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A328642
Positive integers m such that the matrix E_m has order 2^m-1 in GL_m(2) 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
2, 3, 5, 9, 11, 14, 23, 26, 29, 35, 39, 41, 53, 65, 69, 74, 81, 83, 86, 89, 95, 105, 113, 119, 131, 146, 155, 158, 173, 179, 189, 191, 209, 221, 230, 231, 233, 239, 243, 251, 254, 281, 293, 299, 303, 323, 326, 329, 359, 371, 375, 386, 398, 411, 413, 419, 429, 431, 443, 453, 470, 473, 491
OFFSET
1,1
COMMENTS
In this sequence, the (m,m) entry of the matrices will equivalently be 0 since 2 == 0 (mod 2).
The cyclic subgroups of GL_m(q) of order q^m-1 are called Singer cycles.
LINKS
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 2 ^ 5 - 1 = 32 in GL_5(2).
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, 2);
if Determinant(M) <> 0 * Z(2) and Order(M) = 2 ^ 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(is(m, 2), print1(m, ", "))) \\ Andrew Howroyd, Dec 23 2019
CROSSREFS
Cf. A328643.
Sequence in context: A137509 A350916 A014109 * A102940 A157049 A195668
KEYWORD
nonn
AUTHOR
M. Farrokhi D. G., Oct 23 2019
STATUS
approved