o ccur either, else one of its c hildren w ould be r/s , and its n umerator is smaller, the denomi-

nator b eing the same , a con tradiction. If r < s , then r / ( s − r ) doesn’t o ccur either, else one

of its c hildren would b e r/s , and it has a smaller denominator, a con tradiction.

3. No r e duc e d p ositive r ational numb er o c curs at mor e than one vertex. First, the rational

n umber 1 o ccurs only at the top vertex of the tree, for if not, it would b e a child of some

v ertex r /s . But the children of r /s are r / ( r + s ) and ( r + s ) /s , neither of which can b e 1.

Otherwise, among all reduced rationals that o ccur more than once, let r/s hav e the smallest

denominator, and among these, the smallest numerator. If r < s then r /s is a left child of

t wo distinct v ertices, at b oth of whic h r / ( s − r ) liv es, contradicting the minimality of the

denominator. Similarly if r > s . 2

It follows that a list of all p ositiv e rational n umbers, eac h app earing once and only once, can be

made b y writing down 1 / 1, then the fractions on the lev el just b elow the top of the tree, reading

from left to right, then the fractions on the next level down, reading from left to right, etc.

W e claim that if that b e done, then the denominator of eac h fraction is the n umerator of its

successor. This is cle ar if the fraction is a left child and its successor is the right child, of the same

paren t. If the fraction is a right c hild then its denominator is the same as the denominator of its

paren t and the n umerator of its succe ss or is the same as the numerator of the parent of its successor,

hence the result follo ws b y do wnw ard induction on the lev els of the tree. Finally , the rightmost

v ertex of each row has denominator 1, as do es the leftmost vertex of the next row, pro ving the

claim.

Th us, after we make a single sequence of the rationals by reading the successive rows of the tree

as describ ed abov e, the list will be in the form { f ( n ) /f ( n + 1) }

n ≥ 0

, for some f .

No w, as the fractions sit in the tree, the tw o children of f ( n ) /f ( n + 1) are f (2 n + 1) /f (2 n + 2)

and f (2 n + 2) /f (2 n + 3). Hence from the rule of construction of the c hildren of a parent, it m ust

b e that

f (2 n + 1) = f ( n ) and f (2 n + 2) = f ( n ) + f ( n + 1) ( n = 0 , 1 , 2 , . . . ) .

These recurrences eviden tly determine our function f on all nonnegative in tegers.

W e claim that f ( n ) = b ( n ), the n umber of hyperbinary re presentations of n , for all n ≥ 0.

This is true for n = 0, and supp ose true for all integers ≤ 2 n . No w b (2 n + 1) = b ( n ), b ecause if

w e are giv en a h yp erbinary expansion of 2 n + 1, the “1” m ust app ear, hence b y subtracting 1 from

b oth sides and dividing by 2, we’ll get a hyperbinary represen tation of n . Con v ersely , if we hav e

suc h an expansion of n , then double each part and add a 1, to obtain a representation of 2 n + 1.

F urthermore, b (2 n + 2) = b ( n ) + b ( n + 1), for a h yp erbinary expansion of 2 n + 2 might hav e

either t wo 1’s or no 1’s in it. If it has t wo 1’s, then b y deleting them and dividing by 2 w e’ll get an

expansion of n . If it has no 1’s, then we just divide by 2 to get an expansion of n + 1. These maps

are reversible, proving the claim.

It follo ws that b ( n ) and f ( n ) satisfy the same recurrence form ulas and take the same initial

v alues, hence they agree for all nonnegativ e in tegers. W e state the ﬁnal result as follo ws.

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