Is there a term for numbers whose sum of prime factors are “amicable”
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
add a comment |
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
add a comment |
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
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 |
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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
3,370320
add a comment |
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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
3,370320
add a comment |
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
3,370320
add a comment |
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
3,370320
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
3,370320
edited Dec 12 '18 at 22:39
answered Dec 12 '18 at 22:34
plattyplatty
3,370320
answered Dec 12 '18 at 22:34
plattyplatty
3,370320
answered Dec 12 '18 at 22:34
plattyplatty
3,370320
3,370320
add a comment |
add a comment |
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