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Therefore, the following equation holds ( n
>
3):
UB
(
n
)=
UB
(
n
1
)(
2
(
n
1
)
3
)
whence, for n
>
3:
UB
(
n
)=
3
·
5
·...· (
2
(
n
1
)
3
)=
3
·
5
·...· (
2 n
5
) .
Thus, the number UB
(
n
)
given above, for n
>
2, can be concisely expressed by
(denoting n
· (
n
2
) · (
n
4
) ...·
3by n !!):
UB
(
n
)=(
2 n
5
)
!!
Therefore, from Proposition 7.18, we get BT
(
n
)=(
2 n
3
)
!!, which concludes the
proof.
 
 
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