OFFSET
1,1
COMMENTS
LINKS
Michael De Vlieger, Table of n, a(n) for n = 1..10000 (first 5997 terms from Robert Israel)
Eric Weisstein's World of Mathematics, Prime Power.
Eric Weisstein's World of Mathematics, Powerful Number.
FORMULA
Sum_{n>=1} 1/a(n) = zeta(2)*zeta(3)/zeta(6) - Sum_{p prime} 1/(p*(p-1)) - 1 = A082695 - A136141 - 1 = 0.17043976777096407719... - Amiram Eldar, Feb 12 2021
EXAMPLE
-------------------------------
| n | a(n) | prime |
| | | factorization |
|------------------------------
| 1 | 36 | {{2, 2}, {3, 2}} |
| 2 | 72 | {{2, 3}, {3, 2}} |
| 3 | 100 | {{2, 2}, {5, 2}} |
| 4 | 108 | {{2, 2}, {3, 3}} |
| 5 | 144 | {{2, 4}, {3, 2}} |
| 6 | 196 | {{2, 2}, {7, 2}} |
| 7 | 200 | {{2, 3}, {5, 2}} |
| 8 | 216 | {{2, 3}, {3, 3}} |
| 9 | 225 | {{3, 2}, {5, 2}} |
-------------------------------
a(n) = p_1^e_1*p_2^e_2*... : {{p_1, e_1}, {p_2, e_2}, ...}.
MAPLE
N:= 10000:
S:= {1}: P:= {1}:
p:= 1:
do
p:= nextprime(p);
if p^2 > N then break fi;
S:= map(s -> (s, seq(s*p^k, k = 2 .. floor(log[p](N/s)))), S);
P:= P union {seq(p^k, k=2..floor(log[p](N)))}:
od:
sort(convert(S minus P, list)); # Robert Israel, May 14 2017
MATHEMATICA
Select[Range@2750, Min@FactorInteger[#][[All, 2]] > 1 && ! PrimePowerQ[#] &]
(* Second program *)
nn = 2^25; Select[Rest@ Union@ Flatten@ Table[a^2*b^3, {b, nn^(1/3)}, {a, Sqrt[nn/b^3]}], ! PrimePowerQ[#] &] (* Michael De Vlieger, Jun 22 2022 *)
PROG
(Python)
from sympy import primefactors, factorint
print([n for n in range(4, 2745) if len(primefactors(n)) > 1 and min(list(factorint(n).values())) > 1]) # Karl-Heinz Hofmann, Feb 07 2023
(Python)
from math import isqrt
from sympy import integer_nthroot, primepi, mobius
def A286708(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+1+sum(primepi(integer_nthroot(x, k)[0]) for k in range(2, x.bit_length()))
return bisection(f, n, n) # Chai Wah Wu, Sep 10 2024
CROSSREFS
KEYWORD
nonn
AUTHOR
Ilya Gutkovskiy, May 13 2017
STATUS
approved