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It can be shown that a(n) > a(n-1) >= 1 and a(n) <= 2*n - 1 < 2*n (see proofs in LINKSthe Links section).

Ya-Ping Lu, <a href="/A334614/a334614.pdf">Proofs of the two observations in COMMENTSthe Comments section</a>

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It can be shown that a(n) > a(n-1) >= 1 and a(n) <= 2*n - 1 < 2*n (see proofs in LINKS).

Ya-Ping Lu, <a href="/A334614/a334614.pdf">Proofs of the two observations in COMMENTS</a>

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It can be noticed shown that this is an increasing sequence a(n) > a(n-1) >= 1 and the ratio of a(n) to <= 2*n is - 1 < 2. Short proofs of these observations are provided below*n.

Let p_n = prime(n).

i) a(n) > a(n-1) >= 1. a(n) - a(n-1) = (pi(p_n - n) + n) - (pi(p_(n-1) - (n-1)) + (n-1)) = pi(p_n - n) - pi(p_(n-1) - (n-1)) + 1. Since (p_n - n) - (p_(n-1) - (n-1)) = (p_n - p_(n-1) - 1 and p_n - p_(n-1) >= 1, we have pi(p_n -n) - pi(p_(n-1) - (n-1)) >= 0, or a(n) - a(n-1) >= 1. Thus a(n) > a(n - 1) > ... > a(1) = 1.

ii) a(n) < 2*n. It is shown that there is at least one prime in the range (k-pi(k), k) for any integer k > 2 (see comments in A332086). Replacing k by p_n, we know that there should be at least one prime in the range of (p_n - n, p_n), or pi(p_n) - pi(p_n - n) >= 1. Since pi(p_n - n) = n - (pi(p_n) - pi(p_n - n)) <= n - 1, we have a(n)/n = (pi(p_n - n) + n)/n = pi(p_n - n)/n + 1 <= (n - 1)/n + 1 = 2 - 1/n.

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ii) a(n)/n <= 2 - 1/*n. It is shown that there is at least one prime in the range (k-pi(k), k) for any integer k > 2 (see comments in A332086). Replacing k by p_n, we know that there should be at least one prime in the range of (p_n - n, p_n), or pi(p_n) - pi(p_n - n) >= 1. Since pi(p_n - n) = n - (pi(p_n) - pi(p_n - n)) <= n - 1, we have a(n)/n = (pi(p_n - n) + n)/n = pi(p_n - n)/n + 1 <= (n - 1)/n + 1 = 2 - 1/n.

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