000RM.dvi

24.3 Abundant and deficient numbers 633

24.3.1 Appendix: Two enumerations of the rational numbers in
(0,1)

Consider two enumerations of the rational numbers in (0,1).
E 1 lists them by increasing denominator, and for equal denominators,
by increasing numerator. Thus,

E 1 :

1

2

,

1

3

,

2

3

,

1

4

,

3

4

,

1

5

,

2

5

,

3

5

,

4

5

,

1

6

,

5

6

, ....

E 2 , on the other hand, lists them by increasing sum of numerator
and denominator, and for equal sums of terms, by increasing numerator.
Thus,

E 2 :

1

2

,

1

3

,

1

4

,

2

3

,

1

5

,

1

6

,

2

5

,

3

4

,

1

7

,

3

5

,

1

8

, ....

The fractions^12 ,^13 , and^25 occupy the same positions in both se-
quences. More generally, the rational number mn (withm<nand
gcd(m, n)=1) occupies position

∑n−^1

k=2

φ(k)+

∑m

k=1

χ(k, n)

in enumerationE 1 and position

1

2

m∑+n− 1

k=3

φ(k)+

∑m

k=1

χ(k, m+n−k)

in enumerationE 2. Here,

χ(m, n)=

{

1ifgcd(m, n)=1,

0otherwise.

This was Computer Challenge 511 ofJournal of Recreational Math-
ematics. The solution lists 10 of these.^4

Fraction^12132523930731435916367235971974771238513

Position 1 2 7 158 1617 6211 8058 16765 69093 465988

What is the next match? Form+n≤ 20000 , I found four more:

Fraction^172941751922464132397820112174646

Position 5297983 6546724 18588348 38244610

(^4) Journal of Recreational Math., 10 (1977–78) 122–123.