cp's OEIS Frontend

Conjecture: every even semiprime more than 22 is a sum of two semiprimes.

All terms are either 4 or odd.

First occurrence of k-th odd semiprime (A046315): 7, 16, 9, 13, 31, 41, 104, 134, 66, 412, 2769, 447, 1282, 3868, 7003, 3601, 48649, 11016, 5379, 41644, 34575, 83474, 120165, 135566, 21335, 394140, 14899, 876518, 434986, 173914, 691409, 1854580, 3741206, 714807, 1001321, 6427837, 4267513, 14809496, 7795998, 26617567, 2001937, 13958857, 9217135, 18815676, ..., . - Robert G. Wilson v, Apr 26 2014

If a(n) = 4, then 2*prime(n)-4=2*(prime(n)-2) is a semiprime, thus prime(n)-2 is a prime, so prime(n) belongs to A006512 (greater of twin primes). - Michel Marcus, Mar 26 2015

If the sequence has no zeros at least for sufficiently large n, then one can believe that, for a pair of sufficiently large semiprimes of the same parity {r,s}, there is a number k=k(r,s) such that either {r-k, s+k} or {r+k, s-k} is a pair of primes. Then, if a representation 2*n = r+s with, say, min{r,s} > log(n) is considered, then, at least for sufficiently large n, it is reduced to the Goldbach representation 2*n = p+q with primes p,q. It is natural to think that a Goldbach-like conjecture that at least every sufficiently large even number is a sum of two semiprimes could be proved more easily than the classic Goldbach conjecture (cf. Chen's theorem).

If a(n)=2, then prime(n)+2 and prime(n)-2 are both semiprimes; that is, prime(n) belongs to A063643. - Michel Marcus, Mar 26 2015