A359735 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.

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, 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, 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

Offset

0,4

Links

Table of n, a(n) for n=0..99.

Formula

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

Example

f(-1,1) = 2^1 - 1 = 1, is not prime and f(1,1) = 2^1 + 1 = 3 is prime, so a(1) = 1.
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.
f(-1,3) = 2^3 - 3 = 5, is prime and f(1,3) = 2^3 + 3 = 11 is prime, so a(3) = 2.

Prog

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

Crossrefs

Cf. A006127, A048744, A052007, A129962.

Sequence in context: A232243 A034876 A091393 * A284557 A182032 A265245

Adjacent sequences: A359732 A359733 A359734 * A359736 A359737 A359738

Keyword

sign

Author

Jean-Marc Rebert, Jan 12 2023

Status

approved