A069637 Number of prime powers <= n with exponents > 1.
0, 0, 0, 1, 1, 1, 1, 2, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10
Offset: 1
Keywords
References
- H. Sahu, K. Kar and B.S.K.R. Somayajulu, On the average order of pi*(n) - pi(n), Acta Cienc. Indica Math., Vol. 11 (1985), pp. 165-168.
- József Sándor, Dragoslav S. Mitrinovic and Borislav Crstici, Handbook of Number Theory I, Springer Science & Business Media, 2005, Chapter VII, p. 237.
Links
- Daniel Forgues, Table of n, a(n) for n=1..100000.
Programs
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Maple
with(numtheory); A069637 := proc(N) local ct,i; ct:=0; for i from 1 to N do if not isprime(i) and nops(factorset(i))=1 then ct:=ct+1; fi; od; ct; end; # N. J. A. Sloane, Jun 05 2022
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Mathematica
Table[Sum[PrimePi[n^(1/k)], {k, Log[2, n]}]-PrimePi[n],{n,94}] (* Stefano Spezia, Jun 05 2022 *)
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Python
from sympy import primepi, integer_nthroot def A069637(n): return sum(primepi(integer_nthroot(n,k)[0]) for k in range(2,n.bit_length())) # Chai Wah Wu, Aug 15 2024
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SageMath
[A025528(n) - prime_pi(n) for n in (1..100)] # Peter Luschny, Nov 18 2019
Formula
a(n) = A025528(n) - A000720(n) = A000720([n^(1/2)]) + A000720([n^(1/3)]) + ... . - Max Alekseyev, May 11 2009
Sum_{k=1..n} a(k) ~ (4/3) * n^(3/2)/log(n) + O(n^(3/2)/log(n)^2) (Sahu et al., 1985). - Amiram Eldar, Mar 07 2021
Comments