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%I #28 Apr 30 2017 23:13:30
%S 0,0,1,1,0,1,1,1,1,1,3,1,1,2,1,1,1,1,2,1,1,1,1,2,1,1,1,1,0,1,1,1,3,1,
%T 1,2,1,1,1,3,1,2,1,2,1,1,4,1,1,2,1,1,2,4,5,1,1,1,2,1,2,1,1,2,1,2,3,3,
%U 1,2,3,1,1,2,0,1,1,1,2,1,1,1,2,2,1,3,2,1,5
%N Let q_1<q_2<... be primes not dividing 2n+1 such that 2n-q_i + 1 is twice a prime. a(n) is smallest i such that 2n+2q_i+1 is prime or a(n)=0 if there is no such i.
%C 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?
%C The positions of zeros are {1,2,5,29,75} up to 100000. - _Peter J. C. Moses_, Apr 26 2017
%C 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
%C No other zeros up to 5*10^6. - _Michel Marcus_, Apr 29 2017
%o (PARI) a(n) = {i = 0; forprime(p=2, 2*n+1, if ((gcd(p, 2*n+1)==1) && ((2*n-p) % 2) && isprime((2*n-p+1)/2), i++; if (isprime(2*n+2*p+1), return(i)););); return(0);} \\ _Michel Marcus_, Apr 29 2017
%Y Cf. A277688.
%K nonn
%O 1,11
%A _Vladimir Shevelev_, Apr 25 2017
%E More terms from _Peter J. C. Moses_, Apr 25 2017