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. Thus the remaining operations can be picked in ways, with a total of strings.
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, then 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 .
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, we find that Now by using the above relations repeatedly, we find 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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