A072103 -id:A072103 - cp's OEIS frontend

A365930 a(n) = number of pairs {x,y} with (x,y > 1) such that x^y (= terms of A072103) has bit length n.

0, 0, 1, 2, 4, 3, 8, 7, 10, 15, 19, 25, 38, 46, 66, 90, 126, 169, 240, 332, 467, 646, 909, 1270, 1787, 2513, 3529, 4966, 6998, 9853, 13897, 19594, 27644, 39011, 55064, 77741, 109790, 155062, 219049, 309472, 437278, 617928, 873288, 1234268, 1744597, 2466067, 3486093

Offset: 1

Karl-Heinz Hofmann, Sep 23 2023

Number of pairs {x,y} with (x,y > 1) for which applies: 2^(n-1) <= x^y < 2^n-1.
In some special cases different pairs have the same result (see A072103 and the example here) and those multiple representations are counted separately.
There is no need to include 2^n-1 because it is a Mersenne number and it cannot be a power anyway.
Limit_{n->oo} a(n)/a(n-1) = sqrt(2) = A002193.
Terms of A365931 are the partial sums of this sequence.

			For n = 5 the smallest number with bit length 5 is 16 (= 10000 in binary), and the largest number with bit length 5 is 31 (= 11111 in binary). In this range 4 pairs can be found, namely: 2^4 = 16; 4^2 = 16; 5^2 = 25; 3^3 = 27.

Karl-Heinz Hofmann, Table of n, a(n) for n = 1..2000

Cf. A072103, A365931 (partial sums).

Cf. A365932, A126726, A002193.

a[n_] := Sum[Ceiling[2^(n/k)] - Ceiling[2^((n-1)/k)], {k, 2, n}]; Array[a, 50] (* Amiram Eldar, Sep 23 2023 *)

for (blen = 0, 25, my (b1=2^blen, b2=2*b1-1, np=0); for (x = b1, b2, my (m=ispower(x)); if (m>1, np+=(sumdiv(m,y,1)-1), np+=m)); print1 (np,", ")) \\ Hugo Pfoertner, Oct 02 2023

from sympy import integer_nthroot
def A365930(n):
    return sum(integer_nthroot((2**n)-1,y)[0]-integer_nthroot(2**(n-1)-1,y)[0] for y in range(2, n+1))

from sympy import integer_nthroot, integer_log
def A365930(n): # a bit more efficient program
    c, y, a, b = 0, 2, (1<Chai Wah Wu, Oct 16 2023

a(n) = Sum_{y=2..n} (ceiling(2^(n/y)) - ceiling(2^((n-1)/y))).
a(n) = Sum_{y=2..n} (floor((2^n-1)^(1/y)) - floor((2^(n-1)-1)^(1/y))).
a(n) = A365931(n) - A365931(n-1).

A365931 a(n) = number of pairs {x,y} with (x,y > 1) such that x^y (= terms of A072103) has bit length <= n.

Original entry on oeis.org

0, 0, 1, 3, 7, 10, 18, 25, 35, 50, 69, 94, 132, 178, 244, 334, 460, 629, 869, 1201, 1668, 2314, 3223, 4493, 6280, 8793, 12322, 17288, 24286, 34139, 48036, 67630, 95274, 134285, 189349, 267090, 376880, 531942, 750991, 1060463, 1497741, 2115669, 2988957, 4223225, 5967822, 8433889

Offset: 1

Karl-Heinz Hofmann, Oct 07 2023

Number of pairs {x,y} with (x,y > 1) for which x^y < 2^n-1.
In some special cases different pairs have the same result (see A072103 and the example here) and those multiple representations are counted separately.
There is no need to include 2^n-1 because it is a Mersenne number and it cannot be a power anyway.
Limit_{n->oo} a(n)/a(n-1) = sqrt(2) = A002193.
Partial sums of A365930.

			For n = 6: the Mersenne number 2^6-1 = 63 is the largest number with bit length 6 and the upper bound for the following a(6) = 10 powers: 2^2, 2^3, 2^4, 2^5, 3^2, 3^3, 4^2, 5^2, 6^2, 7^2.

Cf. A072103, A002193, A365930 (first differences).

Cf. A017912 (squares), A017981 (cubes).

a[n_] := Sum[Ceiling[2^(n/k)] - 2, {k, 2, n}]; Array[a, 47]

from sympy import integer_nthroot, integer_log
def A365931(n):
    result, nMersenne, new = 0, (1<

a(n) = Sum_{y = 2..n} (ceiling(2^(n/y)) - 2)
a(n) = Sum_{y = 2..n} (floor((2^n-1)^(1/y)) - 1)
a(n) = Sum_{k = 1..n} A365930(k).

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

Original entry on oeis.org

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

A253641 Largest integer b such that n=a^b for some integer a; a(0)=a(1)=1 by convention.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 3, 1, 1, 1, 1, 5, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 0

M. F. Hasler, Jan 25 2015

A000005(a(n))-1 yields the number of times n is listed in A072103, i.e., the number of ways it can be written differently as perfect power.
For any n, a(n) >= 1 since n = n^1.
Integers n = 0 and n = 1 can be written as n^b with arbitrarily large b; to remain consistent with the preceding formula and the comment, the conventional choice a(n) = 1 seemed the best one.
The same as A052409 if the convention is dropped. - R. J. Mathar, Jan 29 2015

			a(4) = 2 since 4 = 2^2. a(64) = 6 since 64 = 2^6 (although also 64 = 4^3 = 8^2).

Antti Karttunen, Table of n, a(n) for n = 0..10000

Cf. A001597, A072103, A175064, A253642, A052409.

a:=proc(n) if n in {0, 1} then 1 else igcd(map(i->i[2], ifactors(n)[2])[]); fi; end: seq(a(n), n=0..120); # Ridouane Oudra, Jun 10 2025

PARI
```
A253641(n)=max(ispower(n),1)
```

from math import gcd
from sympy import factorint
def A253641(n): return gcd(*factorint(n).values()) if n>1 else 1 # Chai Wah Wu, Aug 13 2024

A253642 Number of ways the perfect power A001597(n) can be written as a^b, with a, b > 1.

Original entry on oeis.org

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

Offset: 1

M. F. Hasler, Jan 25 2015

nonn

Run lengths of A072103. Also, the terms a(n) which exceed 1 constitute A175066. - Andrey Zabolotskiy, Aug 17 2016

			a(1)=0 since A001597(1)=1 can be written as a^b for a=1 and any b, but not using a base a > 1.
a(2)=a(3)=a(4)=1 since the following terms 4=2^2, 8=2^3 and 9=3^2 can be written as perfect powers in only one way.
a(5)=2 since A001597(5)=16=a^b for (a,b)=(2,4) and (4,2).

Cf. A001597, A072103, A175064, A253641.

for(n=1,9999,(e=ispower(n))&&print1(numdiv(e)-1,","))

from math import gcd
from sympy import mobius, integer_nthroot, divisor_count, factorint
def A253642(n):
    if n == 1: return 0
    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 divisor_count(gcd(*factorint(kmax).values()))-1 # Chai Wah Wu, Aug 13 2024

a(n) = A000005(A253641(A001597(n))) - 1.

a(n) = A175064(n) - 1.

A089580 Total number of perfect powers > 1 below 10^n, counting multiple representations separately.

Original entry on oeis.org

3, 15, 49, 143, 406, 1174, 3507, 10674, 32965, 102716, 321797, 1011533, 3186389, 10050743, 31730134, 100228040, 316713623, 1001037546, 3164497349, 10004755374, 31632975598, 100021893194, 316274794666, 1000101078148, 3162495003352, 10000467510247, 31623782520064, 100002164895587

Offset: 1

Martin Renner, Dec 29 2003

nonn

From Robert G. Wilson v, Jul 17 2016: (Start)
a(n) ~ sqrt(10^n).
a(n) - A089579(n) = A275358(n).
The four terms which make up the difference a(2) - A089579(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)
This sequence correlates (see Link) to A006880 via a power fit A*x^B. For example, using a(23) through a(29) one obtains (A,B) = (0.047272, 1.96592) with R^2 > 0.999999. This extrapolates A006880(30) as 1.46*10^28. The exponent well may be resolving to 2. - Bill McEachen, Mar 04 2025

			16 = 2^4 = 4^2 counts double, 256 = 2^8 = 4^4 = 16^2 counts three times.

Chai Wah Wu, Table of n, a(n) for n = 1..1998 (n = 1..100 from Robert G. Wilson, n = 101..400 from Karl-Heinz Hofmann)
Karl-Heinz Hofmann, Python program.
Bill McEachen, Plot A089580 vs A006880.

Cf. A001597, A072103, A117453, A275358, A060298.

Cf. A089579 (counting multiple representations only once).

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

Python
```
# see link.
```

a(n) = Sum_{k = 1..n} A060298(k). - Karl-Heinz Hofmann, Sep 18 2023

2 more terms from Martin Renner, Oct 02 2004
More terms from T. D. Noe, Nov 16 2006
More precise name by Hugo Pfoertner, Sep 16 2023

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

A259362 a(1) = 1, for n > 1: a(n) is the number of ways to write n as a nontrivial perfect power.

Original entry on oeis.org

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

Offset: 1

Doug Bell, Jun 24 2015

nonn

a(n) = number of integer pairs (i,j) for distinct values of i where i > 0, j > 1 and n = i^j. Since 1 = 1^r for all real values of r, the requirement for a distinct i causes a(1) = 1 instead of a(1) = infinity.
Alternatively, the sequence can be defined as: a(1) = 1, for n > 1: a(n) = number of pairs (i,j) such that i > 0, j > 1 and n = i^j.
A007916 = n, where a(n) = 0.
A001597 = n, where a(n) > 0.
A175082 = n, where n = 1 or a(n) = 0.
A117453 = n, where n = 1 or a(n) > 1.
A175065 = n, where n > 1 and a(n) > 0 and this is the first occurrence in this sequence of a(n).
A072103 = n repeated a(n) times where n > 1.
A075802 = min(1, a(n)).
A175066 = a(n), where n = 1 or a(n) > 1. This sequence is an expansion of A175066.
A253642 = 0 followed by a(n), where n > 1 and a(n) > 0.
A175064 = a(1) followed by a(n) + 1, where n > 1 and a(n) > 0.
Where n > 1, A001597(x) = n (which implies a(n) > 0), i = A025478(x) and j = A253641(n), then a(n) = A000005(j) - 1, which is the number of factors of j greater than 1. The integer pair (i,j) comprises the smallest value i and the largest value j where i > 0, j > 1 and n = i^j. The a(n) pairs of (a,b) where a > 0, b > 1 and n = a^b are formed with b = each of the a(n) factors of j greater than 1. Examples for n = {8,4096}:
  a(8) = 1, A001597(3) = 8, A025478(3) = 2, A253641(8) = 3, 8 = 2^3 and A000005(3) - 1 = 1 because there is one factor of 3 greater than 1 [3]. The set of pairs (a,b) is {(2,3)}.
  a(4096) = 5, A001597(82) = 4096, A025478(82) = 2, A253641(4096) = 12, 4096 = 2^12 and A000005(12) - 1 = 5 because there are five factors of 12 greater than 1 [2,3,4,6,12]. The set of pairs (a,b) is {(64,2),(16,3),(8,4),(4,6),(2,12)}.
A023055 = the ordered list of x+1 with duplicates removed, where x is the number of consecutive zeros appearing in this sequence between any two nonzero terms.
A070428(x) = number of terms a(n) > 0 where n <= 10^x.
a(n) <= A188585(n).

			a(6) = 0 because there is no way to write 6 as a nontrivial perfect power.
a(9) = 1 because there is one way to write 9 as a nontrivial perfect power: 3^2.
a(16) = 2 because there are two ways to write 16 as a nontrivial perfect power: 2^4, 4^2.
From _Friedjof Tellkamp_, Jun 14 2025: (Start)
n:       1, 2, 3, 4, 5, 6, 7, 8, 9, ...
Squares: 1, 0, 0, 1, 0, 0, 0, 0, 1, ... (A010052)
Cubes:   1, 0, 0, 0, 0, 0, 0, 1, 0, ... (A010057)
...
Sum:    oo, 0, 0, 1, 0, 0, 0, 1, 1, ...
a(1)=1:  1, 0, 0, 1, 0, 0, 0, 1, 1, ... (= this sequence). (End)

Doug Bell, Table of n, a(n) for n = 1..5000

Cf. A001597, A007916, A023055, A025478, A070428, A072103, A075090, A075109, A075802, A117453, A175064, A175066, A175082, A188585, A253641, A253642.

Cf. A010052, A010057, A374016, A089361, A089723, A382691 (alternating).

a[n_] := If[n == 1, 1, Sum[Boole[IntegerQ[n^(1/k)]], {k, 2, Floor[Log[2, n]]}]]; Array[a, 100] (* Friedjof Tellkamp, Jun 14 2025 *)
a[n_] := If[n == 1, 1, DivisorSigma[0, Apply[GCD, Transpose[FactorInteger[n]][[2]]]] - 1]; Array[a, 100] (* Michael Shamos, Jul 06 2025 *)

a(n) = if (n==1, 1, sum(i=2, logint(n, 2), ispower(n, i))); \\ Michel Marcus, Apr 11 2025

a(1) = 1, for n > 1: a(n) = A000005(A253641(n)) - 1.
If n not in A001597, then a(n) = 0, otherwise a(n) = A175064(x) - 1 where A001597(x) = n.
From Friedjof Tellkamp, Jun 14 2025: (Start)
a(n) = A089723(n) - 1, for n > 1.
a(n) = A010052(n) + A010057(n) + A374016(n) + (...), for n > 1.
Sum_{k>=2..n} a(k) = A089361(n), for n > 1.
G.f.: x + Sum_{j>=2, k>=2} x^(j^k).
Dirichlet g.f.: 1 + Sum_{k>=2} zeta(k*s)-1. (End)

A362424 Number of partitions of n into 2 distinct perfect powers (A001597).

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 2, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 2, 1, 1, 2, 1, 0, 0, 2, 2, 0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 2, 1, 0, 0, 0, 2, 1, 1, 0, 1, 0, 1, 0, 2, 0, 0, 2, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 2, 0, 0, 0, 2, 1, 1

Offset: 0

Ilya Gutkovskiy, Apr 19 2023

nonn

Karl-Heinz Hofmann, Table of n, a(n) for n = 0..10000
Eric Weisstein's World of Mathematics, Perfect Power.

Cf. A001597, A072103, A362425.

perfectPowerQ[n_] := n == 1 || GCD @@ FactorInteger[n][[;; , 2]] > 1; a[n_] := Count[IntegerPartitions[n, {2}], ?(AllTrue[#, perfectPowerQ] && UnsameQ @@ # &)]; Array[a, 100, 0] (* _Amiram Eldar, May 05 2023 *)

import numpy
from math import isqrt
A072103 = []
for m in range(2,isqrt(10001)+1):
    k = 2
    while m**k < 10001:
        A072103.append(m**k)
        k += 1
A001597 = sorted(set(A072103)) # eliminates multiples and sorting
A362424 = numpy.zeros(10001+1, dtype="i4")
A001597 = [1] + A001597 # we need a "1" in front of A001597
a = 0
while A001597[a] < 10001 // 2:
    b = a + 1
    while b < len(A001597) and A001597[a] + A001597[b] < 10001:
        A362424[A001597[a] + A001597[b]] += 1
        b += 1
    a += 1
print(list(A362424[0:92])) # Karl-Heinz Hofmann, Sep 16 2023

A337670 Numbers that can be expressed as both Sum x^y and Sum y^x where the x^y are not equal to y^x for any (x,y) pair and all (x,y) pairs are distinct.

Original entry on oeis.org

432, 592, 1017, 1040, 1150, 1358, 1388, 1418, 1422, 1464, 1554, 1612, 1632, 1713, 1763, 1873, 1889, 1966, 1968, 1973, 1990, 2091, 2114, 2190, 2291, 2320, 2364, 2451, 2589, 2591, 2612, 2689, 2697, 2719, 2753, 2775, 2803, 2813, 2883, 3087, 3127, 3141, 3146

Offset: 1

Matej Veselovac, Sep 15 2020

nonn

Numbers m of form m = Sum_{i=1...k} b_i^e_i = Sum_{i=1...k} e_i^b_i such that b_i^e_i != e_i^b_i, b_i > 1, e_i > 1, k = |{{b_i, e_i}, i = 1, 2, ...}|, k > 1.
Terms of the sequence relate to the Diophantine equation Sum_{i=1...k} x_i = 0, k > 1, x_i != 0, where x_i = (b_i^e_i - e_i^b_i) such that b_i > 1, e_i > 1 and (i != j) => ({b_i, e_i} != {b_j, e_j}). That is, we are observing linear combinations of elements from {(r^n - n^r) : n,r > 1} \ {0}, under given conditions.
For sums with k = 20 terms, one infinite family of examples is known: "2^(2t) + t^(4) + 2^(2t+8) + (t+4)^(4) + 2^(2t+16) + (t+8)^(4) + 2^(2t+32) + (t+16)^(4) + 2^(2t+34) + (t+17)^(4) + 4^(t+1) + (2t+2)^(2) + 4^(t+2) + (2t+4)^(2) + 4^(t+10) + (2t+20)^(2) + 4^(t+14) + (2t+28)^(2) + 4^(t+18) + (2t+36)^(2)" is a term of the sequence, for every t > 4.

			17 = 2^3 + 3^2 = 3^2 + 2^3 is not in the sequence because {2,3} = {3,2} are not distinct.
25 = 3^3 + 2^4 = 3^3 + 4^2 is not in the sequence because 3^3 = 3^3 and 2^4 = 4^2 are commutative.
The smallest term of the sequence is:
  a(1) = 432 = 3^2 + 5^2 + 2^6 + 3^4 + 5^3 + 2^7
             = 2^3 + 2^5 + 6^2 + 4^3 + 3^5 + 7^2.
The smallest term that has more than one representation is:
  a(11) = 1554 = 3^2 + 7^2 + 6^3 + 2^8 + 4^5
               = 2^3 + 2^7 + 3^6 + 8^2 + 5^4,
  a(11) = 1554 = 3^2 + 5^2 + 2^6 + 10^2 + 2^7 + 3^5 + 2^8 + 3^6
               = 2^3 + 2^5 + 6^2 + 2^10 + 7^2 + 5^3 + 8^2 + 6^3.
Smallest terms with k = 5, 6, 7, 8, 9, 10 summands are:
  a(9)  = 1422 = 5^2 + 7^2 + 9^2 + 3^5 + 4^5
               = 2^5 + 2^7 + 2^9 + 5^3 + 5^4,
  a(1)  = 432  = 3^2 + 5^2 + 2^6 + 3^4 + 5^3 + 2^7
               = 2^3 + 2^5 + 6^2 + 4^3 + 3^5 + 7^2,
  a(2)  = 592  = 3^2 + 5^2 + 7^2 + 4^3 + 2^6 + 5^3 + 2^8
               = 2^3 + 2^5 + 2^7 + 3^4 + 6^2 + 3^5 + 8^2,
  a(11) = 1554 = 3^2 + 5^2 + 2^6 + 10^2 + 2^7 + 3^5 + 2^8 + 3^6
               = 2^3 + 2^5 + 6^2 + 2^10 + 7^2 + 5^3 + 8^2 + 6^3,
  a(14) = 1713 = 3^2 + 2^5 + 6^2 + 8^2 + 4^3 + 2^7 + 3^5 + 2^9 + 5^4
               = 2^3 + 5^2 + 2^6 + 2^8 + 3^4 + 7^2 + 5^3 + 9^2 + 4^5,
  a(28) = 2451 = 3^2 + 5^2 + 6^2 + 8^2 + 3^4 + 2^7 + 6^3 + 3^5 + 5^4 + 2^10
               = 2^3 + 2^5 + 2^6 + 2^8 + 4^3 + 7^2 + 3^6 + 5^3 + 4^5 + 10^2.

Math StackExchange, Base-Exponent Invariants, 2020.
Matej Veselovac, PYTHON program for A337670
Eric Weisstein's World of Mathematics, Perfect Power.

Cf. A337671 (subsequence for k <= 5).
Cf. A005188 (perfect digital invariants).
Cf. Perfect powers: A001597, A072103.
Cf. Commutative powers: A271936.
Cf. Nonnegative numbers of the form (r^n - n^r), for n,r > 1: A045575.
Cf. Numbers of the form (r^n - n^r): A024012 (r = 2), A024026 (r = 3), A024040 (r = 4), A024054 (r = 5), A024068 (r = 6), A024082 (r = 7), A024096 (r = 8), A024110 (r = 9), A024124 (r = 10), A024138 (r = 11), A024152 (r = 12).

cp's OEIS Frontend

A365930 a(n) = number of pairs {x,y} with (x,y > 1) such that x^y (= terms of A072103) has bit length n.

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A365931 a(n) = number of pairs {x,y} with (x,y > 1) such that x^y (= terms of A072103) has bit length <= n.

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

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A253641 Largest integer b such that n=a^b for some integer a; a(0)=a(1)=1 by convention.

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A253642 Number of ways the perfect power A001597(n) can be written as a^b, with a, b > 1.

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A089580 Total number of perfect powers > 1 below 10^n, counting multiple representations separately.

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