By Herbert S. Wilf

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Suppose we follow the strategy that has so far won the game, that is, writing down the first few xs and trying to guess the pattern. Then we would find that x1 = b1 x0 + c1 , x2 = b2 b1 x0 + b2 c1 + c2 , and we would probably tire rapidly. 28). Though no approach will avoid the unpleasant form of the general answer, the one that we are about to describe at least gives a method that is much simpler than the guessing strategy, for many examples that arise in practice. 28), so a unified method will be a definite asset.

Every number that precedes it is smaller than it is and every number that follows it is larger than it is. What that means is that after sorting the list, the 37 will be in the same place it now occupies, the numbers to its left will have been sorted, but will still be on its left, and the numbers on its right will have been sorted, but will still be on its right. If we are fortunate enough to be given an array that has a ‘splitter,’ like 37, then we can: (a) Sort the numbers to the left of the splitter, and then (b) Sort the numbers to the right of the splitter.

Define d = n mod b. Then d is one of the b permissible digits. By induction, the number n0 = (n − d)/b is uniquely representable, say: n−d = d0 + d1 b + d2 b2 + . . b Then clearly n n−d b b = d + d0 b + d1 b2 + d2 b3 + . . = d+ is a representation of n that uses only the allowed digits. Finally, suppose that n has some other representation in this form also. Then we would have: n = a0 + a 1 b + a 2 b 2 + . . = c 0 + c1 b + c2 b 2 + . . Since a0 and c0 are both equal to n mod b, they are equal to each other.

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