cp's OEIS Frontend

We conjecture that the sequence is infinite. This sequence shows the proximity of A284919 and A277688 based on Goldbach and Lemoine-Levy representations of even and odd numbers respectively.

The differences between the positions of the corresponding terms of A284919 and A277688 are 0,0,0,1,0,2,2,1,1,1,2,3,2,2,2,2,3,1,... .

The differences between these terms are -1,-1,-1,1,-1,-1,1,1,-1,-1,-1,-1,-1,1,-1,1,1,... .

Or, even integers k such that k + p is composite for all primes p, q with p + q = k.

The two initial terms vacuously satisfy the definition, but all even numbers >= 4 are the sum of two primes, according to the Goldbach conjecture.

All odd numbers except for numbers m such that m-2 and m+2 are prime (= A087679) would satisfy the definition. - M. F. Hasler, Apr 05 2017

Conjecture: except for a(4)=6, all terms are coprime to 3. - Bob Selcoe, Apr 06 2017

If E is an even number not divisible by 3, then E is in the sequence unless E-3 and at least one of E+3 and 2E-3 are prime. - Robert Israel, Apr 10 2017

Consider a subsequence with the additional condition: n+odd part of p-1 is composite (for example, for p=19 it is 9). I found that this subsequence begins 0,2,118 and up to 300000 Peter J. C. Moses found only more one term 868. Is this subsequence finite? - Vladimir Shevelev, Apr 12 2017

One can compare the theoretical maxima with the actual sequence numbers of terms. Doing this at powers of 10, we see at powers {2,3,4,5,6} the ratio progression {2.33, 1.51, 1.25, 1.15, 1.096}. This implies that the excluded even numbers become increasingly rare (those coprime to 3). - Bill McEachen, Apr 17 2017

From Robert Israel's comment and the distribution of primes, the proportion of even numbers not divisible by 3 that are in the sequence tends to 1. - Peter Munn, Apr 23 2017

Moreover, If n is not divisible by 3 and 2*n - 3 is composite, then 2*n+p is also composite. Indeed, for these 2*n all primes p such that 2*n-p is prime are in the interval (3, 2*n-3). Then either 2*n-p or 2*n+p should be divisible by 3, but 2*n-p is a prime > 3. So 2*n+p is composite and 2*n is in the sequence. - Vladimir Shevelev, Apr 28 2017

By the Lemoine-Levy conjecture, for every n>=3, there are primes p and q such that 2*n+1=2*p+q. In A277688 were considered the numbers of the form 2n+1+2*p, in this sequence we consider the numbers of the form 2*n+1+2*q. Then to the same condition as in A277688 satisfies an extremely rare set of numbers that contains numbers {2*n+1} for which a(n)=0: {11, 59, 151,...}. Comparing this with our conjecture in A277688, we conjecture here that this set is finite. For the explanation of this conjecture we need not refer to the minimal number of the representation 2*n+1 for large n (most likely, it is, as in the Goldbach presentations, for 2*n, more than c*n/(log n)^2 with some constant c) since we have a prime only among the first several of these representations, as in this sequence. This leaves us with an important question: why does A277688 contain much more terms than the zeros in this sequence?

The positions of zeros are {1,2,5,29,75} up to 100000. - Peter J. C. Moses, Apr 26 2017

The answer on the question follows from the following arguments. Note that if, for a prime p, 2*n+1-2*p is prime, then it is larger than 3 (if not, then p>=n and 2*n+1-2*p<=1). Now if 2*n+1 is not divisible by 3 and 2*n+1-6 is composite, then either 2*n+1-2*p or 2*n+1+2*p is divisible by 3 and, since the first number is prime >3, then 2*n+1+2*p is divisible by 3 and thus such 2*n+1 is in A277688. - Vladimir Shevelev, Apr 28 2017

No other zeros up to 5*10^6. - Michel Marcus, Apr 29 2017