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.

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

Table of n, a(n) for n=1..63.

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

Adjacent sequences: A328639 A328640 A328641 * A328643 A328644 A328645

Keyword

nonn

Author

M. Farrokhi D. G., Oct 23 2019

Status

approved