OFFSET
1,8
COMMENTS
a(n)=0 for n = 1, 2, 3, 7, 19, 31, 139. It is conjectured that there is not any other n for which a(n)=0.
LINKS
Lei Zhou, Table of n, a(n) for n = 1..10000
Lei Zhou, Plot of a(n) for n <= 20000.
EXAMPLE
n=4, 2n=8=3+5. 5-3+1=3 is prime, so a(4)=1;
n=7, 2n=14=3+11. 11-3+1=9 is not prime, so a(7)=0;
...
n=18 2n=36=5+31=7+29=13+23=17+19. 31-5+1=27 is composite, 29-7+1=23 is prime, 23-13+1=11 is prime, 19-17+1=3 is prime: three primes in the form of p2-p1+1 found, so a(18)=3.
MATHEMATICA
Table[e = 2 n; ct = 0; p1 = 1; While[p1 = NextPrime[p1]; p1 < n, p2 = e - p1; If[PrimeQ[p2], If[PrimeQ[p2 - p1 + 1], ct++]]]; ct, {n, 1, 100}]
PROG
(Python)
from sympy import isprime, nextprime
def A254688(n):
....y, x, n2 = 0, 2, 2*n
....while x < n:
........if isprime(n2-x) and isprime(n2-2*x+1):
............y += 1
........x = nextprime(x)
....return y # Chai Wah Wu, Feb 18 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Lei Zhou, Feb 05 2015
STATUS
approved