1971] RESEARCH PROBLEMS 385

and so

2

(6)

Z2

<

I

+

2

-

Since X, Y, Z

are

positive integers, inequality (6)

implies that Z can take only

a

finite number of values.

The same

holds for

Y

(on

using (5)) and X (on using

(3) or the relation

X

< Y+Z).

Since X+ Y+Z =,uA,

A

takes only a finite numb er

of values. So do the

sides

a, b, c,

in

view

of relations

(2). Thus

N(O)

is finite

for each positive X>

2.

If

'X

\2

(i.e.,

,Zb 1), we obtain the same conclusion

on using the second

in-

equality

in

(4)

and

proceeding

as before.

We

shall next

show

that

N(X)

=

0 for

X>

V8

(i.e.,

A>

V2).

For such

a

value

of

A

we

have from

(6)

that Z =1. The

relation (5)

then shows

that

Y=

1. Since

X< Y+Z,

we

now

have

X= 1.

Using relations (3)

and

(2),

one obtains

;=

3,

A=V\3,

a=b=c=2.

REMARKS. The remarks made at the beginning

of this note show that

N(1)

=

5

and

N(2)

=1.

Our theorem shows that

N(3)

=

N(4)

=

*

=

0.

It

would be interesting to consider a similar problem

for

a

quadrilateral,

and

in

general,

a

polygon

of

n sides (n> 2).

RESEARCH

PROBLEMS

EDITED By RICHARD

GuY

In

this

Department

the

Monthly

presents easily

stated

research problems dealing

with

notions

ordinarily encountered in

undergraduate mathematics. Each problem should be

accompanied

by relevant references (if any

are

known

to

the author) and by a brief description

of

known

partial

results.

Material should

be

sent to Richard

Guy, Department of

Mathematics, Statis-

tics,

and

Computing Science,

The

University of Calgary, Calgary

44, Alberta,

Canada.

HOW OFTEN DOES

AN

INTEGER OCCUR AS A BINOMIAL

COEFFICIENT?

DAVID

SINGMASTER, Polytechnic of the

South

Bank,

London

Let N(a) be the number

of times a occurs as a binomial

coefficient, (X). We

have

N(1) =oo, N(2) 1,

N(3) =N(4) N(5) =2, N(6) =3,

etc.

Clearly,

for

a>

1, N(a) <

oo.

Below we establish that

N(a)

=

O(log a). We

conjecture that

N(a)

=

0(1), that is, that the

number of solutions of

()

=a is

bounded for a >

1.

Erd6s,

in

a

private

communication, concurs in this conjecture and

states that it

must be very hard. In a later

communication, he suggests trying

to show N(a)

=

O(log log a).

If

we let

M(k)

be the

first integer

a

such that N(a)

=-k,

we

have: M(1) =2,

M(2)

=

3, M(3)

=

6, M(4)

=

10, M(6)

=

120. The next values would

be interesting

to know.

This content downloaded from 128.235.251.160 on Mon, 22 Dec 2014 21:33:41 PM