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A373396
Matrix inverse of triangle A296548, read by rows.
0
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
0,4
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.
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).
EXAMPLE
Triangle begins
1;
-1, 1;
5, -6, 1;
-113, 140, -28, 1;
10879, -13560, 2800, -120, 1;
-4324129, 5395984, -1120960, 49600, -496, 1;
...
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
CROSSREFS
Sequence in context: A293107 A113106 A171273 * A157832 A330660 A200486
KEYWORD
sign,tabl
AUTHOR
Geoffrey Critzer, Jun 03 2024
STATUS
approved