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A069623 Number of perfect powers <= n. 8

%I

%S 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,

%T 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,

%U 10,10,11,11,11,11,11,11,11,11,11,11,11,11,11,11,11,11,11,12,12,12,12,12

%N Number of perfect powers <= n.

%H Robert Israel, <a href="/A069623/b069623.txt">Table of n, a(n) for n = 1..10000</a>

%H M. A. Nyblom, <a href="http://www.austms.org.au/Publ/Gazette/2006/Nov06/nyblom.pdf">A Counting Function for the Sequence of Perfect Powers</a>, Austral. Math. Soc. Gaz. 33 (2006), 338-343.

%H Stackexchange, <a href="http://math.stackexchange.com/questions/428351/proof-of-formula-of-number-of-power-leq-n">Proof of formula of number of Power <=n</a>, Jun 24 2013

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PerfectPower.html">Perfect Powers</a>.

%F a(n) = n - Sum_{k=1..floor(log_2(n))} mu(k)*[n^(1/k)-1]), where mu = A008683. - _David W. Wilson_, Oct 09 2002

%F a(n) = O(sqrt(n)) (conjectured). a(n) = A076411(n+1) = Sum_{k=1..n} A075802(k). - _Chayim Lowen_, Jul 24 2015

%F 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

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

%p N:= 1000: # to get a(n) for n <= N

%p R:= Vector(N):

%p for p from 2 to ilog2(N) do

%p for i from 1 to floor(N^(1/p)) do

%p R[i^p]:= 1

%p od od:

%p A069623:= map(round,Statistics:-CumulativeSum(R)):

%p convert(A069623,list); # _Robert Israel_, May 19 2014

%p # second Maple program:

%p a:= proc(n) option remember; `if`(n=1, 1, a(n-1)+

%p `if`(igcd(seq(i[2], i=ifactors(n)[2]))>1, 1, 0))

%p end:

%p seq(a(n), n=1..100); # _Alois P. Heinz_, Feb 26 2019

%t 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}]

%t (* Or *) b[n_] := n - Sum[ MoebiusMu[k] * Floor[n^(1/k) - 1], {k, 1, Floor[ Log[2, n]]}]; Table[b[n], {n, 1, 85}]

%o (PARI) a(n) = 1 + sum(k=1, n, ispower(k) != 0); \\ _Michel Marcus_, Jul 25 2015

%o (PARI) a(n)=n-sum(k=1,logint(n,2), moebius(k)*(sqrtnint(n,k)-1)) \\ _Charles R Greathouse IV_, Jul 21 2017

%o (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

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

%Y Cf. A075802 (first differences). - _Chayim Lowen_, Jul 29 2015

%K nonn,easy

%O 1,4

%A _Amarnath Murthy_, Mar 27 2002

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Last modified October 14 18:28 EDT 2019. Contains 328022 sequences. (Running on oeis4.)