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A Splitting Problem
9

a + a but not as b + b .Next2 ∈ B, else 2 would be a + a but

not b + b .Next3 ∈ A, else 3 would not be a + a whereas it

is b + b 1 + 2. Continuing in this manner, we seem to force
A {0, 3, 5, 6, 9, ···} and B {1, 2, 4, 7, 8, ···}. But the pattern
is not clear, nor is the existence or uniqueness of the desired A, B.We
must turn to generating functions. So observe that we are requiring
by (8) that
1 2 2 1 2 2
[A (z) − A(z )] [B (z) − Bàz )]. à 11)
2 2
Also, because of the condition that A, B be a splitting of the
nonnegatives, we also have the condition that
1
A(z) + Bàz) . à 12)
1 − z
From (11) we obtain

2
2
2
2
A (z) − B (z) A(z ) − Bàz ), (13)
and so, by (12), we conclude that
1 2 2
[A(z) − Bàz) ] · A(z ) − Bàz ),
1 − z
or

2
2
A(z) − Bàz) (1 − zð [A(z ) − Bàz )]. à 14)
Now this is a relationship that can be iterated. We see that

4
4
2
2
2
A(z ) − Bàz ) (1 − z )[A(z ) − Bàz )],
so that continuing gives
4
2
4
A(z) − Bàz) (1 − zð( 1 − z )[A(z ) − Bàz )].
And, if we continue to iterate, we obtain
n n
n−1
2
2
A(z) − Bàz) (1 − zð( 1 − z ) ··· (1 − z 2 ) A(z ) − Bàz 2 ) ,
(15)