OFFSET

1,4

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

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).

FORMULA

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))).

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

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

EXAMPLE

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

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 *)

PROG

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

(Haskell)

a183063 = sum . map (1 -) . a247795_row

-- Reinhard Zumkeller, Sep 28 2014, Jan 15 2013, Jan 10 2012

(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

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

(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

CROSSREFS

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

Cf. A247795.

KEYWORD

nonn,easy

AUTHOR

Jaroslav Krizek, Dec 22 2010

EXTENSIONS

Formula corrected by Charles R Greathouse IV, Jul 29 2011

STATUS

approved