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%I #32 Jan 28 2020 11:40:13
%S 1,3,5,9,11,23,29,35,39,41,53,65,69,81,83,89,95,99,105,113,119,131,
%T 155,173,179,189,191,209,221,231,233,239,243,251,281,293,299,303,323,
%U 329,359,371,375,411,413,419,429,431,443,453,491
%N Positive integers m such that the matrix E_m has order 3^m-1 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.
%C The cyclic subgroups of GL_m(q) of order q^m-1 are called Singer cycles.
%H M. Farrokhi D. G., <a href="https://arxiv.org/abs/1910.09844">Lattice paths inside a table: Rows and columns linear combinations</a>, arXiv:1910.09844 [math.CO], 2019.
%H W. M. Kantor, <a href="https://doi.org/10.1016/0021-8693(80)90214-8">Linear groups containing a Singer cycle</a>, J. Algebra 62(1) (1980), 232-234.
%e For n = 3 the a(3) = 5 solution is the matrix E_5 =
%e [ [ 1 1 0 0 0 ],
%e [ 1 1 1 0 0 ],
%e [ 0 1 1 1 0 ],
%e [ 0 0 1 1 1 ],
%e [ 0 0 0 1 2 ] ]
%e since the matrix E_5 has order 3^5 - 1 = 242 in GL_5(3).
%o (GAP)
%o EMatrix := function(n, q)
%o local M, i;
%o M := NullMat(n, n, GF(q));
%o for i in [2..n] do
%o M[i - 1][i - 1] := Z(q) ^ 0;
%o M[i - 1][i] := Z(q) ^ 0;
%o M[i][i - 1] := Z(q) ^ 0;
%o od;
%o M[n][n] := 2 * Z(q) ^ 0;
%o return M;
%o end;
%o for n in [1..100] do
%o M := EMatrix(n, 3);
%o if Determinant(M) <> 0 * Z(3) and Order(M) = 3 ^ n - 1 then
%o Print(n, "\n");
%o fi;
%o od;
%o (PARI)
%o E(m)={matrix(m,m,i,j,(i==m&&j==m) + (abs(i-j)<=1))}
%o is(m,b)={my(ID=matid(m), M=Mod(E(m), b), e=b^m-1); if(M^e==ID, fordiv(e, d, if(d<e && M^d==ID, return(0))); 1, 0)}
%o for(m=1, 100, if(m<>2&&is(m, 3), print1(m, ", "))) \\ _Andrew Howroyd_, Dec 21 2019
%Y Cf. A328642.
%K nonn
%O 1,2
%A _M. Farrokhi D. G._, Oct 23 2019
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