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%I #4 Jun 02 2012 06:36:46
%S 1,2,3,4,5,6,8,10,12,14,15,16,17,18,20,24,26,28,30,32,34,36,40,42,43,
%T 44,46,48,51,52,56,58,60,64,68,70,72,76,78,80,84,85,88,90,96,100,102,
%U 104,112,120,124,126,127,128,130,132,136,140,141,144,145,148,156,160,164
%N Numbers n for which 2^^n == 2^2^n (mod n); for the "^^" notation see A092188.
%H Robert Munafo, <a href="http://www.mrob.com/pub/math/seq-a162002.html">2^^N == 2^(2^N) mod N</a>
%e 3 is in the sequence because 2^2^3 = 2^8 = 256 == 1 mod 3, and 2^^3 = 2^2^2 = 2^4 = 16 == 1 mod 3.
%Y Cf. A092188, A162018, A014221
%K easy,nonn
%O 1,2
%A _Robert Munafo_, Jun 24 2009
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