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%I #24 Oct 21 2018 13:48:17
%S 1,1,3,1,3,9,3,1,9,9,3,9,3,9,27,1,3,9,3,1,27,9,3,33,43,9,27,25,3,9,3,
%T 1,27,9,47,9,3,9,27,1,3,57,3,81,63,9,3,33,31,49,27,81,3,81,67,65,27,9,
%U 3,81,3,9,27,1,113,69,3,81,27,109,3,81,3,9,57,81,75,105,3,1,81,9,3,57,73
%N a(n) = 3^n mod 2n.
%H Vincenzo Librandi, <a href="/A082511/b082511.txt">Table of n, a(n) for n = 1..2000</a>
%e Residues are often also powers of 3, that is, 3^n = k*2*n + 3^j, as is the case for n=1..23. The first terms that are not powers of 3 are a(24)=33 and a(25)=43.
%e a(6)=9: modulus = 2*n = 12; 3^n = 3^6 = 729 = 60*12 + 9 = 720 + a(6).
%t Table[PowerMod[3,n,2n],{n,90}] (* _Harvey P. Dale_, Jan 21 2014 *)
%o (Python) for n in range(1, 80): print(pow(3, n, 2*n), end=" ") # _Stefano Spezia_, Oct 20 2018
%o (PARI) a(n) = lift(Mod(3, 2*n)^n) \\ _Felix Fröhlich_, Oct 20 2018
%Y Cf. A000079, A000244, A002379.
%Y Cf. A083528, A083529, A083530.
%K nonn
%O 1,3
%A _Labos Elemer_, Apr 28 2003
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