A056624 Number of unitary square divisors of n.
1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 2, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 2, 2, 4, 1, 1, 1, 1, 1
Offset: 1
Examples
n=256, it has 5 square divisors of which only 2,{1,256} are unitary, 3 divisors are not. n=124 has 2 (1 and 4) square divisors, both of them unitary a(124) = 2. n=108 has 12 divisors, 4 square divisors: {1,4,9,36} of which 1 and 4 are unitary, 9 and 36 are not. So a(108)=2. The largest unitary square divisor of 108 is 4 with 1 prime divisor so a(108) = 2^1 = 2.
Links
Crossrefs
Programs
-
Maple
isA056624 := (n, d) -> igcd(n, d) = d and igcd(n/d, d) = d and igcd(n/d^2, d) = 1: a := n -> nops(select(k -> isA056624(n, k), [seq(1..n)])): # Peter Luschny, Jun 13 2025
-
Mathematica
Table[DivisorSum[n, 1 &, And[IntegerQ@ Sqrt@ #, CoprimeQ[#, n/#]] &], {n, 105}] (* Michael De Vlieger, Jul 28 2017 *) f[p_, e_] := 2^(1 - Mod[e, 2]); a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Jan 03 2022 *)
-
PARI
a(n) = sumdiv(n, d, if(gcd(d, n/d)==1, issquare(d))); \\ Michel Marcus, Jul 28 2017
-
Python
from sympy import factorint def A056624(n): return 1<
Chai Wah Wu, Aug 03 2024 -
Python
def is_A056624(n, d): return gcd(n, d) == d and gcd(n//d, d) == d and gcd(n//(d*d), d) == 1 def a(n): return len([k for k in range(1, n+1) if is_A056624(n, k)]) print([a(n) for n in range(1, 106)]) # Peter Luschny, Jun 13 2025
-
Scheme
(define (A056624 n) (if (= 1 n) n (* (A000079 (A059841 (A067029 n))) (A056624 (A028234 n))))) ;; Antti Karttunen, Jul 28 2017
Formula
a(n) = 2^r, where r is the number of prime factors of the largest unitary square divisor of n.
Multiplicative with a(p^e) = 2^(1-(e mod 2)). - Vladeta Jovovic, Dec 13 2002
Dirichlet g.f.: zeta(s)*zeta(2*s)/zeta(3*s). - Werner Schulte, Apr 03 2018
Sum_{k=1..n} a(k) ~ n*Pi^2/(6*zeta(3)) + sqrt(n)*zeta(1/2)/zeta(3/2). - Vaclav Kotesovec, Feb 07 2019
a(n) = 2^A162641(n). - Amiram Eldar, Sep 26 2022
Extensions
More terms from Vladeta Jovovic, Dec 13 2002
Comments