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

1,1

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.

Links

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

Example

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.

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}]

Crossrefs

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

Sequence in context: A339506 A178490 A182981 * A003424 A073638 A066464

Adjacent sequences: A234385 A234386 A234387 * A234389 A234390 A234391

Keyword

nonn

Author

Zhi-Wei Sun, Dec 25 2013

Status

approved