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Numbers n such that 2^n == 1 (mod sigma(n)).
2

%I #21 May 14 2019 15:26:02

%S 1,2,8,25,36,50,72,128,200,288,900,1152,1764,1800,2304,3200,3528,7200,

%T 8712,10404,14112,20808,27848,28224,28800,32768,44100,56448,57600,

%U 83232,88200,112896,125316,139392,152100,181476,217800,250632,260100,294912,304200,332928,352800,362952,445568

%N Numbers n such that 2^n == 1 (mod sigma(n)).

%C From _Robert Israel_, Dec 08 2016: (Start)

%C 2^k is in the sequence if and only if k+1 is a power of 2.

%C If k and m are in the sequence with gcd(k,m) = 1 and gcd(sigma(k),sigma(m)) = 1, then k*m is in the sequence. (End)

%H Robert Israel, <a href="/A278836/b278836.txt">Table of n, a(n) for n = 1..1143</a>

%e 8 is a term because sigma(8) = 15 divides 2^8 - 1.

%p N:= 10^7: # to get all terms <= N

%p cands:= [seq(x^2,x=1..floor(sqrt(N))), seq(2*x^2, x=1..floor(sqrt(N/2)))]:

%p sort(select(n -> 2 &^n -1 mod numtheory:-sigma(n) = 0, cands)); # _Robert Israel_, Dec 08 2016

%t {1}~Join~Select[Range[5*10^5], Mod[2^#, DivisorSigma[1, #]] == 1 &] (* _Michael De Vlieger_, Dec 10 2016 *)

%t Join[{1},Select[Range[450000],PowerMod[2,#,DivisorSigma[1,#]]==1&]] (* _Harvey P. Dale_, May 14 2019 *)

%o (PARI) is(n)=Mod(2, sigma(n))^n==1;

%Y Cf. A000203, A000225, A279039. Contains A058891. Contained in A028982.

%K nonn

%O 1,2

%A _Altug Alkan_, Dec 06 2016, following a suggestion from _Michel Marcus_