A089580 -id:A089580 - cp's OEIS frontend

A001597 Perfect powers: m^k where m > 0 and k >= 2.

1, 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 81, 100, 121, 125, 128, 144, 169, 196, 216, 225, 243, 256, 289, 324, 343, 361, 400, 441, 484, 512, 529, 576, 625, 676, 729, 784, 841, 900, 961, 1000, 1024, 1089, 1156, 1225, 1296, 1331, 1369, 1444, 1521, 1600, 1681, 1728, 1764

Offset: 1

N. J. A. Sloane

Might also be called the nontrivial powers. - N. J. A. Sloane, Mar 24 2018
See A175064 for number of ways to write a(n) as m^k (m >= 1, k >= 1). - Jaroslav Krizek, Jan 23 2010
a(1) = 1, for n >= 2: a(n) = numbers m such that sum of perfect divisors of x = m has no solution. Perfect divisor of n is divisor d such that d^k = n for some k >= 1. a(n) for n >= 2 is complement of A175082. - Jaroslav Krizek, Jan 24 2010
A075802(a(n)) = 1. - Reinhard Zumkeller, Jun 20 2011
Catalan's conjecture (now a theorem) is that 1 occurs just once as a difference, between 8 and 9.
For a proof of Catalan's conjecture, see the paper by Metsänkylä. - L. Edson Jeffery, Nov 29 2013
m^k is the largest number n such that (n^k-m)/(n-m) is an integer (for k > 1 and m > 1). - Derek Orr, May 22 2014
From Daniel Forgues, Jul 22 2014: (Start)
a(n) is asymptotic to n^2, since the density of cubes and higher powers among the squares and higher powers is 0. E.g.,
  a(10^1) = 49 (49% of 10^2),
  a(10^2) = 6400 (64% of 10^4),
  a(10^3) = 804357 (80.4% of 10^6),
  a(10^4) = 90706576 (90.7% of 10^8),
  a(10^n) ~ 10^(2n) - o(10^(2n)). (End)
A proper subset of A001694. - Robert G. Wilson v, Aug 11 2014
a(10^n): 1, 49, 6400, 804357, 90706576, 9565035601, 979846576384, 99066667994176, 9956760243243489, ... . - Robert G. Wilson v, Aug 15 2014

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 66.
R. K. Guy, Unsolved Problems in Number Theory, Springer, 1st edition, 1981. See section D9.
René Schoof, Catalan's Conjecture, Springer-Verlag, 2008, p. 1.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

David W. Wilson, Table of n, a(n) for n = 1..10000.
Abdelkader Dendane, Power (Exponential) Calculator.
H. W. Gould, Problem H-170, Fib. Quart., 8 (1970), p. 268, problem H-170.
Werner Hürlimann, A First Digit Theorem for Powers of Perfect Powers, Communications in Mathematics and Applications Vol. 5, No. 3 (2014), pp. 91-99.
Rafael Jakimczuk, On the Distribution of Perfect Powers, Journal of Integer Sequences, Vol. 14 (2011), Article 11.8.5.
Rafael Jakimczuk, Asymptotic Formulae for the n-th Perfect Power, Journal of Integer Sequences, Vol. 15 (2012), Article 12.5.5.
Holly Krieger and Brady Haran, Catalan's Conjecture, Numberphile video (2018).
Tauno Metsänkylä, Catalan's conjecture: another old Diophantine problem solved, Bull. Amer. Math. Soc. (NS), Vol. 41, No. 1 (2004), pp. 43-57.
Preda Mihǎilescu, Primary Cyclotomic Units and a Proof of Catalan's Conjecture, Journal für die Reine und Angewandte Mathematik, vol. 27 (2004), pp. 167-195.
Donald J. Newman, A Problem Seminar, Springer; see Problem #72.
M. A. Nyblom, A Counting Function for the Sequence of Perfect Powers, Austral. Math. Soc. Gaz. 33 (2006), 338-343.
Hugo Pfoertner, 1010196 perfect powers up to 10^12, compressed 7z archive, 3.3 MB (2023).
Alf van der Poorten, Remarks on the sequence of 'perfect' powers.
Michel Waldschmidt, Open Diophantine problems, arXiv:math/0312440 [math.NT], 2003-2004.
Michel Waldschmidt, Lecture on the abc conjecture and some of its consequences, Abdus Salam School of Mathematical Sciences (ASSMS), Lahore, 6th World Conference on 21st Century Mathematics 2013.
Eric Weisstein's World of Mathematics, Perfect Power.

Complement of A007916.
Subsequence of A072103; A072777 is a subsequence.
Union of A075109 and A075090.
Cf. A023055, A023057, A025478, A070428, A072102, A074981, A076404, A239728, A239870, A097054, A089579, A089580.
There are four different sequences which may legitimately be called "prime powers": A000961 (p^k, k >= 0), A246655 (p^k, k >= 1), A246547 (p^k, k >= 2), A025475 (p^k, k=0 and k >= 2), and which are sometimes confused with the present sequence.
First differences give A053289.

import Data.Map (singleton, findMin, deleteMin, insert)
a001597 n = a001597_list !! (n-1)
(a001597_list, a025478_list, a025479_list) =
   unzip3 $ (1, 1, 2) : f 9 (3, 2) (singleton 4 (2, 2)) where
   f zz (bz, ez) m
    | xx < zz = (xx, bx, ex) :
                f zz (bz, ez+1) (insert (bx*xx) (bx, ex+1) $ deleteMin m)
    | xx > zz = (zz, bz, 2) :
                f (zz+2*bz+1) (bz+1, 2) (insert (bz*zz) (bz, 3) m)
    | otherwise = f (zz+2*bz+1) (bz+1, 2) m
    where (xx, (bx, ex)) = findMin m  --  bx ^ ex == xx
-- Reinhard Zumkeller, Mar 28 2014, Oct 04 2012, Apr 13 2012

[1] cat [n : n in [2..1000] | IsPower(n) ];

isA001597 := proc(n)
    local e ;
    e := seq(op(2,p),p=ifactors(n)[2]) ;
    return ( igcd(e) >=2 or n =1 ) ;
end proc:
A001597 := proc(n)
    option remember;
    local a;
    if n = 1 then
        1;
    else
        for a from procname(n-1)+1 do
            if isA001597(a) then
                return a ;
            end if;
         end do;
    end if;
end proc:
seq(A001597(n),n=1..70) ; # R. J. Mathar, Jun 07 2011
N:= 10000: # to get all entries <= N
sort({1,seq(seq(a^b, b = 2 .. floor(log[a](N))), a = 2 .. floor(sqrt(N)))}); # Robert FERREOL, Jul 18 2023

min = 0; max = 10^4;  Union@ Flatten@ Table[ n^expo, {expo, Prime@ Range@ PrimePi@ Log2@ max}, {n, Floor[1 + min^(1/expo)], max^(1/expo)}] (* T. D. Noe, Apr 18 2011; slightly modified by Robert G. Wilson v, Aug 11 2014 *)
perfectPowerQ[n_] := n == 1 || GCD @@ FactorInteger[n][[All, 2]] > 1; Select[Range@ 1765, perfectPowerQ] (* Ant King, Jun 29 2013; slightly modified by Robert G. Wilson v, Aug 11 2014 *)
nextPerfectPower[n_] := If[n == 1, 4, Min@ Table[ (Floor[n^(1/k)] + 1)^k, {k, 2, 1 + Floor@ Log2@ n}]]; NestList[ nextPerfectPower, 1, 55] (* Robert G. Wilson v, Aug 11 2014 *)
Join[{1},Select[Range[2000],GCD@@FactorInteger[#][[All,2]]>1&]] (* Harvey P. Dale, Apr 30 2018 *)

{a(n) = local(m, c); if( n<2, n==1, c=1; m=1; while( cMichael Somos, Aug 05 2009 */

is(n)=ispower(n) || n==1 \\ Charles R Greathouse IV, Sep 16 2015

list(lim)=my(v=List(vector(sqrtint(lim\=1),n,n^2))); for(e=3,logint(lim,2), for(n=2,sqrtnint(lim,e), listput(v,n^e))); Set(v) \\ Charles R Greathouse IV, Dec 10 2019

from sympy import perfect_power
def ok(n): return n==1 or perfect_power(n)
print([m for m in range(1, 1765) if ok(m)]) # Michael S. Branicky, Jan 04 2021

import sympy
class A001597() :
    def _init_(self) :
        self.a = [1]
    def at(self, n):
        if n <= len(self.a):
            return self.a[n-1]
        else:
            cand = self.at(n-1)+1
            while sympy.perfect_power(cand) == False:
                cand += 1
            self.a.append(cand)
            return cand
a001597 = A001597()
for n in range(1,20):
    print(a001597.at(n)) # R. J. Mathar, Mar 28 2023

from sympy import mobius, integer_nthroot
def A001597(n):
    def f(x): return int(n-2+x+sum(mobius(k)*(integer_nthroot(x,k)[0]-1) for k in range(2,x.bit_length())))
    kmin, kmax = 1,2
    while f(kmax) >= kmax:
        kmax <<= 1
    while True:
        kmid = kmax+kmin>>1
        if f(kmid) < kmid:
            kmax = kmid
        else:
            kmin = kmid
        if kmax-kmin <= 1:
            break
    return kmax # Chai Wah Wu, Aug 13 2024

def A001597_list(n) :
    return [k for k in (1..n) if k.is_perfect_power()]
A001597_list(1764) # Peter Luschny, Feb 03 2012

Goldbach showed that Sum_{n >= 2} 1/(a(n)-1) = 1.
Formulas from postings to the Number Theory List by various authors, 2002:
Sum_{i >= 2} Sum_{j >= 2} 1/i^j = 1;
Sum_{k >= 2} 1/(a(k)+1) = Pi^2 / 3 - 5/2;
Sum_{k >= 2} 1/a(k) = Sum_{n >= 2} mu(n)(1- zeta(n)) approx = 0.87446436840494... See A072102.
For asymptotics see Newman.
For n > 1: gcd(exponents in prime factorization of a(n)) > 1, cf. A124010. - Reinhard Zumkeller, Apr 13 2012
a(n) ~ n^2. - Thomas Ordowski, Nov 04 2012
a(n) = n^2 - 2*n^(5/3) - 2*n^(7/5) + (13/3)*n^(4/3) - 2*n^(9/7) + 2*n^(6/5) - 2*n^(13/11) + o(n^(13/11)) (Jakimczuk, 2012). - Amiram Eldar, Jun 30 2023

Minor corrections from N. J. A. Sloane, Jun 27 2010

A070428 Number of perfect powers (A001597) not exceeding 10^n.

Original entry on oeis.org

1, 4, 13, 41, 125, 367, 1111, 3395, 10491, 32670, 102231, 320990, 1010196, 3184138, 10046921, 31723592, 100216745, 316694005, 1001003332, 3164437425, 10004650118, 31632790244, 100021566157, 316274216762, 1000100055684

Offset: 0

Donald S. McDonald, May 15 2002

In the programs for this sequence, 4*n can be replaced by the smaller floor(n*log(10)/log(2)). - T. D. Noe, Nov 17 2006

			a(1) = 4 because the powers 1, 4, 8, 9 do not exceed 10^1.
a(2) = 13 because 1, 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 81 & 100, are the only perfect power numbers less than or equal to 100.

The Dominion (Wellington, NZ), 'wtd sell', 9 Nov. 1991.
sci.math, powers not exceeding n. nz science monthly advt, March 1993, 1:80 integers 1..10000 is perfect square or higher power.

Chai Wah Wu, Table of n, a(n) for n = 0..1999 (terms 0..999 from Robert G. Wilson v)
Eric Weisstein's World of Mathematics, Perfect Power

Cf. A001597.

Cf. A089579, A089580 (number of perfect powers (not including 1) < 10^n).

f[n_] := 1 - Sum[ MoebiusMu[x]*Floor[10^(n/x) - 1], {x, 2, n*Log2[10]}]; Array[f, 25, 0] (* Robert G. Wilson v, May 22 2009; modified Aug 04 2014 *)

for(n=0, 25, print1(sum(x=2, 4*n,-moebius(x)*(floor(10^(n/x)-1)))+1, ", ")); \\ Slightly modified by Jinyuan Wang, Mar 02 2020

from sympy import mobius, integer_nthroot
def A070428(n): return int(1-sum(mobius(x)*(integer_nthroot(10**n,x)[0]-1) for x in range(2,(10**n).bit_length()))) # Chai Wah Wu, Aug 13 2024

a(n) ~ sqrt(10^n).

More terms from Larry Reeves (larryr(AT)acm.org), Oct 03 2002

Edited and extended by Robert G. Wilson v, Oct 11 2002

A089579 Total number of perfect powers > 1 below 10^n.

Original entry on oeis.org

3, 11, 39, 123, 365, 1109, 3393, 10489, 32668, 102229, 320988, 1010194, 3184136, 10046919, 31723590, 100216743, 316694003, 1001003330, 3164437423, 10004650116, 31632790242, 100021566155, 316274216760, 1000100055682, 3162493192563, 10000464300849, 31623776828239, 100002154796112

Offset: 1

Martin Renner, Dec 29 2003

nonn

k is a perfect power <=> there exist integers a and b, b > 1, and k = a^b.
From Robert G. Wilson v, Jul 17 2016: (Start)
Limit_{n->oo} a(n)/sqrt(10^n) = 1.
A089580(n) - a(n) = A275358(n).
The four terms which make up the difference between A089580(2) - a(2) are: 16 = 2^4 = 4^2, 64 = 2^6 = 4^3 = 8^2 and 81 = 3^4 = 9^2; one for 16, two for 64 and one for 81 making a total of 4. See A117453.
(End)

			For n=2, the 11 perfect powers > 1 below 10^2 = 100 are: 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 81. - _Michael B. Porter_, Jul 18 2016

Robert G. Wilson v, Table of n, a(n) for n = 1..100

Cf. A001597, A070428, A089580, A275358.

Table[lim=10^n-1; Sum[ -(Floor[lim^(1/k)]-1)*MoebiusMu[k], {k,2,Floor[Log[2,lim]]}], {n,30}] (* T. D. Noe, Nov 16 2006 *)

from sympy import mobius, integer_nthroot
def A089579(n): return int(sum(mobius(x)*(1-integer_nthroot(10**n,x)[0]) for x in range(2,(10**n).bit_length())))-1 if n>1 else 3 # Chai Wah Wu, Aug 13 2024

def A089579(n):
    gen = (p for p in srange(2, 10^n) if p.is_perfect_power())
    return sum(1 for _ in gen)
print([A089579(n) for n in range(1, 7)])  # Peter Luschny, Sep 15 2023

a(n) = A070428(n) - 2 for n >= 2.

a(9)-a(10) from Martin Renner, Oct 02 2004
More terms from T. D. Noe, Nov 16 2006
More precise name by Hugo Pfoertner, Sep 15 2023

A060298 Number of powers x^y (x,y > 1) with n digits.

Original entry on oeis.org

3, 12, 34, 94, 263, 768, 2333, 7167, 22291, 69751, 219081, 689736, 2174856, 6864354, 21679391, 68497906, 216485583, 684323923, 2163459803, 6840258025, 21628220224, 68388917596, 216252901472, 683826283482, 2162393925204, 6837972506895, 21623315009817

Offset: 1

Michel ten Voorde, Apr 10 2001

Conjectures from Robert G. Wilson v, Aug 29 2012: (Start)
Limit_{n->oo} a(2n)/10^n = 1 - 1/sqrt(10).
Limit_{n->oo} a(2n-1)/10^n = 1/sqrt(10) - 1/10. (End)
These follow from the Formula. - Robert Israel, Apr 29 2020
Limit_{n->oo} a(n)/a(n-1) = sqrt(10). - Bernard Schott, Jan 21 2023

			a(1) = 3 because there are 3 powers with 1 digit: 2^2, 2^3 and 3^2.

Robert Israel, Table of n, a(n) for n = 1..1000
Karl-Heinz Hofmann, Python program

Cf. A001597, A089580 (partial sums).

f:= proc(n) local y;
  add(ceil(10^(n/y))-ceil(10^((n-1)/y)), y=2..floor(n*log[2](10)))
end proc:
f(1):= 3:
map(f, [$1..20]); # Robert Israel, Apr 29 2020

Python
```
# see link
```

from sympy import integer_nthroot, integer_log
def A060298(n):
    if n == 1: return 3
    c, y, a, b, t = 0, 2, 10**n-1, 10**(n-1)-1, (10**n).bit_length()
    while yChai Wah Wu, Oct 16 2023

a(n) = Sum_{y=2..floor(n*log_2(10))} (ceiling(10^(n/y)) - ceiling(10^((n-1)/y))) for n >= 2. - Robert Israel, Apr 29 2020

a(n) = A089580(n+1) - A089580(n) for n > 1. - Karl-Heinz Hofmann, Sep 18 2023

a(10)-a(18) from Donovan Johnson, Dec 14 2009

a(19)-a(27) from Donovan Johnson, Aug 29 2012

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A275358 The difference between A089580(n) and A089579(n).

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A001597 Perfect powers: m^k where m > 0 and k >= 2.

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A070428 Number of perfect powers (A001597) not exceeding 10^n.

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A089579 Total number of perfect powers > 1 below 10^n.

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A060298 Number of powers x^y (x,y > 1) with n digits.

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A379757 a(n) = a(n-1) + 1 with two exceptions: if a(n-1) is prime, a(n) = a(n-2) + a(n-1), or if a(n-1) is a power, a(n) = a(n-1) / (root factor), with initial three terms are 0, 1, 2.

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