A076446 Differences of consecutive powerful numbers (definition 1).

3, 4, 1, 7, 9, 2, 5, 4, 13, 15, 8, 9, 19, 8, 13, 4, 3, 16, 25, 27, 4, 16, 9, 18, 13, 32, 1, 35, 19, 18, 31, 8, 32, 9, 43, 16, 12, 17, 47, 49, 23, 27, 1, 53, 55, 16, 41, 23, 36, 61, 7, 4, 28, 24, 65, 36, 27, 4, 69, 71, 27, 8, 21, 17, 3, 72, 77, 47, 32, 81, 47, 36, 36, 49, 87, 8

Offset: 1

Author

Jud McCranie, Oct 15 2002

Keywords

• nonn

Comments

The term 1 appears infinitely often. Erdős conjectured that two consecutive 1's do not occur. (see Guy).

Examples

			The first two powerful numbers are 1 and 4, their difference is 3, so a(1)=3.
		

References

• R. K. Guy, Unsolved Problems in Number Theory, B16

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
• Eric Weisstein's World of Mathematics, Powerful numbers

Crossrefs

Cf. A001694, A076444.

Programs

Haskell

a076446 n = a076446_list !! (n-1)
a076446_list = zipWith (-) (tail a001694_list) a001694_list
-- Reinhard Zumkeller, Nov 30 2012

Mathematica

Differences[Join[{1},Select[Range[2000],Min[FactorInteger[#][[All, 2]]]>1&]]] (* Harvey P. Dale, Aug 27 2017 *)

Python

from math import isqrt
from sympy import mobius, integer_nthroot
def A076446(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
    def f(x):
        c, l = n+x, 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
    return -(a:=bisection(f,n,n))+bisection(lambda x:f(x)+1,a,a) # Chai Wah Wu, Sep 10 2024

Formula

a(n) = A001694(n+1) - A001694(n).