%N Numbers n such that k*n/(k+n) is prime for some k.
%C A subsequence of a(n) is the prime numbers plus 1 (A008864).
%C a(n) is even for all n > 1. Proof: There are four possibilities for n and k: odd-odd, odd-even, even-even (without loss of generality, even-odd and odd-even are the same). If k and n are odd, then the numerator is odd and the denominator is even. Thus, this will never be an integer or prime. If k and n are even, the numerator is even and the denominator is even. An even divided by an even could be odd or even so primes are a possibility. If one is odd and one is even, the numerator is even and the denominator is odd. The only way this is prime is if it equals 2. Thus, letting k = 2a and n = 2b+1, then 2a*(2b+1)/(2a+2b+1) = 2. Solving this, we get that a=3 and b=1 (meaning k = 6 and n = 3). So, 3 is the only odd number in this sequence.
%C It is believed that numbers in A016742 (except 4) are not included in this sequence.
%e 4*k/(4+k) is prime for some k (let k = 4).
%o (PARI) a(n)=for(k=1,n*(n-1),s=(k*n)/(k+n);if(floor(s)==s,if(ispseudoprime(s),return(k))))
%o n=1;while(n<1000,if(a(n),print1(n,", "));n+=1)
%Y Cf. A008864, A016742.
%A _Derek Orr_, May 27 2014