OFFSET
1,4
COMMENTS
a(n) = number of positive integers m<=n such that A006530(m) <= sqrt(n).
LINKS
Wikipedia, Smooth number
FORMULA
a(n) = n - A241419(n).
If n is in A063539, then a(n)=a(n-1)+1; if n is in A001248, i.e., n=p^2 for prime p, then a(n)=a(n-1)+p; otherwise a(n)=a(n-1).
a(n) = (1 - log(2))*n + O(n/log(n)) as n -> infinity. - Robert Israel, Nov 14 2017
MAPLE
N:= 100: # to get a(1)..a(N)
G:= [0, seq(max(numtheory:-factorset(n)), n=2..N)]:
seq(nops(select(t -> t^2 <= n, G[1..n])), n=1..N); # Robert Israel, Nov 14 2017
a:=[];
for n from 1 to 100 do
c:=0;
for m from 1 to n do
if A006530(m)^2 <= n then c:=c+1; fi; od:
a:=[op(a), c];
od:
a; # (Included because variants of it will apply to related sequences) - N. J. A. Sloane, Apr 10 2020
PROG
(PARI) A295084(n) = my(r=n); forprime(p=sqrtint(n)+1, n, r-=n\p); r;
(Python)
from math import isqrt
from sympy import primerange
def A295084(n): return int(n-sum(n//p for p in primerange(isqrt(n)+1, n+1))) # Chai Wah Wu, Oct 06 2024
CROSSREFS
KEYWORD
nonn,look
AUTHOR
Max Alekseyev, Nov 13 2017
STATUS
approved