2004 AIME I Problems/Problem 15
For all positive integers , let and define a sequence as follows: and for all positive integers . Let be the smallest such that . (For example, and .) Let be the number of positive integers such that . Find the sum of the distinct prime factors of .
We backcount the number of ways. Namely, we start at , which can only be reached if , and then we perform operations that either consist of or . We represent these operations in a string format, starting with the operation that sends and so forth downwards. There are ways to pick the first operations; however, not all of them may be otherwise we return back to , contradicting our assumption that was the smallest value of . Using complementary counting, we see that there are only ways.
Since we performed the operation at least once in the first operations, it follows that , so that we no longer have to worry about reaching again.
However, we must also account for a sequence of or more s in a row, because that implies that at least one of those numbers was divisible by , yet the was never used, contradiction. We must use complementary counting again by determining the number of strings of s of length such that there are s in a row. The first ten are not included since we already accounted for that scenario above, so our string of s must be preceded by a . There are no other restrictions on the remaining seven characters. Letting to denote either of the functions, and to indicate that the character appears times in a row, our bad strings can take the forms:
There are ways to select the operations for the s, and places to place our block. Thus, our answer is , and the answer is .
Note: This solution is quick and most similar to the official solution; however, neither this nor the official solution prove that the final results of these inverted operations are all distinct. A more sophisticated argument, such as the one below, is needed to do so.
We approach the problem by recursion. We partition the positive integers into the sets First, we note that , so by the disjointness of the 's, we know that is not in any of the other sets. Also, we note that for .
We claim that if and , then . To prove this, we show that (the given function) maps bijectively onto . If , then , and , so . Also, , where is the smallest positive integer for which this is true. Therefore, , where is the smallest integer for which this is true. Thus . Also, since on this set, we see that implies that . Hence is an injection. If , then , where , so we know that , and . Therefore, is a surjection, so it must be a bijection. Therefore, if and , then .
We also claim that if , and , then . The proof is the same as in the previous paragraph, though some additional clarification is needed to prove that is a surjection. If , or rather , then if , we see that , and then , not as in the prior argument. However, this does not happen if . It is easy to check that . Therefore, the only time that the above argument could fail is if and . But in every other case, .
Next, we claim that if , then If , then , which is clearly an injective map. Also, , where is the smallest positive integer for which this is true. Therefore, , where is the smallest integer for which this is true. Thus for some . Conversely, if , then , so , so .
Based on these bijections, if we let , then Let . Then by adding the above equations (valid if ), we find that Now by using the above relations repeatedly, we find The first relation will only be valid if . Therefore, for . For smaller values, it is easy to use the relations to compute that the terms are powers of , although we note that because of the failure of the above argument. After this, we can simply use the recurrence relation for , finding Therefore, there are positive integers with . This factors as , where is prime. Thus the answer is .
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