Problem:
Find the number of ordered pairs of integers such that the sequence
is strictly increasing and no set of four (not necessarily consecutive) terms forms an arithmetic progression.
Solution:
We will use complementary counting to find the number of pairs that satisfy the conditions. First, eliminate values of and that would form immediate arithmetic progressions:
or cannot be , since is arithmetic.
or cannot be , since is arithmetic.
This leaves 22 possible values for and in the range from to excluding .
The number of valid increasing pairs is given by the binomial coefficient:
Now, we subtract specific pairs that still form an arithmetic progression with other values in the sequence. Testing, we see that invalid cases include:
forms
forms
forms
So we subtract these 3 cases to get .
The problems on this page are the property of the MAA's American Mathematics Competitions