A183063 Number of even divisors of n.

0, 1, 0, 2, 0, 2, 0, 3, 0, 2, 0, 4, 0, 2, 0, 4, 0, 3, 0, 4, 0, 2, 0, 6, 0, 2, 0, 4, 0, 4, 0, 5, 0, 2, 0, 6, 0, 2, 0, 6, 0, 4, 0, 4, 0, 2, 0, 8, 0, 3, 0, 4, 0, 4, 0, 6, 0, 2, 0, 8, 0, 2, 0, 6, 0, 4, 0, 4, 0, 4, 0, 9, 0, 2, 0, 4, 0, 4, 0, 8, 0, 2, 0, 8, 0, 2

Offset: 1

Author

Jaroslav Krizek, Dec 22 2010

Keywords

• nonn
• easy

Comments

Number of divisors of n that are divisible by 2. More generally, it appears that the sequence formed by starting with an initial set of k-1 zeros followed by the members of A000005, with k-1 zeros between every one of them, can be defined as "the number of divisors of n that are divisible by k", (k >= 1). For example if k = 1 we have A000005 by definition, if k = 2 we have this sequence. Note that if k >= 3 the sequences are not included in the OEIS because the usual OEIS policy is not to include sequences like this where alternate terms are zero; this is an exception. - Omar E. Pol, Oct 18 2011

Number of zeros in n-th row of triangle A247795. - Reinhard Zumkeller, Sep 28 2014

a(n) is also the number of partitions of n into equal parts, minus the number of partitions of n into consecutive parts. - Omar E. Pol, May 04 2017

a(n) is also the number of partitions of n into an even number of equal parts. - Omar E. Pol, May 14 2017

Examples

			For n = 12, set of even divisors is {2, 4, 6, 12}, so a(12) = 4.
On the other hand, there are six partitions of 12 into equal parts: [12], [6, 6], [4, 4, 4], [3, 3, 3, 3], [2, 2, 2, 2, 2, 2] and [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]. And there are two partitions of 12 into consecutive parts: [12] and [5, 4, 3], so a(12) = 6 - 2 = 4, equaling the number of even divisors of 12. - _Omar E. Pol_, May 04 2017
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
• Mircea Merca, Combinatorial interpretations of a recent convolution for the number of divisors of a positive integer, Journal of Number Theory, Volume 160, March 2016, Pages 60-75, function tau_e(n).

Crossrefs

Cf. A000005, A000265, A001227, A007814, A000593, A183064, A136655, A125911, A320111.

Column 2 of A195050. - Omar E. Pol, Oct 19 2011

Cf. A001620, A027750, A083910, A244009.

Cf. A247795.

Programs

Haskell

a183063 = sum . map (1 -) . a247795_row
-- Reinhard Zumkeller, Sep 28 2014, Jan 15 2013, Jan 10 2012

Magma

[IsOdd(n) select 0 else #[d:d in Divisors(n)|IsEven(d)]:n in [1..100]]; // Marius A. Burtea, Dec 16 2019

Maple

A183063 := proc(n)
    if type(n,'even') then
        numtheory[tau](n/2) ;
    else
        0;
    end if;
end proc: # R. J. Mathar, Jun 18 2015

Mathematica

Table[Length[Select[Divisors[n], EvenQ]], {n, 90}] (* Alonso del Arte, Jan 10 2012 *)
a[n_] := (e = IntegerExponent[n, 2]) * DivisorSigma[0, n / 2^e]; Array[a, 100] (* Amiram Eldar, Jul 06 2022 *)

PARI

a(n)=if(n%2,0,numdiv(n/2)) \\ Charles R Greathouse IV, Jul 29 2011

Python

from sympy import divisor_count
def A183063(n): return divisor_count(n>>(m:=(~n&n-1).bit_length()))*m # Chai Wah Wu, Jul 16 2022

Sage

def A183063(n): return len([1 for d in divisors(n) if is_even(d)])
[A183063(n) for n in (1..80)]  # Peter Luschny, Feb 01 2012

Formula

a(n) = A000005(n) - A001227(n).

a(2n-1) = 0; a(2n) = A000005(n).

G.f.: Sum_{d>=1} x^(2*d)/(1 - x^(2*d)) and generally for the number of divisors that are divisible by k: Sum_{d>=1} x^(k*d)/(1 - x^(k*d)). - Geoffrey Critzer, Apr 15 2014

Dirichlet g.f.: zeta(s)^2/2^s and generally for the number of divisors that are divisible by k: zeta(s)^2/k^s. - Geoffrey Critzer, Mar 28 2015

From Ridouane Oudra, Sep 02 2019: (Start)

a(n) = Sum_{i=1..n} (floor(n/(2*i)) - floor((n-1)/(2*i))).

a(n) = 2*A000005(n) - A000005(2n). (End)

Conjecture: a(n) = lim_{x->n} f(Pi*x), where f(x) = sin(x)*Sum_{k>0} (cot(x/(2*k))/(2*k) - 1/x). - Velin Yanev, Dec 16 2019

a(n) = A000005(A000265(n))*A007814(n) - Chai Wah Wu, Jul 16 2022

Sum_{k=1..n} a(k) ~ n*log(n)/2 + (gamma - 1/2 - log(2)/2)*n, where gamma is Euler's constant (A001620). - Amiram Eldar, Nov 27 2022

Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k)/A000005(k) = 1-log(2) (A244009). - Amiram Eldar, Mar 01 2023

Extensions

Formula corrected by Charles R Greathouse IV, Jul 29 2011