A232565 a(n) is the smallest k such that 2^(2^n) - 2^k - 1 is prime, or -1 if no such k exists.

0, 1, 2, 4, 2, 8, 18, 76, 32, 151, 692, 592, 154, 580, 27365, 11267

Offset: 1

Author

Charles R Greathouse IV, Nov 26 2013

Keywords

• nonn
• hard
• more

Comments

Crocker showed that 2^(2^n) - 1 - 2^a - 2^b is not prime (with n > 2) if a and b are distinct. This sequence demonstrates that the theorem is sharp in the sense that distinctness is required.

If n > 2, then the (largest) prime P(n) = 2^(2^n)-2^a(n)-1 is a de Polignac number (A065381); i.e., P(n)-2^m is not prime. It seems that if n > 6, then |P(n)-2^m| is composite for every natural m and P(n)*2^m-1 is composite (by the dual Riesel conjecture). So if n > 6, then the prime P(n) may be a Riesel number (A182296). For example, the prime P(7) = 2^(2^7)-(2^18+1) is the first candidate (note that 2^18+1 is the smallest de Polignac number of form 2^k+1). Also, by Crocker's theorem, the smallest number of form 2^(2^n)-2^m-1, namely 2^(2^n-1)-1 is a de Polignac number (A006285) and for n > 6 may be a dual Riesel number (A101036). For example, the double Mersenne prime 2^(2^7-1)-1 probably is a dual Riesel number. It is not known whether these are Riesel numbers with a covering set. - Thomas Ordowski, Jan 24 2024

Links

• Roger Crocker, "On the sum of a prime and of two powers of two", Pacific Journal of Mathematics 36:1 (1971), pp. 103-107.

Crossrefs

Cf. A156695.

Programs

PARI

a(n)=my(N=2^2^n-1);for(a=1,2^n-1,if(ispseudoprime(N-2^a), return(a)));0

Extensions

a(15) from Charles R Greathouse IV, Dec 02 2013

a(16) from Daniel Suteu, Oct 11 2020

Name edited by Thomas Ordowski, Jan 24 2024