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.