A056811 Number of primes not exceeding square root of n: primepi(sqrt(n)).

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, 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, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4

Offset

1,9

Comments

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).

Number of positive integers <= n with exactly 3 divisors.

Number of squared primes not exceeding n. - Wesley Ivan Hurt, May 24 2013

Maximum number of composite numbers not exceeding n that are all coprime to each other. - Yifan Xie, Jul 07 2024

Links

T. D. Noe, Table of n, a(n) for n = 1..10000

Formula

a(n) = A056170(A003418(n)) = A000720(A000196(n)).

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

G.f.: (1/(1 - x)) * Sum_{k>=1} x^(prime(k)^2). - Ilya Gutkovskiy, Sep 14 2019

a(n) ~ 2*n^(1/2)/log(n), by the prime number theorem. - Harry Richman, Jan 19 2022

Example

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.

Mathematica

Table[PrimePi[Sqrt[n]], {n, 100}] (* T. D. Noe, Mar 13 2013 *)

Prog

(PARI) a(n) = primepi(sqrt(n)); \\ Michel Marcus, Apr 11 2016
(Python)
from math import isqrt
from sympy import primepi
def a(n): return primepi(isqrt(n))
print([a(n) for n in range(1, 88)]) # Michael S. Branicky, Jan 19 2022

Crossrefs

Cf. A000196, A000720, A003418, A056170.

Cf. A069482, A056813.

Sequence in context: A121900 A333525 A373095 * A097430 A054900 A046042

Adjacent sequences: A056808 A056809 A056810 * A056812 A056813 A056814

Keyword

nonn,easy

Author

Labos Elemer, Aug 28 2000

Status

approved