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IMO 2019 Shortlisted Problems

2019 counting and probability

Problem

The infinite sequence of (not necessarily different) integers has the following properties: for all integers , and for all integers . Prove that all integers occur in the sequence (that is, for all , there exists with ).
Solution
We prove by induction on that every initial segment of the sequence, , consists of the following elements (counted with multiplicity, and not necessarily in order), for some with : For we have , which is of this form. Now suppose that for the elements are for some with . It is given that which becomes or, using , that On the other hand, it is well known that and so, by subtracting, we get From this, using the fact that the binomial coefficients are increasing for and decreasing for , we conclude that either or . In either case, is again of the claimed form, which concludes the induction. As a result of this description, any integer appears as a term of the sequence for some .

Techniques

Algebraic properties of binomial coefficientsInduction / smoothing