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Let function f(n) = n^2+1 if it is odd, and n/2 if it is even. How many integers from 1 to 100, inclusive, does f(f...f(n)...)) = 1 for some number of applications of f?

 

I don't need the answer, but a step-by-step guide to this problem will be helpful.\(\)

 Sep 24, 2022
 #1
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My theory is that odd numbers (except 1) will never reach 1, and only powers of 2 will reach 1. Why? First of all after some trial and error, I realized that many of the odd numbers squared + 1 are not powers of 2. If the odd number squared + 1 is not a power of 2, it will redo the process of n^2 + 1. However, if it is a power of 2, then it will reach 1. But trial and error says elsewise and n^2 + 1 won't be a power of 2 (idk if that's a theorem). Thus the only integers from 1 - 100 will be 1, 2, 4, 8, 16, 32, and 64. So 7 of them...

 Sep 28, 2022

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