Suppose I have a line of squares, labelled 0, 1, 2, 3, 4, ... and so on. I place a counter on the square 0. On every turn, I roll a fair die (with the numbers 1 to 6) and move forward as follows: if I was on square before the roll and I get a then I move to square . Let be the chance that I land on the square labelled , where is a positive integer . What is
If the maximum doesn't exist, find the infimum of the set of .
Notation : denotes the floor function .
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For 1 ≤ n ≤ 6 , X n = ∑ k = 1 n N n , k / 6 k where N n , k is the number of writing k numbers (from 1 to 6) so that they add up to n . Since 1 ≤ n ≤ 6 , this is just the number of k-tuples of positive integers ( a 1 , … , a k ) such that a 1 + … + a k = n . So N n , k = ( n − 1 choose k − 1 ) (using the usual stars and bars method).
For n > 6 , note that X n = ∑ k = 1 6 X n − k / 6 . We have determined X n and it is straightforward to show that the maximum is attained at n = 6 with X 6 = 1 6 8 0 7 / 4 6 6 5 6 . For example, it might be helpful to show that X n = ( 2 + r 1 n + … + r 5 n ) / 7 where r 1 , . . . , r 5 are the five complex roots of 6 x 6 − x 5 − x 4 − x 3 − x 2 − x − 1 which are not one, and note that ∣ r i ∣ < 1 .