A359735 - OEIS

%I #39 Jan 28 2023 11:56:10

%S 0,1,-1,2,0,1,0,0,0,2,0,0,0,-1,0,1,0,0,0,-1,0,-1,0,0,0,0,0,0,0,0,0,0,

%T 0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,-1,0,0,0,0,0,0,0,0,0,0,

%U 0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0

%N Let f(s,n) = 2^n + s*n, with s in {-1, 1}. Let c be the number of primes out of the pair f(-1,n), f(1,n). If only f(-1,n) is prime, a(n) = -1, otherwise a(n) = c.

%F a(n) can be = 2 only if n = 6*m + 3 for m >= 0 and m is not congruent to {0, 4} mod 5, not congruent to {2, 4} mod 7, not congruent to {6, 7} mod 11 and not congruent to {3, 9} mod 13. Does a(n) = 2 for n > 9 exist? - _Thomas Scheuerle_, Jan 12 2023

%e f(-1,1) = 2^1 - 1 = 1, is not prime and f(1,1) = 2^1 + 1 = 3 is prime, so a(1) = 1.

%e f(-1,2) = 2^2 - 2 = 2, is prime and f(1,2) = 2^2 + 2 = 6 = 2 * 3 is not prime, so a(2) = -1.

%e f(-1,3) = 2^3 - 3 = 5, is prime and f(1,3) = 2^3 + 3 = 11 is prime, so a(3) = 2.

%o (PARI) f(s,n)=2^n+s*n

%o a(n)=my(a=isprime(f(-1,n)),b=isprime(f(1,n)),c=a+b); if(c==1&&a==1,return(-1),return(c))

%Y Cf. A006127, A048744, A052007, A129962.

%K sign

%O 0,4

%A _Jean-Marc Rebert_, Jan 12 2023