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a(n) = A003474(n)/n.
For n>=2, a(n) = f(n)/(2^(n-1)) where f(n) is the number of Hamiltonian cycles in the 3-ary De 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.
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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]];
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]];
Array[a, 40] (* Jean-François Alcover, Jul 22 2018, after Joerg Arndt *)
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Number of normal bases for GF(3^n) over GF(3). [_- _Joerg Arndt_, Jul 03 2011]
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]. For details on this correspondence, see A192513 [_Dmitrii Pasechnik_, Dec 07 2014].
For details on this correspondence, see A192513. - Dmitrii Pasechnik, Dec 07 2014
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