A020522

[Joshua Zucker]

Proof:
Each of the n elements of the original set has four choices:
it can be in no set, z only, y and z, or x and y and z.
Thus 4^n possibilities.

Take away the 2^n possibilities where x = y = z, and QED.

--Joshua Zucker


On 12/17/05, Ross La Haye <rlahaye at new.rr.com> wrote:
>
> Let x,y,z be elements from some power set P(n).  Define a function f(x,y,z)
> in the following manner:
>
> f(x,y,z) = 1 if x is a subset of y and y is a subset of z and x does not
> equal z;
> f(x,y,z) = 0 if x is not a subset of y or y is not a subset of z or x equals
> z.
>
> Now sum f(x,y,z) for all x,y,z of P(n).  For n = 0,1,2,3,4,5...I get
> 0,2,12,56,240,992...which seems to be A020522, i.e. 4^n - 2^n.  If this is
> so, would someone be interested in submitting a comment accordingly...?
>
> Ross
>
>
>