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
Examples
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.
Links
- T. D. Noe, Table of n, a(n) for n = 1..10000
Programs
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Mathematica
Table[PrimePi[Sqrt[n]], {n, 100}] (* T. D. Noe, Mar 13 2013 *)
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PARI
a(n) = primepi(sqrt(n)); \\ Michel Marcus, Apr 11 2016
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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
Formula
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
Comments