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%I #36 Aug 15 2021 15:01:17
%S 1,3,15,35,119,255,455,1295,2555,2703,3815,3855,4355,5543,6479,8007,
%T 9215,10439,10619,11951,16211,22895,23435,26319,26795,27839,28679,
%U 35207,43055,44099,47519,47879,49679,51119,57239,61919,62567,63167,63935,65535,74447,79055
%N Numbers k such that 2^(k+1) == 1 (mod k).
%C Prime factorizations of the first ten terms: 3, 3*5, 5*7, 7*17, 3*5*17, 5*7*13, 5*7*37, 5*7*73, 3*17*53, 5*7*109.
%H Amiram Eldar, <a href="/A187787/b187787.txt">Table of n, a(n) for n = 1..10000</a>
%e 3 is in the sequence because 2^(3+1) mod 3 = 16 mod 3 = 1.
%p for n from 1 to 100000 do if 2&^(n+1) mod n = 1 then print(n) fi od;
%t m = 1; Join[Select[Range[1, m], Divisible[2^(# + 1), #] &],
%t Select[Range[m + 1, 10^5], PowerMod[2, # + 1, #] == m &]] (* _Robert Price_, Oct 11 2018 *)
%t Join[{1},Select[Range[80000],PowerMod[2,#+1,#]==1&]] (* _Harvey P. Dale_, Aug 19 2019 *)
%o (PARI) for (n=1,10^7, if (Mod(2,n)^(n+1)==1,print1(n,", "))); /* _Joerg Arndt_, Jan 06 2013 */
%Y Cf. A055685, A069927, A112983.
%K nonn
%O 1,2
%A _Franz Vrabec_, Jan 06 2013
%E Term a(1)=1 prepended by _Max Alekseyev_, Nov 29 2014
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