Smallest integers after N divisible by 2, 3, and 4

Python 2, 32 bytes

lambda n:[n+2&-2,n/3*3+3,n+4&-4]

Bit arithmetic for 2 and 4, modular arithmetic for 3.

I found four 7-byte expressions for the next multiple of k above n but none shorter:

n-n%k+k
~n%k-~n
n/k*k+k
~n/k*-k

Any gives 34 bytes when copies for k=2,3,4, and 33 bytes if combined:

[n/2*2+2,n/3*3+3,n/4*4+4]
[n/k*k+k for k in 2,3,4]

But, 2 and 4 are powers of 2 that allow bit tricks to zero out the last 1 or 2 bytes.

n+2&-2
n+4&-4

This gives 6 bytes (instead of 7) for getting the next multiple, for 32 bytes overall, beating the for k in 2,3,4.

Unfortunately, the promising-looking n|1+1 and n|3+1 have the addition done first, so incrementing the output takes parentheses.

Jelly, 8 bytes

~%2r4¤+‘

Try it online! or verify all test cases.

How it works

~%2r4¤+‘  Main link. Argument: n (integer)

~         Bitwise NOT; yield ~n = -(n + 1).
     ¤    Combine the three links to the left into a niladic chain:
  2         Yield 2.
   r4       Yield the range from 2 to 4, i.e., [2, 3, 4].
 %        Yield the remainder of the division of ~n by 2, 3 and 4.
          In Python/Jelly, -(n + 1) % k = k - (n + 1) % k if n, k > 0.
       ‘  Yield n + 1.
      +   Add each modulus to n + 1.

Julia, 16 bytes

n->n-n%(r=2:4)+r

Try it online!