A373396 Matrix inverse of triangle A296548, read by rows.

1, -1, 1, 5, -6, 1, -113, 140, -28, 1, 10879, -13560, 2800, -120, 1, -4324129, 5395984, -1120960, 49600, -496, 1, 6984271295, -8717444064, 1813050624, -80709120, 833280, -2016, 1, -45479775838337, 56768157085760, -11809230892032, 526302695424, -5466697728, 13655040, -8128, 1

Offset: 0

Author

Geoffrey Critzer, Jun 03 2024

Keywords

• sign
• tabl

Comments

Let P_n be the set of n X n idempotent matrices over GF(2) with the ordering: E<=F iff EF=E=FE. Then T(n,k) = Sum mu(0,E) where the sum is taken over the elements in P_n of rank k and mu is the Moebius mu incidence function of P_n.

To obtain the inverse, we regard the triangle as a lower triangular matrix, but then ignore the part above the diagonal.

Examples

			 Triangle begins
         1;
        -1,       1;
         5,      -6,        1;
      -113,     140,      -28,     1;
     10879,  -13560,     2800,  -120,    1;
  -4324129, 5395984, -1120960, 49600, -496, 1;
  ...
		

Crossrefs

Cf. A296548, A002884.

Programs

Mathematica

nn = 6; B[n_] = Product[q^n - q^i, {i, 0, n - 1}] /. q -> 2;e[x_] := Sum[x^n/B[n], {n, 0, nn}];Table[B[n], {n, 0, nn}]*CoefficientList[Series[ e[y x]/e[x], {x, 0, nn}], {x, y}] // Grid

Formula

Sum_{n>=0} Sum_{k=0..n} T(n,k)*y^k*x^n/A002884(n) = e(y*x)/e(x) where e(x) = Sum_{n>=0} x^n/A002884(n).