Interesting question: Describe a set in the simplest words

Define the value of F (N) to split N into several 2's power sum, and each number will not appear more than two times. Required F (0) = 1.
For example, there are five valid solutions to split the numbers into 10: 1 + 1 + 8, 1 + 1 + 4 + 4, 1 + 1 + 2 + 2 + 4, 2 + 4 + 4 and 2 + 8. Therefore, F (10) = 5.
Describe the set {f (N)/F (n-1)} In the simplest sentence )}. Prove your conclusion.

Note: The answer is far more ingenious than a recursive formula. If you find our conclusion, you will deem it as the correct answer at a glance.

Answer: The series {f (N)/F (n-1)} contains all positive rational numbers in the simplest form.

If n is an odd number (equal to 2 m + 1), The number 1 (that is, 2 ^ 0) must appear and can only appear once. The question now is, how many of the two m splitting schemes do not include the number 1? If you think about it, you will immediately find that it is equal to F (M), because all the numbers of all M split schemes are multiplied by 2, which exactly corresponds to the one-to-one 2 m split scheme without 1. Therefore, F (2 m + 1) = f (m)
If n is an even number (equal to 2 m), The number 1 either does not appear or exactly appears twice. For the previous case, we have a possible F (m) solution; for the second case, an F (m-1) solution. Therefore, F (2 m) = f (m) + f (m-1)
In addition, F (k) is obviously a positive number. So F (2k-1) = f (k-1) <F (k-1) + f (K) = F (2 k)
In this way, we can draw the following three conclusions:

Conclusion 1: gcd (f (N), F (n-1) = 1
Proof: mathematical induction of N. Apparently gcd (F (1), F (0) = gcd () = 1
Assume that all the numbers smaller than N are valid. According to the parity of N, one of the following two formulas must be true:
Gcd (f (N), F (n-1) = gcd (F (2 m + 1), F (2 m) = gcd (f (m ), F (m) + f (M) = gcd (f (M), F (M) = 1
Gcd (f (N), F (n-1) = gcd (F (2 m), F (2s-1) = gcd (f (m) + f (S-1 ), m-1) = gcd (f (M), F (S-1) = 1

Conclusion 2: If F (n + 1)/F (n) = f (n '+ 1)/F (n'), then n = N'
Proof: mathematical induction. When Max (N, N') = 0, the conclusion is obvious because n = n' = 0 at this time.
The conclusion is true for all numbers smaller than N. Because F (2k-1) <F (2 k), F (N)/F (n-1) = f (n ')/F (n'-1 ), the parity between N and n' must be the same, so f (m)/F (S-1) = f (m')/F (M'-1) can be introduced ), according to the conclusion, we have M = M', which tells us n = n '.

Conclusion 3: For any rational number R, there is always a positive integer n so that r = f (N)/F (n-1 ).
Proof: write r as the ratio of the number P to the Q of the two elements. We generalize max (p, q.
Obviously, when P = q = 1, the conclusion is true. At this time, n = 1.
If P <q is set, R' is defined as P/(Q-P ). According to the induction hypothesis, there is always a number M that satisfies the requirements of R' = f (m)/F (S-1 ). So r = f (2 m + 1)/F (2 m ). When P> q is used, the same can be proved.

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