Generate some rough numbers

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Background

A number n can be described as B-rough if all of the prime factors of n strictly exceed B.

The Challenge

Given two positive integers B and k, output the first k B-rough numbers.

Examples

Let f(B, k) be a function which returns the set containing the first k B-rough numbers.

> f(1, 10)
1, 2, 3, 4, 5, 6, 7, 8, 9, 10

> f(2, 5)
1, 3, 5, 7, 9

> f(10, 14)
1, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59

code-golf number-theory primes factoring

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asked 1 hour ago

Addison Crump

8,24913279

add a comment | 

up vote
3
down vote

favorite

Background

A number n can be described as B-rough if all of the prime factors of n strictly exceed B.

The Challenge

Given two positive integers B and k, output the first k B-rough numbers.

Examples

Let f(B, k) be a function which returns the set containing the first k B-rough numbers.

> f(1, 10)
1, 2, 3, 4, 5, 6, 7, 8, 9, 10

> f(2, 5)
1, 3, 5, 7, 9

> f(10, 14)
1, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59

code-golf number-theory primes factoring

share|improve this question

asked 1 hour ago

Addison Crump

8,24913279

add a comment | 

up vote
3
down vote

favorite

up vote
3
down vote

favorite

Background

A number n can be described as B-rough if all of the prime factors of n strictly exceed B.

The Challenge

Given two positive integers B and k, output the first k B-rough numbers.

Examples

Let f(B, k) be a function which returns the set containing the first k B-rough numbers.

> f(1, 10)
1, 2, 3, 4, 5, 6, 7, 8, 9, 10

> f(2, 5)
1, 3, 5, 7, 9

> f(10, 14)
1, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59

code-golf number-theory primes factoring

share|improve this question

asked 1 hour ago

Addison Crump

8,24913279

Background

A number n can be described as B-rough if all of the prime factors of n strictly exceed B.

The Challenge

Given two positive integers B and k, output the first k B-rough numbers.

Examples

Let f(B, k) be a function which returns the set containing the first k B-rough numbers.

> f(1, 10)
1, 2, 3, 4, 5, 6, 7, 8, 9, 10

> f(2, 5)
1, 3, 5, 7, 9

> f(10, 14)
1, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59

code-golf number-theory primes factoring

code-golf number-theory primes factoring

share|improve this question

asked 1 hour ago

Addison Crump

8,24913279

share|improve this question

asked 1 hour ago

Addison Crump

8,24913279

share|improve this question

share|improve this question

asked 1 hour ago

Addison Crump

8,24913279

asked 1 hour ago

Addison Crump

8,24913279

asked 1 hour ago

Addison Crump

8,24913279

8,24913279

add a comment | 

add a comment | 

3 Answers
3

active

oldest

votes

up vote
2
down vote

Python 3, 80 bytes

lambda B,k:[*filter(lambda i:all(i%j for j in range(2,B+1)),range(1,-~B*k))][:k]

Try it online!

This assumes that the k'th B-rough number will never exceed $B * k$, which I don't know how to prove, but seems like a fairly safe assumption (and I can't find any counterexamples).

Alternate solution:

Python 2, 83 bytes

B,k=input()
l,i=,1
while k:
 if all(i%j for j in range(2,B+1)):print i;k-=1
 i+=1

Try it online!

This solution does not make the above solution. And is much more efficient.

share|improve this answer

answered 53 mins ago

DJMcMayhem♦

40.5k11143307

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hmm, that assumption is probably verifiable, but an interesting problem nonetheless. I'll bounty for a proof.
  – Addison Crump
  47 mins ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

Perl 6, 35 bytes

(grep (*X%2..$^b).all,1..*)[^$^k]

Try it online!

An anonymous code block that takes two integers and returns a list of integers.

Explanation

 # Anonymous code block
 grep ,1..* # Filter from the positive integers
 * # Is the number
 % # Not divisible by
 ( X ).all # All the numbers 
 2..$^b # From 2 to b
 ( )[^$^k] # And take the first k numbers

share|improve this answer

edited 30 mins ago

answered 55 mins ago

Jo King

18k24199

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What does all do?
  – Addison Crump
  47 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @AddisonCrump all checks if all the elements in the list are truthy. I will be adding an explanation for the whole thing shortly
  – Jo King
  44 mins ago
  

  
  
  
  

add a comment | 

up vote
0
down vote

Jelly, 10 bytes

g³!¤Ịµ⁴#1;

Try it online!

How it works

Coming soon.

share

answered 7 mins ago

Bubbler

4,559749

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

Python 3, 80 bytes

lambda B,k:[*filter(lambda i:all(i%j for j in range(2,B+1)),range(1,-~B*k))][:k]

Try it online!

This assumes that the k'th B-rough number will never exceed $B * k$, which I don't know how to prove, but seems like a fairly safe assumption (and I can't find any counterexamples).

Alternate solution:

Python 2, 83 bytes

B,k=input()
l,i=,1
while k:
 if all(i%j for j in range(2,B+1)):print i;k-=1
 i+=1

Try it online!

This solution does not make the above solution. And is much more efficient.

share|improve this answer

answered 53 mins ago

DJMcMayhem♦

40.5k11143307

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hmm, that assumption is probably verifiable, but an interesting problem nonetheless. I'll bounty for a proof.
  – Addison Crump
  47 mins ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

Python 3, 80 bytes

lambda B,k:[*filter(lambda i:all(i%j for j in range(2,B+1)),range(1,-~B*k))][:k]

Try it online!

This assumes that the k'th B-rough number will never exceed $B * k$, which I don't know how to prove, but seems like a fairly safe assumption (and I can't find any counterexamples).

Alternate solution:

Python 2, 83 bytes

B,k=input()
l,i=,1
while k:
 if all(i%j for j in range(2,B+1)):print i;k-=1
 i+=1

Try it online!

This solution does not make the above solution. And is much more efficient.

share|improve this answer

answered 53 mins ago

DJMcMayhem♦

40.5k11143307

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hmm, that assumption is probably verifiable, but an interesting problem nonetheless. I'll bounty for a proof.
  – Addison Crump
  47 mins ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

Python 3, 80 bytes

lambda B,k:[*filter(lambda i:all(i%j for j in range(2,B+1)),range(1,-~B*k))][:k]

Try it online!

This assumes that the k'th B-rough number will never exceed $B * k$, which I don't know how to prove, but seems like a fairly safe assumption (and I can't find any counterexamples).

Alternate solution:

Python 2, 83 bytes

B,k=input()
l,i=,1
while k:
 if all(i%j for j in range(2,B+1)):print i;k-=1
 i+=1

Try it online!

This solution does not make the above solution. And is much more efficient.

share|improve this answer

answered 53 mins ago

DJMcMayhem♦

40.5k11143307

Python 3, 80 bytes

lambda B,k:[*filter(lambda i:all(i%j for j in range(2,B+1)),range(1,-~B*k))][:k]

Try it online!

This assumes that the k'th B-rough number will never exceed $B * k$, which I don't know how to prove, but seems like a fairly safe assumption (and I can't find any counterexamples).

Alternate solution:

Python 2, 83 bytes

B,k=input()
l,i=,1
while k:
 if all(i%j for j in range(2,B+1)):print i;k-=1
 i+=1

Try it online!

This solution does not make the above solution. And is much more efficient.

share|improve this answer

answered 53 mins ago

DJMcMayhem♦

40.5k11143307

share|improve this answer

share|improve this answer

answered 53 mins ago

DJMcMayhem♦

40.5k11143307

answered 53 mins ago

DJMcMayhem♦

40.5k11143307

answered 53 mins ago

DJMcMayhem♦

40.5k11143307

40.5k11143307

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hmm, that assumption is probably verifiable, but an interesting problem nonetheless. I'll bounty for a proof.
  – Addison Crump
  47 mins ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hmm, that assumption is probably verifiable, but an interesting problem nonetheless. I'll bounty for a proof.
  – Addison Crump
  47 mins ago
  

  
  
  
  

Hmm, that assumption is probably verifiable, but an interesting problem nonetheless. I'll bounty for a proof.
– Addison Crump
47 mins ago

Hmm, that assumption is probably verifiable, but an interesting problem nonetheless. I'll bounty for a proof.
– Addison Crump
47 mins ago

add a comment | 

up vote
1
down vote

Perl 6, 35 bytes

(grep (*X%2..$^b).all,1..*)[^$^k]

Try it online!

An anonymous code block that takes two integers and returns a list of integers.

Explanation

 # Anonymous code block
 grep ,1..* # Filter from the positive integers
 * # Is the number
 % # Not divisible by
 ( X ).all # All the numbers 
 2..$^b # From 2 to b
 ( )[^$^k] # And take the first k numbers

share|improve this answer

edited 30 mins ago

answered 55 mins ago

Jo King

18k24199

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What does all do?
  – Addison Crump
  47 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @AddisonCrump all checks if all the elements in the list are truthy. I will be adding an explanation for the whole thing shortly
  – Jo King
  44 mins ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

Perl 6, 35 bytes

(grep (*X%2..$^b).all,1..*)[^$^k]

Try it online!

An anonymous code block that takes two integers and returns a list of integers.

Explanation

 # Anonymous code block
 grep ,1..* # Filter from the positive integers
 * # Is the number
 % # Not divisible by
 ( X ).all # All the numbers 
 2..$^b # From 2 to b
 ( )[^$^k] # And take the first k numbers

share|improve this answer

edited 30 mins ago

answered 55 mins ago

Jo King

18k24199

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What does all do?
  – Addison Crump
  47 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @AddisonCrump all checks if all the elements in the list are truthy. I will be adding an explanation for the whole thing shortly
  – Jo King
  44 mins ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Perl 6, 35 bytes

(grep (*X%2..$^b).all,1..*)[^$^k]

Try it online!

An anonymous code block that takes two integers and returns a list of integers.

Explanation

 # Anonymous code block
 grep ,1..* # Filter from the positive integers
 * # Is the number
 % # Not divisible by
 ( X ).all # All the numbers 
 2..$^b # From 2 to b
 ( )[^$^k] # And take the first k numbers

share|improve this answer

edited 30 mins ago

answered 55 mins ago

Jo King

18k24199

Perl 6, 35 bytes

(grep (*X%2..$^b).all,1..*)[^$^k]

Try it online!

An anonymous code block that takes two integers and returns a list of integers.

Explanation

 # Anonymous code block
 grep ,1..* # Filter from the positive integers
 * # Is the number
 % # Not divisible by
 ( X ).all # All the numbers 
 2..$^b # From 2 to b
 ( )[^$^k] # And take the first k numbers

share|improve this answer

edited 30 mins ago

answered 55 mins ago

Jo King

18k24199

share|improve this answer

share|improve this answer

edited 30 mins ago

edited 30 mins ago

edited 30 mins ago

answered 55 mins ago

Jo King

18k24199

answered 55 mins ago

Jo King

18k24199

answered 55 mins ago

Jo King

18k24199

18k24199

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What does all do?
  – Addison Crump
  47 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @AddisonCrump all checks if all the elements in the list are truthy. I will be adding an explanation for the whole thing shortly
  – Jo King
  44 mins ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What does all do?
  – Addison Crump
  47 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @AddisonCrump all checks if all the elements in the list are truthy. I will be adding an explanation for the whole thing shortly
  – Jo King
  44 mins ago
  

  
  
  
  

What does all do?
– Addison Crump
47 mins ago

What does all do?
– Addison Crump
47 mins ago

@AddisonCrump all checks if all the elements in the list are truthy. I will be adding an explanation for the whole thing shortly
– Jo King
44 mins ago

@AddisonCrump all checks if all the elements in the list are truthy. I will be adding an explanation for the whole thing shortly
– Jo King
44 mins ago

add a comment | 

up vote
0
down vote

Jelly, 10 bytes

g³!¤Ịµ⁴#1;

Try it online!

How it works

Coming soon.

share

answered 7 mins ago

Bubbler

4,559749

add a comment | 

up vote
0
down vote

Jelly, 10 bytes

g³!¤Ịµ⁴#1;

Try it online!

How it works

Coming soon.

share

answered 7 mins ago

Bubbler

4,559749

add a comment | 

up vote
0
down vote

up vote
0
down vote

Jelly, 10 bytes

g³!¤Ịµ⁴#1;

Try it online!

How it works

Coming soon.

share

answered 7 mins ago

Bubbler

4,559749

Jelly, 10 bytes

g³!¤Ịµ⁴#1;

Try it online!

How it works

Coming soon.

share

answered 7 mins ago

Bubbler

4,559749

share

share

answered 7 mins ago

Bubbler

4,559749

answered 7 mins ago

Bubbler

4,559749

answered 7 mins ago

Bubbler

4,559749

4,559749

add a comment | 

add a comment | 

 

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