A023057 -id:A023057 - 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

A023055 (Apparently) differences between adjacent perfect powers (integers of form a^b, a >= 1, b >= 2).

Original entry on oeis.org

1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 12, 13, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26, 27, 28, 30, 32, 33, 35, 36, 37, 38, 39, 40, 41, 43, 45, 47, 48, 49, 51, 53, 55, 56, 57, 59, 60, 61, 63, 65, 67, 68, 69, 71, 73, 74, 75, 76, 77, 79, 80, 81, 83, 85, 87, 89, 92, 93, 94, 95, 97, 99, 100

Offset: 1

David W. Wilson

Catalan's conjecture (now a theorem) is that 1 occurs just once as a difference, between 8 and 9.

Alf van der Poorten, Remarks on the sequence of 'perfect' powers

Cf. A001597, A023057.

Cf. A189117 (conjectured number of pairs of consecutive perfect powers differing by n).

A074981 Conjectured list of positive numbers which are not of the form r^i - s^j, where r,s,i,j are integers with r>0, s>0, i>1, j>1.

Original entry on oeis.org

6, 14, 34, 42, 50, 58, 62, 66, 70, 78, 82, 86, 90, 102, 110, 114, 130, 134, 158, 178, 182, 202, 206, 210, 226, 230, 238, 246, 254, 258, 266, 274, 278, 290, 302, 306, 310, 314, 322, 326, 330, 358, 374, 378, 390, 394, 398, 402, 410, 418, 422, 426

Offset: 1

Zak Seidov, Oct 07 2002

This is a famous hard problem and the terms shown are only conjectured values.
The terms shown are not the difference of two powers below 10^19. - Don Reble
One can immediately represent all odd numbers and multiples of 4 as differences of two squares. - Don Reble
_Ed Pegg Jr_ remarks (Oct 07 2002) that the techniques of Preda Mihailescu (see MathWorld link) might make it possible to prove that 6, 14, ... are indeed members of this sequence.
Numbers n such that there is no solution to Pillai's equation. - T. D. Noe, Oct 12 2002
The terms shown are not the difference of two powers below 10^27. - Mauro Fiorentini, Jan 03 2020

			Examples showing that certain numbers are not in the sequence: 10 = 13^3 - 3^7, 22 = 7^2 - 3^3, 29 = 15^2 - 14^2, 31 = 2^5 - 1, 52 = 14^2 - 12^2, 54 = 3^4 - 3^3, 60 = 2^6 - 2^2, 68 = 10^2 - 2^5, 72 = 3^4 - 3^2, 76 = 5^3 - 7^2, 84 = 10^2 - 2^4, ... 342 = 7^3 - 1^2, ...

R. K. Guy, Unsolved Problems in Number Theory, Sections D9 and B19.
P. Ribenboim, Catalan's Conjecture, Academic Press NY 1994.
T. N. Shorey and R. Tijdeman, Exponential Diophantine Equations, Cambridge University Press, 1986.

Mauro Fiorentini, Table of n, a(n) for n = 1..138
A. Baker, Review of "Catalan's conjecture" by P. Ribenboim, Bull. Amer. Math. Soc. 32 (1995), 110-112.
M. E. Bennett, On Some Exponential Equations Of S. S. Pillai, Canad. J. Math. 53 (2001), 897-922.
Yu. F. Bilu, Catalan's Conjecture (after Mihailescu)
J. Boéchat and M. Mischler, La conjecture de Catalan racontée a un ami qui a le temps, arXiv:math/0502350 [math.NT], 2005-2006.
C. K. Caldwell, The Prime Glossary, Catalan's Problem
T. Metsankyla, Catalan's Conjecture: Another old Diophantine problem solved, Bull. Amer. Math. Soc. 41 (2004), 43-57.
Alf van der Poorten, Remarks on the sequence of 'perfect' powers
P. Ribenboim, Catalan's Conjecture, Séminaire de Philosophie et Mathématiques, 6 (1994), pp. 1-11.
P. Ribenboim, Catalan's Conjecture, Amer. Math. Monthly, Vol. 103(7) Aug-Sept 1996, pp. 529-538.
Gérard Villemin, Conjecture de Catalan (French)
Eric Weisstein's World of Mathematics, Draft Proof of Catalan's Conjecture Circulated
Eric Weisstein's World of Mathematics, Pillai's Conjecture
Wikipedia, Catalan's conjecture
Wikipedia, Hall's conjecture

Subsequence of A016825 (see second comment of Don Reble).
n such that A076427(n) = 0. [Corrected by Jonathan Sondow, Apr 14 2014]
For a count of the representations of a number as the difference of two perfect powers, see A076427. The numbers that appear to have unique representations are listed in A076438.
For sequence with similar definition, but allowing negative powers, see A066510.
Cf. A001597, A074980, A069586, A023057, A075788-A075791, A053289, A076438, A207079, A219551.

Corrected by Don Reble and Jud McCranie, Oct 08 2002. Corrections were also sent in by Neil Fernandez, David W. Wilson, and Reinhard Zumkeller.

A245713 Sorted imperfect powers b^p with b > 0, p > 2, with multiplicity.

Original entry on oeis.org

1, 8, 16, 27, 32, 64, 64, 81, 125, 128, 216, 243, 256, 256, 343, 512, 512, 625, 729, 729, 1000, 1024, 1024, 1296, 1331, 1728, 2048, 2187, 2197, 2401, 2744, 3125, 3375, 4096, 4096, 4096, 4096, 4913, 5832, 6561, 6561, 6859, 7776, 8000, 8192, 9261

Offset: 1

Anatoly E. Voevudko, Jul 30 2014

No multiple terms for b=1.
This sequence strictly follows requirements of the Beal conjecture.
Less than 550 of these powers satisfy 196 Beal's conjecture equations.

Anatoly E. Voevudko, Table of n, a(n) for n = 1..11539
American Mathematical Society, Beal Prize
Alf van der Poorten, Remarks on the sequence of 'perfect' numbers
Anatoly E. Voevudko, Description of all powers in b245713
Eric W. Weisstein, World of Mathematics, Perfect Power
Wikipedia, Beal's conjecture

Cf. A001597, A023057, A072103, A076467.

N:= 10^5: # to get all terms <= N
L:= [1, seq(seq(b^p, p=3..floor(log[b](N))),b=2..floor(N^(1/3)))]:
sort(L); # Robert Israel, Nov 09 2015

mx = 10000; Join[{1}, Sort@ Flatten@ Table[b^p, {b, 2, Sqrt@ mx}, {p, 3, Log[b, mx]}]] (* Robert G. Wilson v, Nov 09 2015 *)

A245713(lim)={my(L=List(1),lim2=logint(lim,2));for(p=3,lim2, for(b=2,sqrtnint(lim,p),listput(L, b^p);));listsort(L); print(L);} \\ Anatoly E. Voevudko, Sep 21 2015

A189117 Conjectured number of pairs of consecutive perfect powers (A001597) differing by n.

Original entry on oeis.org

1, 1, 2, 3, 1, 0, 2, 1, 3, 1, 2, 1, 3, 0, 2, 1, 5, 2, 3, 1, 1, 0, 1, 2, 1, 2, 1, 3, 0, 1, 0, 1, 1, 0, 2, 1, 1, 1, 3, 1, 1, 0, 1, 0, 1, 0, 3, 1, 2, 0, 1, 0, 2, 0, 2, 1, 1, 0, 1, 2, 1, 0, 1, 0, 3, 0, 2, 2, 1, 0, 2, 0, 2, 1, 1, 1, 1, 0, 3, 1, 1, 0, 1, 0, 1, 0, 1, 0, 3, 0, 0, 1, 1, 1, 2, 0, 2, 0, 1, 5

Offset: 1

T. D. Noe, Apr 16 2011

nonn

Only a(1) is proved. Perfect powers examined up to 10^21. This is similar to A076427, but more restrictive.
Hence, through 10^21, there is only one value in the sequence: Semiprimes which are both one more than a perfect power and one less than another perfect power. This is to perfect powers A001597 approximately as A108278 is to squares. A more exact analogy would be to the set of integers such as 30^2 = 900 since 900-1 = 899 = 29 * 31, and 900+1 = 901 = 17 * 53. A189045 INTERSECTION A189047. a(1) = 26 because 26 = 2 * 13 is semiprime, 26-1 = 25 = 5^2, and 26+1 = 27 = 3^3. - Jonathan Vos Post, Apr 16 2011
Pillai's conjecture is that a(n) is finite for all n. - Charles R Greathouse IV, Apr 30 2012

			1 = 3^2 - 2^3;
2 = 3^3 - 2^5;
3 = 2^2 - 1^2 = 2^7 - 5^3;
4 = 2^3 - 2^2 = 6^2 - 2^5 = 5^3 - 11^2.

Cf. A023056 (least k such that k and k+n are consecutive perfect powers).

Cf. A023057 (conjectured n such that a(n)=0).

nn = 10^12; pp = Join[{1}, Union[Flatten[Table[n^i, {i, 2, Log[2, nn]}, {n, 2, nn^(1/i)}]]]]; d = Select[Differences[pp], # <= 100 &]; Table[Count[d, n], {n, 100}]

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

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A023055 (Apparently) differences between adjacent perfect powers (integers of form a^b, a >= 1, b >= 2).

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A074981 Conjectured list of positive numbers which are not of the form r^i - s^j, where r,s,i,j are integers with r>0, s>0, i>1, j>1.

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A245713 Sorted imperfect powers b^p with b > 0, p > 2, with multiplicity.

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A189117 Conjectured number of pairs of consecutive perfect powers (A001597) differing by n.

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A074980 Numbers which are not of the form m^p - n^q where p = 2 or 3, q = 2 or 3.

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A066510 Conjectured list of positive numbers which are not of the form r^i-s^j, where r,s,i,j are integers with i>1, j>1.

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A077286 Primes which are not the difference between two successive perfect powers (A001597).

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