A061142 Replace each prime factor of n with 2: a(n) = 2^bigomega(n), where bigomega = A001222, number of prime factors counted with multiplicity.

1, 2, 2, 4, 2, 4, 2, 8, 4, 4, 2, 8, 2, 4, 4, 16, 2, 8, 2, 8, 4, 4, 2, 16, 4, 4, 8, 8, 2, 8, 2, 32, 4, 4, 4, 16, 2, 4, 4, 16, 2, 8, 2, 8, 8, 4, 2, 32, 4, 8, 4, 8, 2, 16, 4, 16, 4, 4, 2, 16, 2, 4, 8, 64, 4, 8, 2, 8, 4, 8, 2, 32, 2, 4, 8, 8, 4, 8, 2, 32, 16, 4, 2, 16, 4, 4, 4, 16, 2, 16, 4, 8, 4, 4, 4

Offset

1,2

Comments

The inverse Möbius transform of A162510. - R. J. Mathar, Feb 09 2011

Links

R. Zumkeller, Table of n, a(n) for n = 1..10005

R. J. Mathar, Survey of Dirichlet Series of Multiplicative Arithmetic Functions, arXiv:1106.4038 [math.NT], 2011-2012. See eq. (2.12).

Index entries for sequences computed from exponents in factorization of n

Formula

a(n) = Sum_{d divides n} 2^(bigomega(d)-omega(d)) = Sum_{d divides n} 2^(A001222(d) - A001221(d)). - Benoit Cloitre, Apr 30 2002

a(n) = A000079(A001222(n)), i.e., a(n)=2^bigomega(n). - Emeric Deutsch, Feb 13 2005

Totally multiplicative with a(p) = 2. - Franklin T. Adams-Watters, Oct 04 2006

Dirichlet g.f.: Product_{p prime} 1/(1-2*p^(-s)). - Ralf Stephan, Mar 28 2015

a(n) = A001316(A156552(n)). - Antti Karttunen, May 29 2017

Dirichlet g.f.: zeta(s)^2 * Product_{p prime} 1/(1 - 1/(p^s - 1)^2). - Vaclav Kotesovec, Mar 14 2023

Example

a(100)=16 since 100=2*2*5*5 and so a(100)=2*2*2*2.

Maple

with(numtheory): seq(2^bigomega(n), n=1..95);

Mathematica

Table[2^PrimeOmega[n], {n, 1, 95}] (* Jean-François Alcover, Jun 08 2013 *)

Prog

(PARI) a(n)=direuler(p=1, n, 1/(1-2*X))[n] /* Ralf Stephan, Mar 28 2015 */
(PARI) a(n) = 2^bigomega(n); \\ Michel Marcus, Aug 08 2017

Crossrefs

Cf. A000079, A001222, A001316, A034444, A069205 (partial sums), A123667, A124508, A156552.

Sequence in context: A318316 A328721 A165872 * A318312 A318474 A326306

Adjacent sequences: A061139 A061140 A061141 * A061143 A061144 A061145

Keyword

easy,nonn,mult

Author

Henry Bottomley, May 29 2001

Status

approved