JavaScript (Node.js), 86 bytes

f=(a,b=2,c=(a,b=2)=>a-b?a%b&&c(a,b+1):1,d=a=>c(++a)?a:d(a))=>!c(b)|d(b)-b<a?f(a,b+1):b

Try it online!

posted almost 4 years ago

CC0

Hakerh400‭

1591 reputation 4 55 164 63

Dyalog APL Extended, 14 bytes

{¯4⍭4⍭⍣(⍵≤-)2}

Try it online!

{¯4⍭4⍭⍣(⍵≤-)2}  Monadic dfn
            2   start with 2
      ⍣         Repeat
    4⍭          the function "next prime"
       (⍵≤-)    until the difference from the previous one is ≥ the input
 ¯4⍭            previous prime of that

posted almost 4 years ago

CC BY-SA 4.0

dzaima‭

759 reputation 6 10 89 14

Husk, 8 bytes

Ψḟo≥⁰≠İp

Try it online! or Verify first 8 values

It is always a good day when you get to use Ψ in your program.

Explanation

Ψḟo≥⁰≠İp
      İp to the infinite list of prime numbers,
Ψ        apply this higher order function on overlapping pairs
 ḟo      first element where
     ≠   absolute difference
   ≥⁰    is greater than or equal to the input.

posted almost 4 years ago

CC BY-SA 4.0

4y ago

Razetime‭

2860 reputation 42 65 321 83

Raku, 33 bytes

{1+(3...{($^a...&is-prime)>=$_})}

Try it online!

Anonymous code block that takes a number and returns a number.

Explanation

{                               }  # Anonymous code block
   (3...{                     })   # Increment from 3 until
         (               )>=$_     # The input is less than or equal to
          $^a...&is-prime          # The difference between the current number and the next prime plus 1
 1+                                # And add one to the length of this list

posted almost 4 years ago

CC BY-SA 4.0

4y ago

Jo King‭

251 reputation 0 6 25 3

Japt, 15 14 bytes

@§XnÈj}aX+j}a2

Try it or run all test cases

@§XnÈj}aX+j}a2     :Implicit input of integer U
@                  :Function taking an integer X as argument
 §                 :  Is U <= ...
  Xn               :  Subtract X from
    È              :    Function taking an integer as input
     j             :      Is prime?
      }            :    End function
       a           :    Get the first integer that returns true when passed through that function starting at
        X+j        :      X plus is X prime
           }       :End function
            a2     :Get the first integer that returns true when passed through that function starting at 2

posted over 3 years ago

CC BY-SA 4.0

3y ago

Shaggy‭

2111 reputation 4 115 214 166

Ruby, 56 bytes

->n{require'prime';Prime.each_cons(2).find{_2-_1>=n}[0]}

Attempt This Online!

posted almost 3 years ago

CC BY-SA 4.0

radarek‭

199 reputation 0 31 19 42

C (gcc), 126 129 bytes

N=9999;f(n){int p[N],i,j,P;memset(p,1,N);for(i=P=2;i*i<N;i++)if(p[i]){for(j=i*i;j<N;j+=i)p[j]=0;i-P>=n?j=N:(P=i);}e:return P;}

Try it online!

This is an integer input/output function solution. The upper limit of prime number supported is the square root ofN, so currently it counts prime numbers up to 99 and prints nonsense if needed to go beyond that, but it can be extended up to sqrt(INT_MAX) long as the stack can handle the VLA p.

I'm still a rookie at this, quite likely the algorithm itself (Sieve of Eratosthenes) is naive for code golfing purposes. I'm also quite sure that this could be rewritten with recursion somehow to shave off a bit of loop syntax overhead...

posted almost 4 years ago

CC BY-SA 4.0

4y ago

Lundin‭

1149 reputation 17 43 134 70

JavaScript (Node.js), 81 bytes

d=(p,i=2)=>i<p?!(p%i)||d(p,i+1):0
f=(n,a=2,p=a+1)=>d(p)?f(n,a,p+1):p-a<n?f(n,p):a

Try it online!

Explanation:

d is a helper function that returns true if p is not prime.

f=(n, a=2, p=a+1) =>
    d(p)?f(n, a, p + 1)  // Recurse until p is prime
                         // p starts at a+1 so it gets the prime after a
    :p-a<n?f(n, p):a     // If a, p doesn't work, we already computed the next prime so it's easy to recurse

posted almost 4 years ago

CC BY-SA 4.0

Moshi‭

1722 reputation 17 58 204 80