OFFSET
1,2
COMMENTS
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
MATHEMATICA
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 *)
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladeta Jovovic, Jun 07 2004
EXTENSIONS
Terms > 5064320 by Joerg Arndt, Jul 03 2011
STATUS
approved