cp's OEIS Frontend

Positions of second occurrences of n in A165634: A165634(a(n)) = n. [Reinhard Zumkeller, Sep 23 2009]

a(n) = b(n)-th highest positive integer not equal to any a(k), 1 <= k <= n-1, where b(n) = primes = A000040(n). a(1) = 2, a(n) = a(n-1) + A000040(n) - A000040(n-1) + 1 for n >= 2. a(1) = 2, a(n) = a(n-1) + A001223(n-1) + 1 for n >= 2. a(n) = A014688(n) - 1. [Jaroslav Krizek, Oct 28 2009]

Comment from N. J. A. Sloane, Mar 28 2024 (Start):

On March 23 2024, Davide Rotondo sent me an email with the following conjecture. (I've simplified it a bit.)

For a positive integer n, define a sequence b by b(0) = n; b(i) = n - pi(b(i-1)) for i >= 1, where pi(x) = number of primes <= x.

The conjecture is that after some initial terms, b becomes periodic with period length 1 or 2, and the n for which the period is 2 are 3 together with the present sequence, that is, 2, 3, 4, 7, 10, 15, 18, 23, 26, 31, 38, 41, 48, ... (End)

Proof from Robert Israel, Mar 26 2024 (Start):

This is simply a consequence of the fact that if x < y, 0 <= pi(y) - pi(x) <= y - x and the inequality on the right is strict if y-x > 1 except for the case of 1 and 3.

Thus we start with b(0) - b(1) = pi(n). While |b(i) - b(i+1)| > 2 we get |b(i+1) - b(i+2)| = |pi(b(i+1)) - pi(b(i+2))| < |b(i) - b(i+1)|.

Eventually we must either reach |b(j+1) - b(j)| = 0 or |b(j+1) - b(j)| = 1.

If we reach 0, i.e. b(j+1) = b(j), then clearly b(k) = b(j) for all k > j.

If b(j+1) = b(j) + 1 = n - pi(b(j)), then b(j+2) = n - pi(b(j)+1) = b(j+1) or b(j+1)-1.

If b(j+1) = b(j) - 1, then b(j+2) = n - pi(b(j)-1) = b(j+1) or b(j+1)+1.

Thus from this point on we either get a 2-cycle or a 1-cycle. (End)