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

Are 14 and 21 the only “interesting” numbers?Origin of well-ordering proof of uniqueness in the FToArithmeticProof - Fundamental Theorem of Arithmetic using Euclid's LemmaAdding a power of two to a composite odd numberEvery integer $n > 1$ can be written in one and only one way with a certain propertyCan a carmichael number have consecutive prime factors?Show that if $n$ is a positive integer with $r$ distinct odd prime factors, then $2^r mid varphi(n)$First $k$ numbers with given prime factorsPrime Factorization of $m^2$Is this proof for a and b being perfect squares correct?A lemma in a proof there are infinitely many primes.

12

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 $n=10000$, 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 22 mins ago

Parcly Taxel

43.3k1372102

asked 2 hours ago

Pruthviraj HajariPruthviraj Hajari

714

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
  1 hour 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
  1 hour ago
  

  
  
  
  

add a comment | 

12

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 $n=10000$, 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 22 mins ago

Parcly Taxel

43.3k1372102

asked 2 hours ago

Pruthviraj HajariPruthviraj Hajari

714

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
  1 hour 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
  1 hour 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 $n=10000$, 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 22 mins ago

Parcly Taxel

43.3k1372102

asked 2 hours ago

Pruthviraj HajariPruthviraj Hajari

714

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 22 mins ago

Parcly Taxel

43.3k1372102

asked 2 hours ago

Pruthviraj HajariPruthviraj Hajari

714

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 22 mins ago

Parcly Taxel

43.3k1372102

asked 2 hours ago

Pruthviraj HajariPruthviraj Hajari

714

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 22 mins ago

Parcly Taxel

43.3k1372102

edited 22 mins ago

Parcly Taxel

43.3k1372102

asked 2 hours ago

Pruthviraj HajariPruthviraj Hajari

714

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 2 hours ago

Pruthviraj HajariPruthviraj Hajari

714

1 Answer
1

active

oldest

votes

13

$begingroup$

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 least 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 least 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.

---

*Technically we have to repeat the argument for other possible least prime factors $k$ of $n$ or $n+1$ – and the upper bound $N_k$ of the solution to the inequalities in $n$ increases accordingly, each 3 replaced with $k$. However, the least composite number with least prime factor $k$ is $k^2$, and this increases much faster than $N_k$ (which is $simfrac k2$). Indeed, $5^2$ already exceeds $N_5$ for both inequalities.

The method I use above has very strong similarities to the method I used in my most famous answer of all. It is sheer coincidence that 21 is a solution to both the problems I answered.

share|cite|improve this answer

edited 30 mins ago

answered 1 hour ago

Parcly TaxelParcly Taxel

43.3k1372102

$endgroup$

add a comment | 

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13

$begingroup$

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 least 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 least 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.

---

*Technically we have to repeat the argument for other possible least prime factors $k$ of $n$ or $n+1$ – and the upper bound $N_k$ of the solution to the inequalities in $n$ increases accordingly, each 3 replaced with $k$. However, the least composite number with least prime factor $k$ is $k^2$, and this increases much faster than $N_k$ (which is $simfrac k2$). Indeed, $5^2$ already exceeds $N_5$ for both inequalities.

The method I use above has very strong similarities to the method I used in my most famous answer of all. It is sheer coincidence that 21 is a solution to both the problems I answered.

share|cite|improve this answer

edited 30 mins ago

answered 1 hour ago

Parcly TaxelParcly Taxel

43.3k1372102

$endgroup$

add a comment | 

13

$begingroup$

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 least 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 least 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.

---

*Technically we have to repeat the argument for other possible least prime factors $k$ of $n$ or $n+1$ – and the upper bound $N_k$ of the solution to the inequalities in $n$ increases accordingly, each 3 replaced with $k$. However, the least composite number with least prime factor $k$ is $k^2$, and this increases much faster than $N_k$ (which is $simfrac k2$). Indeed, $5^2$ already exceeds $N_5$ for both inequalities.

The method I use above has very strong similarities to the method I used in my most famous answer of all. It is sheer coincidence that 21 is a solution to both the problems I answered.

share|cite|improve this answer

edited 30 mins ago

answered 1 hour ago

Parcly TaxelParcly Taxel

43.3k1372102

$endgroup$

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 least 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 least 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.

---

*Technically we have to repeat the argument for other possible least prime factors $k$ of $n$ or $n+1$ – and the upper bound $N_k$ of the solution to the inequalities in $n$ increases accordingly, each 3 replaced with $k$. However, the least composite number with least prime factor $k$ is $k^2$, and this increases much faster than $N_k$ (which is $simfrac k2$). Indeed, $5^2$ already exceeds $N_5$ for both inequalities.

The method I use above has very strong similarities to the method I used in my most famous answer of all. It is sheer coincidence that 21 is a solution to both the problems I answered.

share|cite|improve this answer

edited 30 mins ago

answered 1 hour ago

Parcly TaxelParcly Taxel

43.3k1372102

$endgroup$

share|cite|improve this answer

edited 30 mins ago

answered 1 hour ago

Parcly TaxelParcly Taxel

43.3k1372102

answered 1 hour ago

Parcly TaxelParcly Taxel

43.3k1372102

answered 1 hour ago

Parcly TaxelParcly Taxel

43.3k1372102