%I #30 Jul 22 2018 08:42:24

%S 1,2,6,8,32,54,208,256,1458,2560,10648,17496,70304,151424,629856,

%T 819200,5064320,9565938,40781104,65536000,331619184,623589472,

%U 2728756984,3673320192,22315420160,32127240704,188286357654,321009188864,1577709824480,2975389355520,13283298844816,17626562560000

%N a(n) = A003474(n)/n.

%C Number of normal bases for GF(3^n) over GF(3). - _Joerg Arndt_, Jul 03 2011

%C For n>=2, a(n) = f(n)/(2^(n-1)) where f(n) is the number of Hamiltonian cycles in the 3-ary de Bruijn graph (i.e., graph with 3*n nodes {0..3*n-1} and edges from each i to 3*i (mod 3*n), 3*i+1 (mod 3*n), and 3*i+2 (mod 3*n); cf. A192513). - _Joerg Arndt_, Jul 03 2011.

%C For details on this correspondence, see A192513. - _Dmitrii Pasechnik_, Dec 07 2014

%t p = 3; numNormalp[n_] := Module[{r, i, pp = 1}, Do[r = MultiplicativeOrder[p, d]; i = EulerPhi[d]/r; pp *= (1 - 1/p^r)^i, {d, Divisors[n]}]; Return[pp]];

%t a[1] = 1; a[n_] := Module[{t = 1, q = n, pp}, While[0 == Mod[q, p], q /= p; t += 1]; pp = numNormalp[q]; pp *= p^n/n; Return[pp]];

%t Array[a, 40] (* _Jean-François Alcover_, Jul 22 2018, after _Joerg Arndt_ *)

%o (PARI) a(n)=if(n==1,return(1));my(r,i,t=3^n/n);fordiv(n/3^valuation(n,3), d, r=znorder(Mod(3,d)); i=eulerphi(d)/r; t*=(1-1/3^r)^i);t \\ _Charles R Greathouse IV_, Jan 03 2013

%K nonn

%O 1,2

%A _Vladeta Jovovic_, Jun 07 2004

%E Terms > 5064320 by _Joerg Arndt_, Jul 03 2011