A289538 Expected dimension of the null space of a random linear operator on an n-dimensional vector space over the field with two elements as n -> infinity.

8, 5, 0, 1, 7, 9, 8, 3, 0, 8, 7, 3, 9, 7, 9, 3, 3, 2, 8, 7, 6, 0, 6, 3, 2, 8, 1, 4, 9, 3, 5, 9, 1, 8, 7, 8, 8, 4, 0, 4, 2, 6, 7, 2, 5, 9, 7, 3, 2, 0, 2, 7, 2, 5, 9, 8, 7, 3, 5, 8, 0, 5, 2, 5, 5, 6, 3, 0, 9, 5, 9, 4, 1, 1, 8, 3, 3, 1, 3, 4, 4, 3, 6, 3, 0, 4, 1, 0, 6, 7, 0, 8, 8, 5, 9, 3, 5, 6, 5, 8

Offset: 0

Author

Geoffrey Critzer, Jul 10 2017

Keywords

• cons
• nonn

Comments

More precisely, let X:L(V) -> {0,1,2,...,n} be the random variable that assigns to each linear operator T on n-dimensional vector space V over F_2, the integer j in {0,1,2,...,n} such that the dimension of the null space of T = j. Then E(X) = 0.850179183...

Crossrefs

Cf. A002884, A182176, A289537, A289539, A289541, A289542.

Programs

Mathematica

nn = 300; q := 2;A[x_] := Sum[1/(FunctionExpand[QFactorial[j, q]] (q - 1)^j q^Binomial[j, 2]) Product[1 - 1/q^i, {i, j + 1, \[Infinity]}] x^j, {j, 0, nn}];RealDigits[
  N[Normal[Series[D[A[x], x] /. x -> 1, {x, 0, nn}]], 100]][[1]]

Formula

Let A(x) = Sum_{n>=0} Product_{i>=n+1} (1-1/2^i)*x^n/A002884(n). Then A'(1) = 0.85017983...