A336175 Numbers k such that there are no powerful numbers between k^2 and (k+1)^2.

1, 3, 4, 6, 7, 9, 12, 13, 17, 21, 23, 24, 26, 27, 30, 32, 34, 35, 38, 40, 43, 47, 49, 54, 60, 61, 64, 66, 68, 69, 71, 75, 80, 81, 85, 86, 91, 95, 97, 99, 105, 106, 108, 112, 120, 123, 125, 128, 131, 133, 136, 137, 139, 142, 143, 151, 153, 154, 159, 160, 162, 163

Offset: 1

Author

Amiram Eldar, Jul 10 2020

Keywords

• nonn

Comments

Positions of 0's in A119241.

Shiu (1980) proved that this sequence has an asymptotic density 0.2759... A more accurate calculation using his formula gives 0.275965511407718981...

Examples

			1 is a term since the two numbers between 1^2 = 1 and (1+1)^2 = 4, 2 and 3, are not powerful.
2 is not a term since there is a powerful number, 8 = 2^3, between 2^2 = 4 and (2+1)^2 = 9.
		

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, A336176, A336177, A336178.

Programs

Mathematica

powQ[n_] := (n == 1) || Min @@ FactorInteger[n][[;; , 2]] > 1; Select[Range[150], ! AnyTrue[Range[#^2 + 1, (# + 1)^2 - 1], powQ] &]

PARI

is(n)=forfactored(k=n^2+1,n^2+2*n, if(ispowerful(k), return(0))); 1 \\ Charles R Greathouse IV, Oct 31 2022

Python

from functools import lru_cache
from math import isqrt
from sympy import mobius, integer_nthroot
def A336175(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+1:
                c -= 1
            a = b
        return c
    return bisection(f,n,n) # Chai Wah Wu, Sep 14 2024

Formula

a(n) ~ k*n, where k = 3.623641211175... is the inverse of the density, see Shiu link. - Charles R Greathouse IV, Oct 31 2022