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%I #40 Jan 20 2022 10:33:13
%S 0,0,0,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3,
%T 3,3,3,3,3,3,3,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,
%U 4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4
%N Number of primes not exceeding square root of n: primepi(sqrt(n)).
%C Number of primes among factors of LCM(1,...,n) whose exponent is > 1, i.e., number of non-unitary prime factors of LCM(1,...,n).
%C Number of positive integers <= n with exactly 3 divisors.
%C Number of squared primes not exceeding n. - _Wesley Ivan Hurt_, May 24 2013
%H T. D. Noe, <a href="/A056811/b056811.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = A056170(A003418)) = A000720(A000196(n)).
%F For k = 1, 2, ..., repeat k A069482(k) (that is, prime(k+1)^2 - prime(k)^2) times, and add 0 three times at the beginning (or begin the preceding by k = 0, with prime(0) set to 1). - _Jean-Christophe Hervé_, Oct 30 2013
%F G.f.: (1/(1 - x)) * Sum_{k>=1} x^(prime(k)^2). - _Ilya Gutkovskiy_, Sep 14 2019
%F a(n) ~ 2*n^(1/2)/log(n), by the prime number theorem. - _Harry Richman_, Jan 19 2022
%e If n=169,...,288 = p()^2,...,p(7)^2-1, then only the first 6 primes have exponents larger than 1, resulting in powers: 128, 81, 125, 49, 121, 169. So a(n)=6 for as much as 288-169+1 = 120 values of n.
%t Table[PrimePi[Sqrt[n]], {n, 100}] (* _T. D. Noe_, Mar 13 2013 *)
%o (PARI) a(n) = primepi(sqrt(n)); \\ _Michel Marcus_, Apr 11 2016
%o (Python)
%o from math import isqrt
%o from sympy import primepi
%o def a(n): return primepi(isqrt(n))
%o print([a(n) for n in range(1, 88)]) # _Michael S. Branicky_, Jan 19 2022
%Y Cf. A000196, A000720, A003418, A056170.
%Y Cf. A069482, A056813.
%K nonn,easy
%O 1,9
%A _Labos Elemer_, Aug 28 2000
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