A234388 Primes of the form 2^k*(2^{phi(m)} - 1) + 1, where k and m are positive integers, and phi(.) is Euler's totient function.

3, 5, 7, 13, 17, 31, 61, 97, 127, 193, 241, 257, 769, 1009, 1021, 2017, 4093, 7681, 8161, 8191, 12289, 15361, 16369, 16381, 32257, 61441, 64513, 65521, 65537, 131041, 131071, 523777, 524287, 786433, 1032193, 1048573, 4194301, 8257537, 8380417, 16515073, 16760833, 16776961, 16777153, 16777213, 67043329, 132120577, 134215681, 268369921, 536870401, 1073479681, 2013265921, 2113929217, 2146959361, 2147483137, 2147483647, 3221225473, 4293918721, 17175674881, 34359214081, 34359738337

Offset: 1

Author

Zhi-Wei Sun, Dec 25 2013

Keywords

• nonn

Comments

Conjecture: (i) Any integer n > 1 can be written as k + m with k > 0 and m > 0 such that 2^k*(2^{phi(m)} - 1) + 1 is prime.

(ii) Each integer n > 2 can be written as k + m with k > 0 and m > 0 such that 2^k*(2^{phi(m)} - 1) - 1 is prime.

Part (i) of the conjecture implies that this sequence has infinitely many terms. See also A234399.

Note that the sequence contains all Fermat primes and Mersenne primes since 2^k + 1 = 2^k*(2^{phi(1)} - 1) + 1 and 2^p - 1 = 2*(2^{phi(p)} - 1) + 1, where k is a positive integer and p is a prime.

Examples

			a(1) = 3 since 2*(2^{phi(1)} - 1) + 1 = 3 is prime.
a(2) = 5 since 2^2*(2^{phi(1)} - 1) + 1 = 5 is prime.
a(3) = 7 since 2*(2^{phi(3)} - 1) + 1 = 7 is prime.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A000010, A000040, A000079, A000668, A019434, A152449, A234309, A234310, A234337, A234344, A234346, A234347, A234359, A234360, A234361, A234399

Programs

Mathematica

S:=Intersection[Union[Table[EulerPhi[k],{k,1,5000}]],Table[k,{k,1,500}]]
n=0;Do[If[MemberQ[S,k]&&PrimeQ[2^m-2^(m-k)+1],n=n+1;Print[n," ",2^m-2^(m-k)+1]],{m,1,500},{k,1,m-1}]