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Largest odd divisor of tau(n): a(n) = A000265(A000005(n)).
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%I #44 Nov 04 2022 05:07:14

%S 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,

%T 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,

%U 1,1,1,3,1,1,3,3,1,1,1,5,5,1,1,3,1,1,1,1

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

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

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

%C Multiplicative because A000005 is multiplicative and A000265 is completely multiplicative. - _Andrew Howroyd_, Aug 01 2018

%C a(n) = 1 iff the number of divisors of n is a power of 2 (A036537). - _Bernard Schott_, Nov 04 2022

%H Antti Karttunen, <a href="/A212181/b212181.txt">Table of n, a(n) for n = 1..10000</a>

%H <a href="/index/Eu#epf">Index entries for sequences computed from exponents in factorization of n</a>.

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

%F From _Antti Karttunen_, Jan 14 2020: (Start)

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

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

%F (End)

%F 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

%e 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.

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

%o (PARI) a(n) = my(nd = numdiv(n)); nd/2^valuation(nd, 2); \\ _Michel Marcus_, Jul 19 2017

%o (Python)

%o from sympy import divisor_count, divisors

%o def a(n): return [i for i in divisors(divisor_count(n)) if i%2][-1]

%o print([a(n) for n in range(1, 101)]) # _Indranil Ghosh_, Jul 19 2017

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

%K nonn,mult

%O 1,4

%A _Matthew Vandermast_, Jun 04 2012