Return to Answer

added 9 characters in body

Source Link

edited Oct 14, 2020 at 7:31

131.4k
13
144
384

05AB1E, 1515 14 bytes

∞+.ΔαD3mαy¿yp≠*ΔαD3mαy¿yp+

Inspired by @Giuseppe's Gaia answer, so make sure to upvote him as well!

Try it online Try it online or verify all test cases verify all test cases.

Explanation:

∞            # Push an infinite positive list: [1,2,3,...]
 +           # Add the (implicit) input `n` to each: [n+1,n+2,n+3,...]
  .Δ         # Find the first `n+b` which is truthy for:
    α        #  Take the absolute difference with the (implicit) input: [1,2,3,...]
     D       #  Duplicate this `b`
      3m     #  Cube it: b³
        α    #  Take the absolute difference with the `b` we've duplicated: |b-b³|
         y¿  #  Get the greatest common divisor with the current `b+n`: gcd(|b-b³|,b+n)
    yp≠yp       #  Check thatwhether `b+n` is NOT a prime number: !isPrime(b+n1 if prime; 0 if not)
    *+        #  MultiplyAdd them together: gcd(|b-b³|,b+n)*!isPrime+isPrime(b+n)
             #  (NOTE: only 1 is truthy in 05AB1E)
             # (after which the result is output implicitly as result)

05AB1E, 15 bytes

∞+.ΔαD3mαy¿yp≠*

Inspired by @Giuseppe's Gaia answer, so make sure to upvote him as well!

Try it online or verify all test cases.

Explanation:

∞            # Push an infinite positive list: [1,2,3,...]
 +           # Add the (implicit) input `n` to each: [n+1,n+2,n+3,...]
  .Δ         # Find the first `n+b` which is truthy for:
    α        #  Take the absolute difference with the (implicit) input: [1,2,3,...]
     D       #  Duplicate this `b`
      3m     #  Cube it: b³
        α    #  Take the absolute difference with the `b` we've duplicated: |b-b³|
         y¿  #  Get the greatest common divisor with the current `b+n`: gcd(|b-b³|,b+n)
    yp≠      #  Check that `b+n` is NOT a prime number: !isPrime(b+n)
    *        #  Multiply them together: gcd(|b-b³|,b+n)*!isPrime(b+n)
             #  (NOTE: only 1 is truthy in 05AB1E)
             # (after which the result is output implicitly as result)

05AB1E, 15 14 bytes

∞+.ΔαD3mαy¿yp+

Inspired by @Giuseppe's Gaia answer, so make sure to upvote him as well!

Try it online or verify all test cases.

Explanation:

∞            # Push an infinite positive list: [1,2,3,...]
 +           # Add the (implicit) input `n` to each: [n+1,n+2,n+3,...]
  .Δ         # Find the first `n+b` which is truthy for:
    α        #  Take the absolute difference with the (implicit) input: [1,2,3,...]
     D       #  Duplicate this `b`
      3m     #  Cube it: b³
        α    #  Take the absolute difference with the `b` we've duplicated: |b-b³|
         y¿  #  Get the greatest common divisor with the current `b+n`: gcd(|b-b³|,b+n)
    yp       #  Check whether `b+n` is a prime number (1 if prime; 0 if not)
    +        #  Add them together: gcd(|b-b³|,b+n)+isPrime(b+n)
             #  (NOTE: only 1 is truthy in 05AB1E)
             # (after which the result is output implicitly as result)

Source Link

created Oct 14, 2020 at 7:21

Kevin Cruijssen

131.4k
13
144
384

05AB1E, 15 bytes

∞+.ΔαD3mαy¿yp≠*

Inspired by @Giuseppe's Gaia answer, so make sure to upvote him as well!

Try it online or verify all test cases.

Explanation:

∞            # Push an infinite positive list: [1,2,3,...]
 +           # Add the (implicit) input `n` to each: [n+1,n+2,n+3,...]
  .Δ         # Find the first `n+b` which is truthy for:
    α        #  Take the absolute difference with the (implicit) input: [1,2,3,...]
     D       #  Duplicate this `b`
      3m     #  Cube it: b³
        α    #  Take the absolute difference with the `b` we've duplicated: |b-b³|
         y¿  #  Get the greatest common divisor with the current `b+n`: gcd(|b-b³|,b+n)
    yp≠      #  Check that `b+n` is NOT a prime number: !isPrime(b+n)
    *        #  Multiply them together: gcd(|b-b³|,b+n)*!isPrime(b+n)
             #  (NOTE: only 1 is truthy in 05AB1E)
             # (after which the result is output implicitly as result)