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%I #12 Dec 25 2013 19:42:37
%S 3,5,7,13,17,31,61,97,127,193,241,257,769,1009,1021,2017,4093,7681,
%T 8161,8191,12289,15361,16369,16381,32257,61441,64513,65521,65537,
%U 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
%N 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.
%C 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.
%C (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.
%C Part (i) of the conjecture implies that this sequence has infinitely many terms. See also A234399.
%C 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.
%H Zhi-Wei Sun, <a href="/A234388/b234388.txt">Table of n, a(n) for n = 1..1000</a>
%e a(1) = 3 since 2*(2^{phi(1)} - 1) + 1 = 3 is prime.
%e a(2) = 5 since 2^2*(2^{phi(1)} - 1) + 1 = 5 is prime.
%e a(3) = 7 since 2*(2^{phi(3)} - 1) + 1 = 7 is prime.
%t S:=Intersection[Union[Table[EulerPhi[k],{k,1,5000}]],Table[k,{k,1,500}]]
%t 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}]
%Y Cf. A000010, A000040, A000079, A000668, A019434, A152449, A234309, A234310, A234337, A234344, A234346, A234347, A234359, A234360, A234361, A234399
%K nonn
%O 1,1
%A _Zhi-Wei Sun_, Dec 25 2013
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