

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



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
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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

ZhiWei 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^m2^(mk)+1], n=n+1; Print[n, " ", 2^m2^(mk)+1]], {m, 1, 500}, {k, 1, m1}]


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

ZhiWei Sun, Dec 25 2013


STATUS

approved



