A336178 Numbers k such that there are exactly three powerful numbers between k^2 and (k+1)^2.

31, 36, 67, 93, 132, 140, 145, 161, 166, 189, 192, 220, 223, 265, 280, 290, 296, 311, 316, 322, 364, 384, 407, 468, 537, 576, 592, 602, 623, 639, 644, 656, 659, 661, 670, 690, 722, 769, 771, 793, 828, 883, 888, 890, 896, 950, 961, 981, 984, 987, 992, 995, 1018

Offset: 1

Author

Amiram Eldar, Jul 10 2020

Keywords

• nonn

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...

Examples

			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.
		

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.

Crossrefs

Cf. A001694, A119241, A119242, A336175, A336176, A336177.

Programs

Mathematica

powQ[n_] := (n == 1) || Min @@ FactorInteger[n][[;; , 2]] > 1; Select[Range[1000], Count[Range[#^2 + 1, (# + 1)^2 - 1], _?powQ] == 3 &]

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