A175065 -id:A175065 - cp's OEIS frontend

A175064 a(1) = 1; for n >= 2, a(n) = number of ways h to write the n-th perfect power A001597(n) as m^k with m >= 2 and k >= 1.

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

Offset: 1

Jaroslav Krizek, Jan 23 2010

nonn

Perfect powers with first occurrence of h >= 2: 4, 16, 64, 65536, 4096, ... [The perfect power corresponding to h is A175065(h) = 2^A005179(h). - Jianing Song, Oct 27 2024]

			For n = 11: A001597(11) = 64; there are 4 ways to write 64 as m^k: 64^1 = 8^2 = 4^3 = 2^6.

Cf. A253641, A253642, A000005, A001597.

from math import gcd
from sympy import mobius, integer_nthroot, divisor_count, factorint
def A175064(n):
    if n == 1: return 1
    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())) # Chai Wah Wu, Aug 13 2024

a(n) = A000005(A253641(A001597(n))) = A253642(n)+1. - M. F. Hasler, Jan 25 2015

Extended by T. D. Noe, Apr 21 2011

Definition clarified by Jonathan Sondow, Nov 30 2012

A175066 a(1) = 1, for n >= 2: a(n) = number of ways h to write perfect powers A117453(n) as m^k (m >= 2, k >= 2).

Original entry on oeis.org

1, 2, 3, 2, 3, 2, 2, 3, 3, 2, 2, 5, 3, 2, 2, 3, 3, 2, 2, 2, 3, 2, 3, 2, 3, 4, 2, 2, 3, 2, 2, 2, 2, 5, 2, 2, 3, 2, 5, 2, 2, 2, 2, 3, 5, 2, 2, 2, 3, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 2, 3, 2, 3, 3, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 7, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 3, 2

Offset: 1

Jaroslav Krizek, Jan 23 2010

nonn

Perfect powers with first occurrence of h >= 2: 16, 64, 65536, 4096, ... (A175065)
a(n) for n>1 is the subsequence of A253642 formed by the terms which exceed 1; equivalently, a(n)+1 for n>1 is the subsequence of A175064 formed by the terms which exceed 2. Also, sum of a(n)-1 over such n that A117453(n)<10^m gives A275358(m). - Andrey Zabolotskiy, Aug 16 2016
Numbers n such that a(n) is nonprime are 1, 26, 110, ... - Altug Alkan, Aug 22 2016

			For n = 12, A117453(12) = 4096 and a(12)=5 since there are 5 ways to write 4096 as m^k: 64^2 = 16^3 = 8^4 = 4^6 = 2^12.
729=27^2=9^3=3^6 and 1024=32^2=4^5=2^10 yield a(8)=a(9)=3. - _R. J. Mathar_, Jan 24 2010

Cf. A117453.

lista(nn) = {print1(1, ", "); for (i=2, nn, if (po = ispower(i), np = sum(j=2, po, ispower(i, j)); if (np>1, print1(np, ", "));););} \\ Michel Marcus, Mar 20 2013

from math import gcd
from sympy import mobius, integer_nthroot, factorint, divisor_count, primerange
def A175066(n):
    if n == 1: return 1
    def bisection(f,kmin=0,kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        while kmax-kmin > 1:
            kmid = kmax+kmin>>1
            if f(kmid) <= kmid:
                kmax = kmid
            else:
                kmin = kmid
        return kmax
    def f(x): return int(n+sum(mobius(k)*(integer_nthroot(x,k)[0]-1+sum(integer_nthroot(x,p*k)[0]-1 for p in primerange((x//k).bit_length()))) for k in range(1,x.bit_length())))
    return divisor_count(gcd(*factorint(bisection(f,n,n)).values()))-1 # Chai Wah Wu, Nov 24 2024

If A117453(n) = m^k with k maximal, then a(n) = tau(k) - 1. - Charlie Neder, Mar 02 2019

Corrected and extended by R. J. Mathar, Jan 24 2010

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)

cp's OEIS Frontend

A175064 a(1) = 1; for n >= 2, a(n) = number of ways h to write the n-th perfect power A001597(n) as m^k with m >= 2 and k >= 1.

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A175066 a(1) = 1, for n >= 2: a(n) = number of ways h to write perfect powers A117453(n) as m^k (m >= 2, k >= 2).

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

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