Vanishing Integer Points
The domain of a function f is the set of Natural Numbers. The function is defined as follows:
Where denotes the nearest integer less than or equal to k. Consider the following sequence
where represents the function obtained by composing f with itself k times and m is any natural number.
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1 solution
f ( 1 ) = 2 , f ( 2 ) = 3 , f ( 3 ) = 4 , f ( 4 ) = 6 , f ( 5 ) = 7 , f ( 6 ) = 8 , f ( 7 ) = 9 , f ( 8 ) = 1 0 , f ( 9 ) = 1 2 , f ( 1 0 ) = 1 3 , f ( 1 1 ) = 1 4 , f ( 1 2 ) = 1 5 , f ( 1 3 ) = 1 6 , f ( 1 4 ) = 1 7 , f ( 1 5 ) = 1 8 , f ( 1 6 ) = 2 0 , f ( 1 7 ) = 2 1 , f ( 1 8 ) = 2 2 , f ( 1 9 ) = 2 3 , f ( 2 0 ) = 2 4 , f ( 2 1 ) = 2 5 , f ( 2 2 ) = 2 6 , f ( 2 3 ) = 2 7 , f ( 2 4 ) = 2 8 , f ( 2 5 ) = 3 0 , f ( 2 6 ) = 3 1 , f ( 2 7 ) = 3 2
Notice the type of integers that are missed in the sequence. The sequence misses integers that are exactly of the type n 2 + n − 1 . We can prove this by a simple argument. Replace n by ( n − 1 ) 2 and then by n 2 and see the integer that is missed. Also notice that only integer of the type n 2 + n − 1 are missed as all other values occur consecutively. The function is certainly one-to-one. And for the Natural numbers excluding the type n 2 + n − 1 , f is also onto. so f is bijective. The sequence in the Question runs indefinitely and increases without bounds. [As the functional value never takes a number of the form n 2 + n − 1 . At the most it can be n 2 + n ] Now consider f − 1 ( n 2 . This must exist by our previous Discussion. Indeed it exists and is equal to n 2 − n + 1 . Also f − 1 ( ( n 2 − n + 1 ) must also exist. If f k ( t ) is a perfect square then the sequence stops[Assuming the contra-positive statement to be true] but the sequence runs indefinitely just as we proved above. Now f k ( n ) must be a perfect square for some n as f ( t 2 − t + 1 ) must exist for some t. This is also true since f k ( n ) can and will take values of the form t 2 − t + 1 ) .
* Note that f ( n ) only misses integers that are of the form m 2 + m − 1 . Therefore it always(and MUST) take all other integral values. *