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A056671
1 + the number of unitary and squarefree divisors of n = number of divisors of reduced squarefree part of n.
13
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
Steven R. Finch, Unitarism and Infinitarism, February 25, 2004. [Cached copy, with permission of the author]
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
KEYWORD
mult,nonn
AUTHOR
Labos Elemer, Aug 10 2000
STATUS
approved