A natural number i has dubs in basen if the last two digits are the same. The dubs predicate in scheme: (lambda (i n) (= (remainder i n) (remainder (floor (/ i n)) n)))
For every natural number i, where i > 2, there exists ``trivial dubs'' xxx..11 in basei-1
A natural number i is a ``dubless number'' if it only has trivial dubs. In other words, it doesn't have dubs in any other base than basei-1
Bateman conjecture:
The set of dubless numbers is infinite