A PARTY TRICK FEATURING THE TRIANGULAR NUMBERS

SHAILESH A SHIRALI
Take any number n. (In this article, number means positive integer.) Replace it by
any two smaller numbers, say a and b, whose sum is n. Record their product ab. Repeat the
same step for each of the two smaller numbers. Iteratively continue this till all the numbers
left are 1s. Now compute the sum of all the products. Here is the surprise: whichever route
one takes to reach the nal string of 1s, the sum obtained will be exactly the same!
For example, starting with n = 8 one could trace the following trajectory, in which the
nested brackets in the central column show the way the numbers get split, and the products
are recorded in the third column:
Step Partition Product
1 5+3 15
2 (3+2)+3 6
3 (3+2)+(2+1) 2
4 ((2+1)+2)+(2+1) 2
5 ((2+1)+2)+((1+1)+1) 1
6 (((1+1)+1)+2)+((1+1)+1) 1
7 (((1+1)+1)+(1+1))+((1+1)+1) 1
The sum of the products is 15+6+2+2+1+1+1=28. You are invited to repeat the exercise,
following some different trajectory. Check that you get the same sum.
The invariance of the answer invites us to present the property suitably garbed as a party
trick or game.
We shall prove this property in two different ways. Indeed, we shall show: If the initial
number is n, then the sum of the products is {
n
2
). The sum is thus a triangular number, which
explains the title of this article.
I. AN INDUCTIVE PROOF
If we write Let f (n) denote the sum of the products . . . , then the onus will be on us to
show that f is a well-dened function, as this is by no means obvious. Thus, care is needed
in writing the proof. At the outset the only obvious fact is that the number of steps taken to
complete the game is 1 less than the starting number.
2 SHAILESH A SHIRALI
But we do indeed start with just that sentence. Let f (n) denote the sum of the products
obtained from n. We shall show inductively that the function is well dened. It is natural
to put f (1) = 0, since the process of splitting n into smaller numbers cannot even be begun
for n = 1. By experimentation we nd that f (n) does exist for n = 2 and n = 3, with f (2) = 1
and f (3) = 3. Thus, f (n) is well dened for n = 1, 2, 3.
Let us inductively assume the following: f (n) exists for 1 n m1, and f (n) = {
n
2
) for
these values of n. We shall now show that f (m) exists, and is equal to {
m
2
).
Let m be partitioned as m= k +(mk) where 1 k m1. By the inductive hypothesis,
f (k) and f (mk) are well dened, and the sum associated with this partition is
k(mk)+_
k
2
_+_
mk
2
_.
If we can show that the above sum is independent of k, then it will mean that f (m) is
well-dened. We have, now:
k(mk)+_
k
2
_+_
mk
2
_
= _
k(mk)
2
+
k(k 1)
2
_+_
k(mk)
2
+
(mk)(mk 1)
2
_
=
k(m1)
2
+
(mk)(m1)
2
=
m(m1)
2
= _
m
2
_.
The claim follows.
II. A COMBINATORIAL PROOF
The compactness of the formula for f (n) invites us to look for a combinatorial proof.
Here is one such. We use the basic identity: if n = a+b where 1 a, b n1, then
_
n
2
_ = ab+_
a
2
_+_
b
2
_.
This is easier to justify by counting than by algebra: if an n-element set S is written as AB,
then a pair of elements of S must be of one of three types: it may have one element each
from A and B; both its elements may be from A; or both its elements may be from B. Now
write a = |A| and b = |B| and count the pairs of each kind . . . .
Recursively applying the identity to the binomial coefcients {
a
2
) and {
b
2
), we obtain the
desired result.
TRIANGULAR NUMBERS 3
III. EXTENSIONS
We may devise various party games of this kind using elementary algebraic identities.
For example, from the identity
(a+b)
3
a
3
b
3
= 3ab(a+b)
we extract the following trick: Take any number n. Replace it by any two smaller numbers,
say a and b, whose sum is n. Record the quantity ab(a+b). Repeat the same step for each
of the two summands. Iteratively continue this till all the numbers left are 1s. Now compute
the sum of all the recorded quantities. The sum will be the same, whichever route one takes
to reach the nal string of 1s.
From the quoted identity we infer that the sum will be
1
3
(n
3
n). Here is an example of
a game trajectory for n = 6:
Step Partition Quantity
1 4+2 86 = 48
2 (3+1)+2 34 = 12
3 (3+1)+(1+1) 12 = 2
4 ((2+1)+1)+(1+1) 23 = 6
5 (((1+1)+1)+1)+(1+1) 12 = 2
The sum of the quantities in the last column is 48+12+2+6+2 = 70 =
1
3
{6
3
6), as claimed.
The identities
(a+b)
5
a
5
b
5
= 5ab(a+b){a
2
+ab+b
2
),
(a+b)
7
a
7
b
7
= 7ab(a+b){a
2
+ab+b
2
)
2
yield more entertainments. We conclude with a relation concerning binomial coefcients:
(a+b)! = a! b! _
a+b
a
_.
This yields the following: Take any number n. Replace it by any two smaller numbers, say
a and b, whose sum is n. Record the quantity {
a+b
a
). Repeat the same step for each of the
two summands. Iteratively continue this till all the numbers left are 1s. Now compute the
product of all the recorded quantities. The product will be the same, whichever route one
takes to reach the nal string of 1s. (The product will be n!.)
RISHI VALLEY SCHOOL, RISHI VALLEY 517 352, ANDHRA PRADESH, INDIA
E-mail address: shailesh.shirali@gmail.com