A320966 Powerful numbers A001694 divisible by a cube > 1.
8, 16, 27, 32, 64, 72, 81, 108, 125, 128, 144, 200, 216, 243, 256, 288, 324, 343, 392, 400, 432, 500, 512, 576, 625, 648, 675, 729, 784, 800, 864, 968, 972, 1000, 1024, 1125, 1152, 1296, 1323, 1331, 1352, 1372, 1568, 1600, 1728, 1800, 1936, 1944, 2000, 2025, 2048, 2187, 2197, 2304, 2312, 2401, 2500
Offset: 1
Keywords
Links
- Hugo Pfoertner, Table of n, a(n) for n = 1..10000
Programs
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Mathematica
Select[Range[2500], (m = MinMax[FactorInteger[#][[;; , 2]]])[[1]] > 1 && m[[2]] > 2 &] (* Amiram Eldar, Jun 25 2022 *)
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PARI
isA001694(n)=n=factor(n)[, 2]; for(i=1, #n, if(n[i]==1, return(0))); 1 \\ from Charles R Greathouse IV isA046099(n)=n=factor(n)[, 2]; for(i=1, #n, if(n[i]>2, return(1)));0 for (k=1,2500,if(isA001694(k)&&isA046099(k),print1(k,", ")))
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Python
from math import isqrt from sympy import mobius, integer_nthroot def A320966(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+squarefreepi(isqrt(x))-squarefreepi(integer_nthroot(x,3)[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) return c+l return bisection(f,n,n) # Chai Wah Wu, Sep 15 2024
Formula
Sum_{n>=1} 1/a(n) = zeta(2)*zeta(3)/zeta(6) - 15/Pi^2 = 0.4237786821... . - Amiram Eldar, Jun 25 2022
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