n!*Fibonacci(n)

[Brendan McKay]

* Karol <penson at lptl.jussieu.fr> [050327 17:11]:
>    Dear Brendan,
>     Do you have also another version where you have not cheated ?
>     Thanks in advance ,    Karol

Let me define "not cheating".  This would be to find n!*Fib(n)
as the size of some natural set which cannot be decomposed into
X x Y where X and Y are natural sets of size n! and Fib(n)
respectively.  Or at least where such decomposition is non-trivial
to see or prove. Here "natural" is a technical word meaning "lovely".

The answer is: no, I don't know how to do it but it would be lovely.
 
Brendan.
 
 
> Brendan McKay wrote:
> >* Karol <penson at lptl.jussieu.fr> [050327 08:35]:
> >
> >>     I wonder if  a combinatorial ( or other ) interpretation of the 
> >>sequences
> >>          n!*Fibonacci(n),   n!*Pell(n)...   etc . is known.
> >
> >Suppose you have a lot of bricks. Some have height 1 and are
> >labelled with a single positive integer. The others have height 2
> >and are labelled with an ordered pair of two positive integers.
> >Then n!*Fibonacci(n) is the number of ways of stacking up some
> >bricks of total height n such that the integers 1,2,...,n each
> >appear exactly once.
> >
> >I hope you can all see that I cheated!!
> >
> >Brendan.