login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

n/[largest power of squarefree kernel] equals 1; perfect powers of sqf-kernels (or sqf-numbers).
19

%I #29 Aug 19 2024 13:16:52

%S 2,3,4,5,6,7,8,9,10,11,13,14,15,16,17,19,21,22,23,25,26,27,29,30,31,

%T 32,33,34,35,36,37,38,39,41,42,43,46,47,49,51,53,55,57,58,59,61,62,64,

%U 65,66,67,69,70,71,73,74,77,78,79,81,82,83,85,86,87,89,91,93,94,95,97,100

%N n/[largest power of squarefree kernel] equals 1; perfect powers of sqf-kernels (or sqf-numbers).

%C The sequence contains numbers m such that the exponents e are identical for all prime power factors p^e | m. It is clear from this alternate definition that m / K^E = 1 iff E is an integer. - _Michael De Vlieger_, Jun 24 2022

%H Michael De Vlieger, <a href="/A062770/b062770.txt">Table of n, a(n) for n = 1..10000</a>

%F A062760(a(n)) = 1, i.e., a(n)/(A007947(a(n))^A051904(a(n))) = a(n)/A062759(a(n)) = 1.

%F a(n) = A072774(n+1). - _Chai Wah Wu_, Aug 19 2024

%e Primes, squarefree numbers and perfect powers are here.

%e From _Michael De Vlieger_, Jun 24 2022 (Start):

%e 144 cannot be in the sequence, since the exponents of its prime power factors differ. The squarefree kernel of 144 = 2^4 * 3^2 is 2*3 = 6. The largest power of 6 less than 144 is 36. 144/36 = 4, so it is not in the sequence.

%e 216 is in the sequence because 216 = 2^3 * 3^3 is 2*3 = 6. But 216 = 6^3, hence 6^3 / 6^3 = 1. (End)

%t Select[Range[2, 2^16], Length@ Union@ FactorInteger[#][[All, -1]] == 1 &] _Michael De Vlieger_, Jun 24 2022

%o (PARI) is(n)=ispower(n,,&n); issquarefree(n) && n>1 \\ _Charles R Greathouse IV_, Sep 18 2015

%o (PARI) is(n)=#Set(factor(n)[,2])==1 \\ _Charles R Greathouse IV_, Sep 18 2015

%o (Python)

%o from math import isqrt

%o from sympy import mobius, integer_nthroot

%o def A062770(n):

%o def g(x): return int(sum(mobius(k)*(x//k**2) for k in range(1, isqrt(x)+1)))

%o def f(x): return n-2+x+(y:=x.bit_length())-sum(g(integer_nthroot(x,k)[0]) for k in range(1,y))

%o kmin, kmax = 1,2

%o while f(kmax) >= kmax:

%o kmax <<= 1

%o while True:

%o kmid = kmax+kmin>>1

%o if f(kmid) < kmid:

%o kmax = kmid

%o else:

%o kmin = kmid

%o if kmax-kmin <= 1:

%o break

%o return kmax # _Chai Wah Wu_, Aug 19 2024

%Y Cf. A062759, A062760, A007947, A003557, A051904, A005117, A001597, A072774.

%K nonn

%O 1,1

%A _Labos Elemer_, Jul 18 2001

%E Offset corrected by _Charles R Greathouse IV_, Sep 18 2015