Problems & Puzzles: Puzzles

 Puzzle 314. f(f(n))=prime(n) Joseph L. Pe, poses the following puzzle: I'd like to propose the following problem for your website. I don't know of any solution, or whether the problem is easy or hard. Maybe you'll have better luck than me.   Questions: Is there an integer-valued function f defined on integers, such that f(f(n)) = prime(n)? If there is, write down an explicit definition. If there is none, prove your assertion.

A solution came from Luke Pebody.

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Luke wrote:

Yes there is such a function.

Define integers z(m,n) as follows iteratively as follows:
z(m,1) is the mth non-prime number (z(1,1)=1, z(2,1)=4, z(3,1)=6, ...)
z(m,n+1) is the z(m,n)'th prime

Define f by f(z(2m+1,k))=z(2m+2,k) and
f(z(2m+2,k))=z(2m+1,k+1).

Theorem: For every positive integer n, there exists a,b such that n=z(a,b)

Proof:
Suppose not. Then there is a smallest positive integer n which fails to be of this form.
Then either n is prime, in which case it is the tth prime for some 1<=t<n. Then t=z(a,b) for some a,b, and hence n=z(a,b+1), which is a contradiction.
Or n is non-prime, in which case it is the tth non-prime for some 1<=t<=n, whence n=z(t,1).
QED

Corollary: For all integers n, f(f(n)) is the nth prime

Proof:
n=z(a,b). Either a is even, in which case f(n)=z(a-1,b+1) and f(f(n))=z(a,b+1)=nth prime, or a is odd, in which case f(n)=z(a+1,b) and f(f(n))=z(a,b+1)=nth prime.
QED

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