If a figure can be tiled by N 2x3 rectangles it is trivial to place the same number of U pentominoes in that figure.

Find the smallest N such that at least N+1 U pentominos can be placed in a figure tiled by N 2x3 rectangles.

(In reply to

re: posible solution 2 by Jer)

Congratulations! That is the smallest I could find. I expect this to be the smallest solution. It seems that all the U pentominos need to be joined into interlocking pairs in order to see any reduction in the space needed for the U pentominos. In the big H solution, it takes four such pairs to create two sockets for the fifth pair to slide into, totaling 10 U pentominos.