A272143 For a given n, and any m less than n-1, the total number of primes of the form 2^n-2^m-1.

0, 1, 1, 2, 2, 3, 0, 4, 4, 3, 1, 5, 1, 4, 0, 3, 2, 8, 1, 11, 4, 5, 0, 7, 1, 2, 0, 1, 5, 4, 0, 7, 5, 1, 1, 9, 0, 6, 0, 7, 1, 6, 0, 4, 7, 2, 1, 10, 3, 3, 1, 2, 1, 6, 0, 4, 3, 0, 1, 8, 3, 3, 0, 3, 1, 8, 1, 2, 2, 3, 0, 9, 1, 5, 2, 5, 8, 3, 0, 10

Offset: 1

Author

Hans Havermann, Apr 21 2016

Keywords

• nonn

Comments

For the first 12000 terms the average is ~3.8 with a maximum of 25 at a(11520).

Essentially the same as A095058. - R. J. Mathar, Apr 24 2016

Examples

			For n=1, m<0, so there are no solutions. For n=2 there is one solution: m=0, yielding prime 2. For n=3, one solution: m=1, yielding prime 5. For n=4 there are two solutions: m=2 and m=1, yielding primes 11 and 13 respectively. The primes so formed are terms of A095078.
		

Links

• Hans Havermann, Table of n, a(n) for n = 1..12000
• Hans Havermann, Table of n, {m1, m2, ...} for primes of the form 2^n-2^m-1, m

Crossrefs

Cf. A095078.

Programs

Mathematica

Table[Length[Select[Table[2^n - 2^m - 1, {m, 0, n - 2}], PrimeQ[#] & ]], {n, 1, 100}] (* Robert Price, Apr 21 2016 *)

Python

from sympy import isprime
def a(n): return sum(1 for i in range(n-1) if isprime(2**n-1-2**i))
print([a(n) for n in range(1, 81)]) # Michael S. Branicky, Nov 09 2023