Maximum number of distinct prime factors of numbers below $2^{64}$
0
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What is the maximum number of distinct prime factors of numbers below $2^{64}$? I'm interested in the exact count, not just an estimate.
In other words, what is the largest $omega(n)$, where $n < 2^{64}$?
number-theory prime-numbers prime-factorization
asked Dec 19 '18 at 9:18
Ecir HanaEcir Hana
417414
$endgroup$
add a comment |
0
$begingroup$
What is the maximum number of distinct prime factors of numbers below $2^{64}$? I'm interested in the exact count, not just an estimate.
In other words, what is the largest $omega(n)$, where $n < 2^{64}$?
number-theory prime-numbers prime-factorization
asked Dec 19 '18 at 9:18
Ecir HanaEcir Hana
417414
$endgroup$
add a comment |
0
0
0
$begingroup$
What is the maximum number of distinct prime factors of numbers below $2^{64}$? I'm interested in the exact count, not just an estimate.
In other words, what is the largest $omega(n)$, where $n < 2^{64}$?
number-theory prime-numbers prime-factorization
asked Dec 19 '18 at 9:18
Ecir HanaEcir Hana
417414
$endgroup$
What is the maximum number of distinct prime factors of numbers below $2^{64}$? I'm interested in the exact count, not just an estimate.
In other words, what is the largest $omega(n)$, where $n < 2^{64}$?
number-theory prime-numbers prime-factorization
number-theory prime-numbers prime-factorization
asked Dec 19 '18 at 9:18
Ecir HanaEcir Hana
417414
asked Dec 19 '18 at 9:18
Ecir HanaEcir Hana
417414
asked Dec 19 '18 at 9:18
Ecir HanaEcir Hana
417414
asked Dec 19 '18 at 9:18
Ecir HanaEcir Hana
417414
asked Dec 19 '18 at 9:18
Ecir HanaEcir Hana
417414
417414
add a comment |
add a comment |
1 Answer
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5
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The number with the most distinct prime factors will have the form
$$2cdot 3cdot 5cdot 7cdots$$
So all you need is to multiply all primes until you reach $2^{64}$.
answered Dec 19 '18 at 9:20
5xum5xum
91.3k394161
$endgroup$
$begingroup$
It's 15, in case anyone's wondering... Thanks!
$endgroup$
– Ecir Hana
Dec 19 '18 at 9:26
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How did you get 15 @EcirHana ?
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– rsadhvika
Dec 19 '18 at 9:36
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@rsadhvika the product of 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 is 614889782588491410, which is less than 614889782588491410 ($2^{64}$). Including 53 it is 32589158477190044730.
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– Ecir Hana
Dec 19 '18 at 9:45
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@EcirHana Is that a typo?
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– Szeto
Dec 19 '18 at 9:47
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@Szeto Sorry, what is a typo?
$endgroup$
– Ecir Hana
Dec 19 '18 at 9:48
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
5
$begingroup$
The number with the most distinct prime factors will have the form
$$2cdot 3cdot 5cdot 7cdots$$
So all you need is to multiply all primes until you reach $2^{64}$.
answered Dec 19 '18 at 9:20
5xum5xum
91.3k394161
$endgroup$
$begingroup$
It's 15, in case anyone's wondering... Thanks!
$endgroup$
– Ecir Hana
Dec 19 '18 at 9:26
$begingroup$
How did you get 15 @EcirHana ?
$endgroup$
– rsadhvika
Dec 19 '18 at 9:36
$begingroup$
@rsadhvika the product of 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 is 614889782588491410, which is less than 614889782588491410 ($2^{64}$). Including 53 it is 32589158477190044730.
$endgroup$
– Ecir Hana
Dec 19 '18 at 9:45
$begingroup$
@EcirHana Is that a typo?
$endgroup$
– Szeto
Dec 19 '18 at 9:47
$begingroup$
@Szeto Sorry, what is a typo?
$endgroup$
– Ecir Hana
Dec 19 '18 at 9:48
|
show 1 more comment
5
$begingroup$
The number with the most distinct prime factors will have the form
$$2cdot 3cdot 5cdot 7cdots$$
So all you need is to multiply all primes until you reach $2^{64}$.
answered Dec 19 '18 at 9:20
5xum5xum
91.3k394161
$endgroup$
$begingroup$
It's 15, in case anyone's wondering... Thanks!
$endgroup$
– Ecir Hana
Dec 19 '18 at 9:26
$begingroup$
How did you get 15 @EcirHana ?
$endgroup$
– rsadhvika
Dec 19 '18 at 9:36
$begingroup$
@rsadhvika the product of 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 is 614889782588491410, which is less than 614889782588491410 ($2^{64}$). Including 53 it is 32589158477190044730.
$endgroup$
– Ecir Hana
Dec 19 '18 at 9:45
$begingroup$
@EcirHana Is that a typo?
$endgroup$
– Szeto
Dec 19 '18 at 9:47
$begingroup$
@Szeto Sorry, what is a typo?
$endgroup$
– Ecir Hana
Dec 19 '18 at 9:48
|
show 1 more comment
5
5
5
$begingroup$
The number with the most distinct prime factors will have the form
$$2cdot 3cdot 5cdot 7cdots$$
So all you need is to multiply all primes until you reach $2^{64}$.
answered Dec 19 '18 at 9:20
5xum5xum
91.3k394161
$endgroup$
The number with the most distinct prime factors will have the form
$$2cdot 3cdot 5cdot 7cdots$$
So all you need is to multiply all primes until you reach $2^{64}$.
answered Dec 19 '18 at 9:20
5xum5xum
91.3k394161
answered Dec 19 '18 at 9:20
5xum5xum
91.3k394161
answered Dec 19 '18 at 9:20
5xum5xum
91.3k394161
answered Dec 19 '18 at 9:20
5xum5xum
91.3k394161
91.3k394161
$begingroup$
It's 15, in case anyone's wondering... Thanks!
$endgroup$
– Ecir Hana
Dec 19 '18 at 9:26
$begingroup$
How did you get 15 @EcirHana ?
$endgroup$
– rsadhvika
Dec 19 '18 at 9:36
$begingroup$
@rsadhvika the product of 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 is 614889782588491410, which is less than 614889782588491410 ($2^{64}$). Including 53 it is 32589158477190044730.
$endgroup$
– Ecir Hana
Dec 19 '18 at 9:45
$begingroup$
@EcirHana Is that a typo?
$endgroup$
– Szeto
Dec 19 '18 at 9:47
$begingroup$
@Szeto Sorry, what is a typo?
$endgroup$
– Ecir Hana
Dec 19 '18 at 9:48
|
show 1 more comment
$begingroup$
It's 15, in case anyone's wondering... Thanks!
$endgroup$
– Ecir Hana
Dec 19 '18 at 9:26
$begingroup$
How did you get 15 @EcirHana ?
$endgroup$
– rsadhvika
Dec 19 '18 at 9:36
$begingroup$
@rsadhvika the product of 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 is 614889782588491410, which is less than 614889782588491410 ($2^{64}$). Including 53 it is 32589158477190044730.
$endgroup$
– Ecir Hana
Dec 19 '18 at 9:45
$begingroup$
@EcirHana Is that a typo?
$endgroup$
– Szeto
Dec 19 '18 at 9:47
$begingroup$
@Szeto Sorry, what is a typo?
$endgroup$
– Ecir Hana
Dec 19 '18 at 9:48
$begingroup$
It's 15, in case anyone's wondering... Thanks!
$endgroup$
– Ecir Hana
Dec 19 '18 at 9:26
$begingroup$
It's 15, in case anyone's wondering... Thanks!
$endgroup$
– Ecir Hana
Dec 19 '18 at 9:26
$begingroup$
How did you get 15 @EcirHana ?
$endgroup$
– rsadhvika
Dec 19 '18 at 9:36
$begingroup$
How did you get 15 @EcirHana ?
$endgroup$
– rsadhvika
Dec 19 '18 at 9:36
$begingroup$
@rsadhvika the product of 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 is 614889782588491410, which is less than 614889782588491410 ($2^{64}$). Including 53 it is 32589158477190044730.
$endgroup$
– Ecir Hana
Dec 19 '18 at 9:45
$begingroup$
@rsadhvika the product of 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 is 614889782588491410, which is less than 614889782588491410 ($2^{64}$). Including 53 it is 32589158477190044730.
$endgroup$
– Ecir Hana
Dec 19 '18 at 9:45
$begingroup$
@EcirHana Is that a typo?
$endgroup$
– Szeto
Dec 19 '18 at 9:47
$begingroup$
@EcirHana Is that a typo?
$endgroup$
– Szeto
Dec 19 '18 at 9:47
$begingroup$
@Szeto Sorry, what is a typo?
$endgroup$
– Ecir Hana
Dec 19 '18 at 9:48
$begingroup$
@Szeto Sorry, what is a typo?
$endgroup$
– Ecir Hana
Dec 19 '18 at 9:48
|
show 1 more comment
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