A328643 - OEIS

A328643

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.

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^m-1 are called Singer cycles.

LINKS

Table of n, a(n) for n=1..51.

M. Farrokhi D. G., Lattice paths inside a table: Rows and columns linear combinations, arXiv:1910.09844 [math.CO], 2019.

W. M. Kantor, Linear groups containing a Singer cycle, J. Algebra 62(1) (1980), 232-234.

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(i-j)<=1))}

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)}

for(m=1, 100, if(m<>2&&is(m, 3), print1(m, ", "))) \\ Andrew Howroyd, Dec 21 2019

CROSSREFS

Cf. A328642.

Sequence in context: A092917 A256220 A163778 * A160358 A319084 A120806

Adjacent sequences: A328640 A328641 A328642 * A328644 A328645 A328646

KEYWORD

nonn

AUTHOR

M. Farrokhi D. G., Oct 23 2019

STATUS

approved