Prove this function changes parity.

[e2theipi2026]

Given the prime number sequence $$p_n$$ and the function $$f(n)$$ which counts all composite $$k\le n$$ such that $$k$$ and $$k+2$$ are both composite, prove that $$f(p_n)$$ changes parity an infinite number of times.
Can there be such a proof?

 

[jvicens]

Yes, there can be a proof for this statement. Here is one possible approach to prove this:

First, we need to define the function f(n) more precisely. Let's say that f(n) is the number of pairs of consecutive composite numbers (k, k+2) where both k and k+2 are less than or equal to n. For example, for n=10, the pairs would be (4,6), (6,8), and (8,10), so f(10)=3.

Now, let's consider the prime number sequence p_n. This sequence contains all the prime numbers in ascending order, starting from 2. We can see that the sequence starts with an even number (2) and then alternates between even and odd numbers (3, 5, 7, etc.).

Next, let's look at the values of f(p_n) for the first few values of n:

- f(p_2) = f(2) = 0 (since there are no pairs of consecutive composite numbers less than or equal to 2)
- f(p_3) = f(3) = 0 (again, no pairs of consecutive composite numbers)
- f(p_4) = f(5) = 1 (the pair (4,6) is the only one that satisfies the conditions)
- f(p_5) = f(7) = 2 (the pairs (4,6) and (6,8) satisfy the conditions)
- f(p_6) = f(11) = 3 (the pairs (4,6), (6,8), and (8,10) satisfy the conditions)

We can see that as n increases, the value of f(p_n) also increases. However, the pattern of this increase is not consistent. For example, we have f(p_4) = 1 and f(p_5) = 2, so the parity (whether it is odd or even) changes from one value to the next. This change in parity continues as n increases, as shown by the increasing values of f(p_n).

Now, let's assume that at some point, the parity of f(p_n) stops changing and remains constant. This would mean that from that point on, the values of f(p_n) would either all be even or all be odd. However, this would also mean that there are no more pairs of