Is there a term for numbers whose sum of prime factors are “amicable”
$begingroup$
Amicable numbers are two different numbers so related that the sum of the proper divisors of each is equal to the other number.
Is there a name for two different numbers who sum of prime factors is equal to the other number?
number-theory prime-numbers prime-factorization
asked Dec 12 '18 at 22:30
lifebalancelifebalance
1014
$endgroup$
add a comment |
$begingroup$
Amicable numbers are two different numbers so related that the sum of the proper divisors of each is equal to the other number.
Is there a name for two different numbers who sum of prime factors is equal to the other number?
number-theory prime-numbers prime-factorization
asked Dec 12 '18 at 22:30
lifebalancelifebalance
1014
$endgroup$
add a comment |
$begingroup$
Amicable numbers are two different numbers so related that the sum of the proper divisors of each is equal to the other number.
Is there a name for two different numbers who sum of prime factors is equal to the other number?
number-theory prime-numbers prime-factorization
asked Dec 12 '18 at 22:30
lifebalancelifebalance
1014
$endgroup$
Amicable numbers are two different numbers so related that the sum of the proper divisors of each is equal to the other number.
Is there a name for two different numbers who sum of prime factors is equal to the other number?
number-theory prime-numbers prime-factorization
number-theory prime-numbers prime-factorization
asked Dec 12 '18 at 22:30
lifebalancelifebalance
1014
asked Dec 12 '18 at 22:30
lifebalancelifebalance
1014
asked Dec 12 '18 at 22:30
lifebalancelifebalance
1014
asked Dec 12 '18 at 22:30
lifebalancelifebalance
1014
asked Dec 12 '18 at 22:30
lifebalancelifebalance
1014
1014
add a comment |
add a comment |
1 Answer
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I suspect not, for the following reason: The sum of the prime factors of any composite number greater than $5$ is always strictly less than that number. I believe a proof by induction works here, but trying it out with a number of small examples should indicate that this sum grows more slowly than the product. So no such pairs would exist.
Here is the proof:
Let $f(n)$ be the sum of the prime factors of $n$. The claim can then be stated to be, for all $n geq 5$, if $n$ is composite, then $f(n) < n$.
Let $n geq 5$ be composite, and suppose (IH) that any composite number at most $n-1$ satisfies the above property. Let $p$ be the smallest prime factor of $n$; it suffices to show that $p + f(n/p) < n$. But $n/p geq p$ (otherwise, $n$ would have a smaller prime factor), so we have:
$$
p + fleft(frac{n}{p}right) < p + frac{n}{p} leq 2left( frac{n}{p}right) leq p left( frac{n}{p}right) = n
$$
where the last inequality comes from the fact that $p$ is prime (and therefore at least $2$).
answered Dec 12 '18 at 22:34
plattyplatty
3,350320
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1 Answer
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1 Answer
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$begingroup$
I suspect not, for the following reason: The sum of the prime factors of any composite number greater than $5$ is always strictly less than that number. I believe a proof by induction works here, but trying it out with a number of small examples should indicate that this sum grows more slowly than the product. So no such pairs would exist.
Here is the proof:
Let $f(n)$ be the sum of the prime factors of $n$. The claim can then be stated to be, for all $n geq 5$, if $n$ is composite, then $f(n) < n$.
Let $n geq 5$ be composite, and suppose (IH) that any composite number at most $n-1$ satisfies the above property. Let $p$ be the smallest prime factor of $n$; it suffices to show that $p + f(n/p) < n$. But $n/p geq p$ (otherwise, $n$ would have a smaller prime factor), so we have:
$$
p + fleft(frac{n}{p}right) < p + frac{n}{p} leq 2left( frac{n}{p}right) leq p left( frac{n}{p}right) = n
$$
where the last inequality comes from the fact that $p$ is prime (and therefore at least $2$).
answered Dec 12 '18 at 22:34
plattyplatty
3,350320
$endgroup$
add a comment |
$begingroup$
I suspect not, for the following reason: The sum of the prime factors of any composite number greater than $5$ is always strictly less than that number. I believe a proof by induction works here, but trying it out with a number of small examples should indicate that this sum grows more slowly than the product. So no such pairs would exist.
Here is the proof:
Let $f(n)$ be the sum of the prime factors of $n$. The claim can then be stated to be, for all $n geq 5$, if $n$ is composite, then $f(n) < n$.
Let $n geq 5$ be composite, and suppose (IH) that any composite number at most $n-1$ satisfies the above property. Let $p$ be the smallest prime factor of $n$; it suffices to show that $p + f(n/p) < n$. But $n/p geq p$ (otherwise, $n$ would have a smaller prime factor), so we have:
$$
p + fleft(frac{n}{p}right) < p + frac{n}{p} leq 2left( frac{n}{p}right) leq p left( frac{n}{p}right) = n
$$
where the last inequality comes from the fact that $p$ is prime (and therefore at least $2$).
answered Dec 12 '18 at 22:34
plattyplatty
3,350320
$endgroup$
add a comment |
$begingroup$
I suspect not, for the following reason: The sum of the prime factors of any composite number greater than $5$ is always strictly less than that number. I believe a proof by induction works here, but trying it out with a number of small examples should indicate that this sum grows more slowly than the product. So no such pairs would exist.
Here is the proof:
Let $f(n)$ be the sum of the prime factors of $n$. The claim can then be stated to be, for all $n geq 5$, if $n$ is composite, then $f(n) < n$.
Let $n geq 5$ be composite, and suppose (IH) that any composite number at most $n-1$ satisfies the above property. Let $p$ be the smallest prime factor of $n$; it suffices to show that $p + f(n/p) < n$. But $n/p geq p$ (otherwise, $n$ would have a smaller prime factor), so we have:
$$
p + fleft(frac{n}{p}right) < p + frac{n}{p} leq 2left( frac{n}{p}right) leq p left( frac{n}{p}right) = n
$$
where the last inequality comes from the fact that $p$ is prime (and therefore at least $2$).
answered Dec 12 '18 at 22:34
plattyplatty
3,350320
$endgroup$
I suspect not, for the following reason: The sum of the prime factors of any composite number greater than $5$ is always strictly less than that number. I believe a proof by induction works here, but trying it out with a number of small examples should indicate that this sum grows more slowly than the product. So no such pairs would exist.
Here is the proof:
Let $f(n)$ be the sum of the prime factors of $n$. The claim can then be stated to be, for all $n geq 5$, if $n$ is composite, then $f(n) < n$.
Let $n geq 5$ be composite, and suppose (IH) that any composite number at most $n-1$ satisfies the above property. Let $p$ be the smallest prime factor of $n$; it suffices to show that $p + f(n/p) < n$. But $n/p geq p$ (otherwise, $n$ would have a smaller prime factor), so we have:
$$
p + fleft(frac{n}{p}right) < p + frac{n}{p} leq 2left( frac{n}{p}right) leq p left( frac{n}{p}right) = n
$$
where the last inequality comes from the fact that $p$ is prime (and therefore at least $2$).
answered Dec 12 '18 at 22:34
plattyplatty
3,350320
edited Dec 12 '18 at 22:39
answered Dec 12 '18 at 22:34
plattyplatty
3,350320
answered Dec 12 '18 at 22:34
plattyplatty
3,350320
answered Dec 12 '18 at 22:34
plattyplatty
3,350320
3,350320
add a comment |
add a comment |
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