

A328643


Positive integers m such that the matrix E_m has order 3^m1 in GL_m(3) where E_m is the m X m invertible tridiagonal matrix with all nonzero entries equal to 1 except for the (m,m) entry that is equal to 2.


1



1, 3, 5, 9, 11, 23, 29, 35, 39, 41, 53, 65, 69, 81, 83, 89, 95, 99, 105, 113, 119, 131, 155, 173, 179, 189, 191, 209, 221, 231, 233, 239, 243, 251, 281, 293, 299, 303, 323, 329, 359, 371, 375, 411, 413, 419, 429, 431, 443, 453, 491
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OFFSET

1,2


COMMENTS

The cyclic subgroups of GL_m(q) of order q^m1 are called Singer cycles.


LINKS



EXAMPLE

For n = 3 the a(3) = 5 solution is the matrix E_5 =
[ [ 1 1 0 0 0 ],
[ 1 1 1 0 0 ],
[ 0 1 1 1 0 ],
[ 0 0 1 1 1 ],
[ 0 0 0 1 2 ] ]
since the matrix E_5 has order 3^5  1 = 242 in GL_5(3).


PROG

(GAP)
EMatrix := function(n, q)
local M, i;
M := NullMat(n, n, GF(q));
for i in [2..n] do
M[i  1][i  1] := Z(q) ^ 0;
M[i  1][i] := Z(q) ^ 0;
M[i][i  1] := Z(q) ^ 0;
od;
M[n][n] := 2 * Z(q) ^ 0;
return M;
end;
for n in [1..100] do
M := EMatrix(n, 3);
if Determinant(M) <> 0 * Z(3) and Order(M) = 3 ^ n  1 then
Print(n, "\n");
fi;
od;
(PARI)
E(m)={matrix(m, m, i, j, (i==m&&j==m) + (abs(ij)<=1))}
is(m, b)={my(ID=matid(m), M=Mod(E(m), b), e=b^m1); if(M^e==ID, fordiv(e, d, if(d<e && M^d==ID, return(0))); 1, 0)}
for(m=1, 100, if(m<>2&&is(m, 3), print1(m, ", "))) \\ Andrew Howroyd, Dec 21 2019


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



