OFFSET
1,1
COMMENTS
Positions of 3's in A119241.
Shiu (1980) proved that this sequence has an asymptotic density = 0.0770... A more accurate calculation using his formula gives 0.0770742722233...
REFERENCES
József Sándor, Dragoslav S. Mitrinovic and Borislav Crstici, Handbook of Number Theory I, Springer Science & Business Media, 2005, chapter VI, p. 226.
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000
P. Shiu, On the number of square-full integers between successive squares, Mathematika, Vol. 27, No. 2 (1980), pp. 171-178.
EXAMPLE
31 is a term since there are exactly three powerful numbers, 968 = 2^3 * 11^2, 972 = 2^2 * 3^5 and 1000 = 2^3 * 5^3 between 31^2 = 961 and (31+1)^2 = 1024.
MATHEMATICA
powQ[n_] := (n == 1) || Min @@ FactorInteger[n][[;; , 2]] > 1; Select[Range[1000], Count[Range[#^2 + 1, (# + 1)^2 - 1], _?powQ] == 3 &]
PROG
(Python)
from functools import lru_cache
from math import isqrt
from sympy import mobius, integer_nthroot
def A336178(n):
def squarefreepi(n): return int(sum(mobius(k)*(n//k**2) for k in range(1, isqrt(n)+1)))
def bisection(f, kmin=0, kmax=1):
while f(kmax) > kmax: kmax <<= 1
while kmax-kmin > 1:
kmid = kmax+kmin>>1
if f(kmid) <= kmid:
kmax = kmid
else:
kmin = kmid
return kmax
@lru_cache(maxsize=None)
def g(x):
c, l = 0, 0
j = isqrt(x)
while j>1:
k2 = integer_nthroot(x//j**2, 3)[0]+1
w = squarefreepi(k2-1)
c += j*(w-l)
l, j = w, isqrt(x//k2**3)
c += squarefreepi(integer_nthroot(x, 3)[0])-l
return c
def f(x):
c, a = n+x, 1
for k in range(1, x+1):
b = g((k+1)**2)
if b == a+4:
c -= 1
a = b
return c
return bisection(f, n, n) # Chai Wah Wu, Sep 14 2024
CROSSREFS
KEYWORD
nonn
AUTHOR
Amiram Eldar, Jul 10 2020
STATUS
approved