A346412 Triangular array read by rows: T(n,k) is the number of nilpotent n X n matrices over GF(2) having rank k, 0 <= k <= n-1, n >= 1.

1, 1, 3, 1, 21, 42, 1, 105, 1470, 2520, 1, 465, 32550, 390600, 624960, 1, 1953, 605430, 36325800, 406848960, 629959680, 1, 8001, 10417302, 2768025960, 155009453760, 1680102466560, 2560156139520, 1, 32385, 172741590, 192779614440, 47809344381120, 2590958018073600, 27636885526118400, 41781748196966400

Offset: 1

Author

Geoffrey Critzer, Jul 15 2021

Keywords

• nonn
• tabl

Examples

			Array begins
  1;
  1,    3;
  1,   21,     42;
  1,  105,   1470,     2520;
  1,  465,  32550,   390600,    624960;
  1, 1953, 605430, 36325800, 406848960, 629959680
T(2,0) = 1 because the zero matrix has rank 0.
T(2,1) = 3 because {{0,0},{1,0}}, {{0,1},{0,0}}, {{1,1},{1,1}} have rank 1.
		

References

• G. Lusztig, A note on counting nilpotent matrices of fixed rank, Bull. London Math. Soc. v.8 (1976), no. 1, 77--80; MR0407050.

Links

• Jason Fulman, A probabilistic approach toward the finite general linear and unitary group, arxiv.org, Dec 1997, page 12.
• J. Kung, The cycle structure of a linear transformation over a finite field,Linear Algebra and its Applications, vol 36, March 1981,141-155.
• Kent E. Morrison, Integer Sequences and Matrices Over Finite Fields, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.1.

Crossrefs

Cf. A134057 (column k=1), A083402 (main diagonal), A053763 (row sums).

Programs

Mathematica

nn = 10; q = 2; b[p_, i_] := Count[p, i];d[p_, i_] :=Sum[j b[p, j], {j, 1, i}] + i Sum[b[p, j], {j, i + 1, Total[p]}]; aut[deg_, p_] :=  Product[Product[
   q^(d[p, i] deg) - q^((d[p, i] - k) deg), {k, 1, b[p, i]}], {i, 1, Total[p]}];
g[u_, v_] := Total[Map[v^(Total[#] - Length[#]) u^Total[#]/aut[1, #] &,
   Level[Table[IntegerPartitions[n], {n, 0, nn}], {2}]]];Map[Select[#, # > 0 &] &,Drop[Table[Product[q^n - q^i, {i, 0, n - 1}], {n, 0, nn}] CoefficientList[
     Series[g[u, v], {u, 0, nn}], {u, v}], 1]] // Grid

Formula

T(n,n-k) = A002884(n)*Product_{i=k..n-1}(1-1/2^i)/(A002884(k)*2^(n-k)*Product_{i=1..n-k}(1-1/2^i)) Theorem 6 in Fulman link. - Geoffrey Critzer, Dec 23 2024