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A061142
Replace each prime factor of n with 2: a(n) = 2^bigomega(n), where bigomega = A001222, number of prime factors counted with multiplicity.
49
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
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
KEYWORD
easy,nonn,mult
AUTHOR
Henry Bottomley, May 29 2001
STATUS
approved