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%I #26 Nov 14 2020 08:37:17
%S 5,41,439,4523,46411,470303,4751053,47868179,481346903,4834216447,
%T 48507378197,486425324491,4875521711327,48850585043591,
%U 489323461383733,4900321995437591,49065251868835099,491200019422325489,4916868348839249987,49212253916770634411
%N Nearest prime to the median of the primes below 10^n.
%H Kim Walisch, <a href="https://github.com/kimwalisch/primecount">primecount</a>.
%F a(n) = prime(floor(A006880(n) / 2) + 1). - _David A. Corneth_, Mar 19 2019
%e From _David A. Corneth_, Mar 19 2019: (Start)
%e There are 25 primes <= 10^2 = 100 so a(2) is the (25 + 1)/2 = 13th prime, being 41.
%e There are 168 primes <= 10^3 = 1000 so a(3) is the prime closest to the average of prime(84) = 433 and prime(85) = 439. That average is 436, equally distant from both primes. So the largest is chosen. This disables the need to compute the median and we can immediately say that a(3) = prime(168 / 2 + 1) = 439. (End)
%t a[n_] := Prime[Floor[PrimePi[10^n]/2] + 1]; Array[a, 10] (* _Amiram Eldar_, Mar 20 2019 *)
%o (Julia) using Statistics, Primes
%o function MedianPrimes(n)
%o med = Int64(round(median(primes(10^n))))
%o isprime(med) && return med
%o prevmed = prevprime(med); nextmed = nextprime(med)
%o abs(med - prevmed) < abs(med - nextmed) ? prevmed : nextmed end
%o println([MedianPrimes(n) for n in 1:12])
%o (PARI) a(n) = pi = primepi(10^n); prime(pi \ 2 + 1) \\ _David A. Corneth_, Mar 19 2019
%Y Cf. A000040, A000720, A006880.
%K nonn,more
%O 1,1
%A _Peter Luschny_, Mar 19 2019
%E a(11) from _David A. Corneth_, Mar 19 2019
%E a(12)-a(19) using Kim Walisch's primecount, from _Amiram Eldar_, Mar 20 2019
%E a(20) from _Jinyuan Wang_, Nov 14 2020
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