A056671 1 + the number of unitary and squarefree divisors of n = number of divisors of reduced squarefree part of n.

1, 2, 2, 1, 2, 4, 2, 1, 1, 4, 2, 2, 2, 4, 4, 1, 2, 2, 2, 2, 4, 4, 2, 2, 1, 4, 1, 2, 2, 8, 2, 1, 4, 4, 4, 1, 2, 4, 4, 2, 2, 8, 2, 2, 2, 4, 2, 2, 1, 2, 4, 2, 2, 2, 4, 2, 4, 4, 2, 4, 2, 4, 2, 1, 4, 8, 2, 2, 4, 8, 2, 1, 2, 4, 2, 2, 4, 8, 2, 2, 1, 4, 2, 4, 4, 4, 4, 2, 2, 4, 4, 2, 4, 4, 4, 2, 2, 2, 2, 1, 2, 8, 2, 2, 8

Offset

1,2

Comments

Note that 1 is regarded as free of squares of primes and is also a square number and a unitary divisor.

Links

Antti Karttunen, Table of n, a(n) for n = 1..10000

Steven R. Finch, Unitarism and Infinitarism, February 25, 2004. [Cached copy, with permission of the author]

Index entries for sequences computed from exponents in factorization of n

Formula

a(n) = A000005(A055231(n)) = A000005(A007913(n)/A055229(n)).

Multiplicative with a(p) = 2 and a(p^e) = 1 for e > 1. a(n) = 2^A056169(n). - Vladeta Jovovic, Nov 01 2001

a(n) = A034444(n) - A056674(n). - Antti Karttunen, Jul 19 2017

From Vaclav Kotesovec, Feb 11 2023: (Start)

Dirichlet g.f.: zeta(s) * Product_{primes p} (1 + 1/p^s - 1/p^(2*s)).

Dirichlet g.f.: zeta(s)^2 * Product_{primes p} (1 - 2/p^(2*s) + 1/p^(3*s)), (with a product that converges for s=1).

Let f(s) = Product_{primes p} (1 - 2/p^(2*s) + 1/p^(3*s)), then Sum_{k=1..n} a(k) ~ n * (f(1) * (log(n) + 2*gamma - 1) + f'(1)), where f(1) = Product_{primes p} (1 - 2/p^2 + 1/p^3) = A065464 = 0.42824950567709444021876..., f'(1) = f(1) * Sum_{primes p} (4*p-3) * log(p) / (p^3 - 2*p + 1) = 0.808661108949590913395... and gamma is the Euler-Mascheroni constant A001620. (End)

a(n) = Sum_{d|n, gcd(d,n/d)=1} mu(d)^2. - Wesley Ivan Hurt, May 25 2023

a(n) = Sum_{d|n} A343443(d)*mu(n/d). - Ridouane Oudra, Dec 18 2023

Example

n = 252 = 2*2*3*3*7 has 18 divisors, 8 unitary and 8 squarefree divisors of which 2 are unitary and squarefree, divisors {1,7};
n = 2520 = 2*2*2*3*3*5*7 has 48 divisors, 16 unitary and 16 squarefree divisors of which {1,5,7,35} are both, thus a(2520) = 4.
a(2520) = a(2^3*3^2*5*7) = a(2^3)*a(3^2)*a(5)*a(7) = 1*1*2*2 = 4.

Mathematica

Array[DivisorSigma[0, #] &@ Denominator[#/Apply[Times, FactorInteger[#][[All, 1]]]^2] &, 105] (* or *)
Table[DivisorSum[n, 1 &, And[SquareFreeQ@ #, CoprimeQ[#, n/#]] &], {n, 105}] (* Michael De Vlieger, Jul 19 2017 *)
f[p_, e_] := If[e==1, 2, 1]; a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, May 14 2019 *)

Prog

(PARI)
A057521(n) = { my(f=factor(n)); prod(i=1, #f~, if(f[i, 2]>1, f[i, 1]^f[i, 2], 1)); } \\ Charles R Greathouse IV, Aug 13 2013
A055231(n) = n/A057521(n);
A056671(n) = numdiv(A055231(n));
\\ Or:
A055229(n) = { my(c=core(n)); gcd(c, n/c); }; \\ This function from Charles R Greathouse IV, Nov 20 2012
A056671(n) = numdiv(core(n)/A055229(n)); \\ Antti Karttunen, Jul 19 2017
(PARI) for(n=1, 100, print1(direuler(p=2, n, (1 + X - X^2)/(1-X))[n], ", ")) \\ Vaclav Kotesovec, Feb 11 2023
(Scheme) (define (A056671 n) (if (= 1 n) n (* (if (= 1 (A067029 n)) 2 1) (A056671 (A028234 n))))) ;; (After the given multiplicative formula) - Antti Karttunen, Jul 19 2017
(Python)
from sympy import factorint, prod
def a(n): return 1 if n==1 else prod([2 if e==1 else 1 for p, e in factorint(n).items()])
print([a(n) for n in range(1, 51)]) # Indranil Ghosh, Jul 19 2017

Crossrefs

Cf. A000005, A007913, A034444, A055229, A055231, A056169, A056674.

Cf. A343443.

Sequence in context: A304649 A228441 A156260 * A354349 A278763 A278762

Adjacent sequences: A056668 A056669 A056670 * A056672 A056673 A056674

Keyword

mult,nonn

Author

Labos Elemer, Aug 10 2000

Status

approved