A069623 Number of perfect powers <= n.

1, 1, 1, 2, 2, 2, 2, 3, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12

Offset: 1

Author

Amarnath Murthy, Mar 27 2002

Keywords

• nonn
• easy

Examples

			a(27) = 7 as the perfect powers <= 27 are 1, 4, 8, 9, 16, 25 and 27.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• M. A. Nyblom, A Counting Function for the Sequence of Perfect Powers, Austral. Math. Soc. Gaz. 33 (2006), 338-343.
• Stackexchange, Proof of formula of number of Power <=n, Jun 24 2013
• Eric Weisstein's World of Mathematics, Perfect Powers.

Crossrefs

Perfect powers are A001597. Cf. A053289. A076411(n) = a(n-1) is another version.

Cf. A075802 (first differences). - Chayim Lowen, Jul 29 2015

Cf. A002321.

Programs

Maple

N:= 1000:  # to get a(n) for n <= N
R:= Vector(N):
for p from 2 to ilog2(N) do
  for i from 1 to floor(N^(1/p)) do
      R[i^p]:= 1
od od:
A069623:= map(round,Statistics:-CumulativeSum(R)):
convert(A069623,list); # Robert Israel, May 19 2014
# second Maple program:
a:= proc(n) option remember; `if`(n=1, 1, a(n-1)+
     `if`(igcd(seq(i[2], i=ifactors(n)[2]))>1, 1, 0))
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Feb 26 2019

Mathematica

a[1] = 1; a[n_] := If[ !PrimeQ[n] && GCD @@ Last[Transpose[FactorInteger[n]]] > 1, a[n - 1] + 1, a[n - 1]]; Table[a[n], {n, 1, 85}]
(* Or *) b[n_] := n - Sum[ MoebiusMu[k] * Floor[n^(1/k) - 1], {k, 1, Floor[ Log[2, n]]}]; Table[b[n], {n, 1, 85}]

PARI

a(n) = 1 + sum(k=1, n, ispower(k) != 0); \\ Michel Marcus, Jul 25 2015

PARI

a(n)=n-sum(k=1,logint(n,2), moebius(k)*(sqrtnint(n,k)-1)) \\ Charles R Greathouse IV, Jul 21 2017

PARI

a(n)=my(s=n); forsquarefree(k=1,logint(n,2), s-=(sqrtnint(n,k[1])-1)*moebius(k)); s \\ Charles R Greathouse IV, Jan 08 2018

Python

from sympy import mobius, integer_nthroot
def A069623(n): return int(n+sum(mobius(k)*(1-integer_nthroot(n,k)[0]) for k in range(1,n.bit_length()))) # Chai Wah Wu, Aug 13 2024

Formula

a(n) = n - Sum_{k=1..floor(log_2(n))} mu(k)*floor(n^(1/k)-1), where mu = A008683. - David W. Wilson, Oct 09 2002

a(n) = O(sqrt(n)) (conjectured). a(n) = A076411(n+1) = Sum_{k=1..n} A075802(k). - Chayim Lowen, Jul 24 2015

The conjecture is true: The number of squares < n is n^(1/2) + O(1). The number of higher powers < n is nonnegative and less than n^(1/3) log_2(n). Thus a(n) = n^(1/2) + O(n^(1/3) log n). - Robert Israel, Jul 31 2015

a(n) = n - Sum_{k=2..n} M(floor(log_k(n))), where M is Mertens's function A002321. - Ridouane Oudra, Dec 30 2020