A380254 Number of powerful numbers (in A001694) that do not exceed primorial A002110(n).

1, 1, 2, 7, 22, 85, 330, 1433, 6450, 31555, 172023, 964560, 5891154, 37807505, 248226019, 1702890101, 12401685616, 95277158949, 744210074157, 6091922351106, 51332717836692, 438592279944173, 3898316990125822, 35515462315592564, 335052677538616216, 3299888425002527366

Offset: 0

Author

Michael De Vlieger, Jan 19 2025

Keywords

• nonn
• hard

Comments

In other words, A001694(a(n)) is the largest powerful number less than or equal to A002110(n).

Examples

			Let P = A002110 and let s = A001694.
a(0) = 1 since P(0) = 1, and the set s(1) = {1} contains k that do not exceed 1.
a(1) = 1 since P(1) = 2, and the set s(1) = {1} contains k <= 2.
a(2) = 2 since P(2) = 6, and the set s(1..2) = {1, 4} contains k <= 6.
a(3) = 7 since P(3) = 30, and the set s(1..7) = {1, 4, 8, 9, 16, 25, 27} contains k <= 30.
a(4) = 22 since P(4) = 210, and the set s(1..19) = {1, 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 72, 81, 100, 108, 121, 125, 128, 144, 169, 196, 200} contains k <= 210, etc.
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 0..26

Crossrefs

Cf. A001694, A002110, A062762, A118896, A380337.

Programs

Mathematica

f[x_] := Sum[If[SquareFreeQ[ii], Floor[Sqrt[x/ii^3]], 0], {ii, x^(1/3)}];
Table[f[#[[k + 1]]], {k, 0, Length[#] - 1}] &[
  FoldList[Times, 1, Prime[Range[12] ] ] ] (* function f after Robert G. Wilson v at A118896 *)

Python

from math import isqrt
from sympy import primorial, integer_nthroot, mobius
def A380254(n):
    def squarefreepi(n): return int(sum(mobius(k)*(n//k**2) for k in range(1, isqrt(n)+1)))
    if n == 0: return 1
    m = primorial(n)
    c, l, j = squarefreepi(integer_nthroot(m, 3)[0]), 0, isqrt(m)
    while j>1:
        k2 = integer_nthroot(m//j**2,3)[0]+1
        w = squarefreepi(k2-1)
        c += j*(w-l)
        l, j = w, isqrt(m//k2**3)
    return c-l # Chai Wah Wu, Jan 24 2025

Extensions

a(18)-a(25) from Chai Wah Wu, Jan 24 2025