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(MAGMAMagma) [ NthPrime(NthPrime(n)): n in [1..51] ]; // Jason Kimberley, Apr 02 2010
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Ernest G. Hibbs, <a href="https://www.proquest.com/openview/4012f0286b785cd732c78eb0fc6fce80">Component Interactions of the Prime Numbers</a>, Ph. D. Thesis, Capitol Technology Univ. (2022), see p. 33.
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Lim_Limit_{n->infinity} a(n)/(n*(log(n))^2) = 1. Proof: By Cipolla's asymptotic formula, prime(n) ~ L(n) + R(n), where L(n)/n = log(n) + log(log(n)) - 1 and R(n)/n decreases logarithmically to 0. Hence, for large n, a(n) = prime(prime(n)) ~ L(L(n)+R(n)) + R(L(n)+R(n)) = n*(log(n))^2 + r(n), where r(n) grows as O(n*log(n)*log(log(n))). The rest of the proof is trivial. The convergence is very slow: for k = 1,2,3,4,5,6, sqrt(a(10^k)/10^k)/log(10^k) evaluates to 2.055, 1.844, 1.695, 1.611, 1.545, and 1.493, respectively. - Stanislav Sykora, Dec 09 2015
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(Python)
from sympy import prime
def a(n): return prime(prime(n))
print([a(n) for n in range(1, 52)]) # Michael S. Branicky, Aug 11 2021
(Python) # much faster version for initial segment of sequence
from sympy import nextprime, isprime
def aupton(terms):
alst, p, pi = [], 2, 1
while len(alst) < terms:
if isprime(pi): alst.append(p)
p, pi = nextprime(p), pi+1
return alst
print(aupton(10000)) # Michael S. Branicky, Aug 11 2021
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