

513

New contributor

Pruthviraj Hajari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 9 mins ago

Especially Lime

22.3k22858

edited 9 mins ago

Especially Lime

22.3k22858

asked 1 hour ago

Pruthviraj HajariPruthviraj Hajari

513

New contributor

Pruthviraj Hajari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 1 hour ago

Pruthviraj HajariPruthviraj Hajari

513

1 Answer
1

active

oldest

votes

9

$begingroup$

This is a very interesting question…

$newcommand{sopfr}{operatorname{sopfr}}$
$f(n)=frac{n(n+1)}2$ and $g(n)=sopfr(n)$, the sum of prime factors of $n$ with repeats (OEIS A001414). We want $n$ such that $g(f(n))=n$ or
$$sopfrleft(frac{n(n+1)}2right)=ntag1$$
which can be split into two cases due to the property $sopfr(ab)=sopfr(a)+sopfr(b)$.

• If $n$ is even, then $sopfrleft(frac n2right)+sopfr(n+1)=n$. We know that $sopfr(n)le n$, so $sopfrleft(frac n2right)lefrac n2$ and consequently $sopfr(n+1)gefrac n2$. Either $n+1$ is a prime, in which case the LHS of $(1)$ is greater than $n$ and so the equality cannot hold, or $n+1$ is odd composite and so has a prime factor at least 3, yielding $sopfr(n+1)le3+frac{n+1}3$ and thus
  $$frac n2le3+frac{n+1}3$$
  which is only true for $nle20$. Checking those $n$ reveals no solutions to $(1)$.


• If $n$ is odd, the reasoning is similar: $sopfrleft(frac{n+1}2right)+sopfr(n)=n$, where $sopfrleft(frac{n+1}2right)lefrac{n+1}2$ and so $sopfr(n)gefrac{n-1}2$. Since $n$ is odd, either it is prime and the LHS of $(1)$ is greater than $n$, or it has a prime factor at least 3 and $sopfr(n)le3+frac n3$, giving
  $$frac{n-1}2le3+frac n3$$
  which only holds for $nle21$. 21 is the solution to $(1)$ pointed out in the original question; we have just shown it is the only one.

share|cite|improve this answer

answered 15 mins ago

Parcly TaxelParcly Taxel

43.2k1372101

$endgroup$

add a comment | 

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9

$begingroup$

This is a very interesting question…

$newcommand{sopfr}{operatorname{sopfr}}$
$f(n)=frac{n(n+1)}2$ and $g(n)=sopfr(n)$, the sum of prime factors of $n$ with repeats (OEIS A001414). We want $n$ such that $g(f(n))=n$ or
$$sopfrleft(frac{n(n+1)}2right)=ntag1$$
which can be split into two cases due to the property $sopfr(ab)=sopfr(a)+sopfr(b)$.

• If $n$ is even, then $sopfrleft(frac n2right)+sopfr(n+1)=n$. We know that $sopfr(n)le n$, so $sopfrleft(frac n2right)lefrac n2$ and consequently $sopfr(n+1)gefrac n2$. Either $n+1$ is a prime, in which case the LHS of $(1)$ is greater than $n$ and so the equality cannot hold, or $n+1$ is odd composite and so has a prime factor at least 3, yielding $sopfr(n+1)le3+frac{n+1}3$ and thus
  $$frac n2le3+frac{n+1}3$$
  which is only true for $nle20$. Checking those $n$ reveals no solutions to $(1)$.


• If $n$ is odd, the reasoning is similar: $sopfrleft(frac{n+1}2right)+sopfr(n)=n$, where $sopfrleft(frac{n+1}2right)lefrac{n+1}2$ and so $sopfr(n)gefrac{n-1}2$. Since $n$ is odd, either it is prime and the LHS of $(1)$ is greater than $n$, or it has a prime factor at least 3 and $sopfr(n)le3+frac n3$, giving
  $$frac{n-1}2le3+frac n3$$
  which only holds for $nle21$. 21 is the solution to $(1)$ pointed out in the original question; we have just shown it is the only one.

share|cite|improve this answer

answered 15 mins ago

Parcly TaxelParcly Taxel

43.2k1372101

$endgroup$

add a comment | 

9

$begingroup$

This is a very interesting question…

$newcommand{sopfr}{operatorname{sopfr}}$
$f(n)=frac{n(n+1)}2$ and $g(n)=sopfr(n)$, the sum of prime factors of $n$ with repeats (OEIS A001414). We want $n$ such that $g(f(n))=n$ or
$$sopfrleft(frac{n(n+1)}2right)=ntag1$$
which can be split into two cases due to the property $sopfr(ab)=sopfr(a)+sopfr(b)$.

• If $n$ is even, then $sopfrleft(frac n2right)+sopfr(n+1)=n$. We know that $sopfr(n)le n$, so $sopfrleft(frac n2right)lefrac n2$ and consequently $sopfr(n+1)gefrac n2$. Either $n+1$ is a prime, in which case the LHS of $(1)$ is greater than $n$ and so the equality cannot hold, or $n+1$ is odd composite and so has a prime factor at least 3, yielding $sopfr(n+1)le3+frac{n+1}3$ and thus
  $$frac n2le3+frac{n+1}3$$
  which is only true for $nle20$. Checking those $n$ reveals no solutions to $(1)$.


• If $n$ is odd, the reasoning is similar: $sopfrleft(frac{n+1}2right)+sopfr(n)=n$, where $sopfrleft(frac{n+1}2right)lefrac{n+1}2$ and so $sopfr(n)gefrac{n-1}2$. Since $n$ is odd, either it is prime and the LHS of $(1)$ is greater than $n$, or it has a prime factor at least 3 and $sopfr(n)le3+frac n3$, giving
  $$frac{n-1}2le3+frac n3$$
  which only holds for $nle21$. 21 is the solution to $(1)$ pointed out in the original question; we have just shown it is the only one.

share|cite|improve this answer

answered 15 mins ago

Parcly TaxelParcly Taxel

43.2k1372101

$endgroup$

add a comment | 

$begingroup$

This is a very interesting question…

$newcommand{sopfr}{operatorname{sopfr}}$
$f(n)=frac{n(n+1)}2$ and $g(n)=sopfr(n)$, the sum of prime factors of $n$ with repeats (OEIS A001414). We want $n$ such that $g(f(n))=n$ or
$$sopfrleft(frac{n(n+1)}2right)=ntag1$$
which can be split into two cases due to the property $sopfr(ab)=sopfr(a)+sopfr(b)$.

• If $n$ is even, then $sopfrleft(frac n2right)+sopfr(n+1)=n$. We know that $sopfr(n)le n$, so $sopfrleft(frac n2right)lefrac n2$ and consequently $sopfr(n+1)gefrac n2$. Either $n+1$ is a prime, in which case the LHS of $(1)$ is greater than $n$ and so the equality cannot hold, or $n+1$ is odd composite and so has a prime factor at least 3, yielding $sopfr(n+1)le3+frac{n+1}3$ and thus
  $$frac n2le3+frac{n+1}3$$
  which is only true for $nle20$. Checking those $n$ reveals no solutions to $(1)$.


• If $n$ is odd, the reasoning is similar: $sopfrleft(frac{n+1}2right)+sopfr(n)=n$, where $sopfrleft(frac{n+1}2right)lefrac{n+1}2$ and so $sopfr(n)gefrac{n-1}2$. Since $n$ is odd, either it is prime and the LHS of $(1)$ is greater than $n$, or it has a prime factor at least 3 and $sopfr(n)le3+frac n3$, giving
  $$frac{n-1}2le3+frac n3$$
  which only holds for $nle21$. 21 is the solution to $(1)$ pointed out in the original question; we have just shown it is the only one.

share|cite|improve this answer

answered 15 mins ago

Parcly TaxelParcly Taxel

43.2k1372101

$endgroup$

share|cite|improve this answer

answered 15 mins ago

Parcly TaxelParcly Taxel

43.2k1372101

answered 15 mins ago

Parcly TaxelParcly Taxel

43.2k1372101

answered 15 mins ago

Parcly TaxelParcly Taxel

43.2k1372101