A212181 Largest odd divisor of tau(n): a(n) = A000265(A000005(n)).

1, 1, 1, 3, 1, 1, 1, 1, 3, 1, 1, 3, 1, 1, 1, 5, 1, 3, 1, 3, 1, 1, 1, 1, 3, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 3, 3, 1, 1, 5, 3, 3, 1, 3, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 3, 7, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 3, 3, 1, 1, 1, 5, 5, 1, 1, 3, 1, 1, 1, 1

Offset: 1

Author

Matthew Vandermast, Jun 04 2012

Keywords

• nonn
• mult

Comments

Completely determined by the exponents >=2 in the prime factorization of n (cf. A212172).

Not the same as the number of odd divisors of n (A001227(n)); see example.

Multiplicative because A000005 is multiplicative and A000265 is completely multiplicative. - Andrew Howroyd, Aug 01 2018

a(n) = 1 iff the number of divisors of n is a power of 2 (A036537). - Bernard Schott, Nov 04 2022

Examples

			48 has a total of 10 divisors (1, 2, 3, 4, 6, 8, 12, 16, 24 and 48). Since the largest odd divisor of 10 is 5, a(48) = 5.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..10000
• Index entries for sequences computed from exponents in factorization of n.

Crossrefs

Cf. A000005, A000079, A000265, A036537, A108951, A212172, A295664, A331286 (applied to primorial inflation of n).

Programs

Mathematica

Table[Block[{nd=DivisorSigma[0, n]}, nd/2^IntegerExponent[nd, 2]], {n, 100}] (* Indranil Ghosh, Jul 19 2017, after PARI code *)

PARI

a(n) = my(nd = numdiv(n)); nd/2^valuation(nd, 2); \\ Michel Marcus, Jul 19 2017

Python

from sympy import divisor_count, divisors
def a(n): return [i for i in divisors(divisor_count(n)) if i%2][-1]
print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Jul 19 2017

Formula

a(n) = A000265(A000005(n)).

From Antti Karttunen, Jan 14 2020: (Start)

a(n) = A000005(n) / A000079(A295664(n)).

a(A108951(n)) = A331286(n).

(End)

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Product_{p odd prime} ((1 - 1/p)*(1 + Sum_{k>=1} a(k+1)/p^k)) = 2.076325817863586... . - Amiram Eldar, Oct 15 2022