American Mathematical Monthly
Problem 11406 by A. A. Dzhumadil’daeva (edited) For any positive integer n, we define an integer n!! by n!! = Q
k∈{1,2,...,n};
k≡nmod 2
k.
(For instance, 1!! = 1, 2!! = 2, 3!! = 1·3 = 3,4!! = 2·4 = 8,7!! = 1·3·5·7 = 105, and 8!! = 2·4·6·8 = 384.)
Also, set (−1)!! = 1 and 0!! = 1.
For any positive integer n, prove that
n
X
i=0
n i
(2i−1)!! (2 (n−i)−1)!! = 2nn!.
Solution by Darij Grinberg.
For any integer i≥0,we can easily see that (2i−1)!! = (2i)!
2ii! (1)
1. Substituting n−i fori in (1), we see that
(2 (n−i)−1)!! = (2 (n−i))!
2n−i(n−i)! (2)
for any integers n and i with n−i≥0.
1In fact, fori= 0,this follows from
(2i−1)!! = (2·0−1)!! = (−1)!! = 1 = 1
1·1 =(2·0)!
20·0! =(2i)!
2ii!, and fori >0, this follows from
(2i−1)!! = Y
k∈{1,2,...,2i−1};
k≡2i−1 mod 2
k= Y
k∈{1,2,...,2i−1};
kis odd
k (sincek≡2i−1 mod 2 holds if and only ifkis odd)
= Y
k∈{1,2,...,2i};
kis odd
k
since, for an odd number k, the conditionsk∈ {1,2, ...,2i−1}
andk∈ {1,2, ...,2i} are equivalent (because 2iis not odd)
= Q
k∈{1,2,...,2i}
k
Q
k∈{1,2,...,2i};
kis even
k = (2i)!
2ii!
(since Q
k∈{1,2,...,2i}
k= (2i)! and Y
k∈{1,2,...,2i};
kis even
k= Y
t∈{1,2,...,i}
(2t) (here we substituted 2tfork)
= 2i Y
t∈{1,2,...,i}
t= 2ii!
).
1
A famous formula ([1], paragraph 5.3, formula (5.39), also proven in [2]) states that
n
X
i=0
2i i
2 (n−i) n−i
= 4n. (3)
2 Now,
n
X
i=0
n i
(2i−1)!! (2 (n−i)−1)!! =
n
X
i=0
n i
· (2i)!
2ii! · (2 (n−i))!
2n−i(n−i)! (by (1) and (2))
=
n
X
i=0
n!
i!·(n−i)!· (2i)!
2ii! · (2 (n−i))!
2n−i(n−i)!
since
n i
= n!
i!·(n−i)!
=
n
X
i=0
n!
2i·2n−i
| {z }
=
n!
2n
· (2i)!
i!·i!
| {z }
=
2i i
· (2 (n−i))!
(n−i)!·(n−i)!
| {z }
=
2 (n−i) n−i
= n!
2n ·
n
X
i=0
2i i
2 (n−i) n−i
= n!
2n ·4n (by (3))
= n!
2n ·(2n)2 = 2nn!, qed.
Remark. We more or less showed that the problem is equivalent to (3), an identity notorious for the difficulty of finding a bijective proof (the one given in [2] is 5 pages long). However, a bijective solution of the given problem does not automatically yield a bijective proof of (3), so the given problem could turn out easier to solve bijectively.
Still, my attempts at finding such a proof did not show much success.
References
[1] R. L. Graham, D. E. Knuth, O. Patashnik, Concrete Mathematics, 2nd Edition 1994, Addison-Wesley.
[2] David Callan, Bijections for the identity 4n =
n
P
k=0
2k k
2 (n−k) n−k
. http://www.stat.wisc.edu/~callan/notes/
2Internet inhabitants can find some more proofs of (3) in
http://www.mathlinks.ro/Forum/viewtopic.php?t=40150
2
