OFFSET
1,1
COMMENTS
A subsequence of a(n) is the prime numbers plus 1 (A008864).
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.
It is believed that numbers in A016742 (except 4) are not included in this sequence.
EXAMPLE
4*k/(4+k) is prime for some k (let k = 4).
PROG
(PARI) a(n)=for(k=1, n*(n-1), s=(k*n)/(k+n); if(floor(s)==s, if(ispseudoprime(s), return(k))))
n=1; while(n<1000, if(a(n), print1(n, ", ")); n+=1)
CROSSREFS
KEYWORD
nonn
AUTHOR
Derek Orr, May 27 2014
STATUS
approved