OFFSET
1,11
COMMENTS
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
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladimir Shevelev, Apr 25 2017
EXTENSIONS
More terms from Peter J. C. Moses, Apr 25 2017
STATUS
approved