A295664 Exponent of the highest power of 2 dividing number of divisors of n: a(n) = A007814(A000005(n)); 2-adic valuation of tau(n).

0, 1, 1, 0, 1, 2, 1, 2, 0, 2, 1, 1, 1, 2, 2, 0, 1, 1, 1, 1, 2, 2, 1, 3, 0, 2, 2, 1, 1, 3, 1, 1, 2, 2, 2, 0, 1, 2, 2, 3, 1, 3, 1, 1, 1, 2, 1, 1, 0, 1, 2, 1, 1, 3, 2, 3, 2, 2, 1, 2, 1, 2, 1, 0, 2, 3, 1, 1, 2, 3, 1, 2, 1, 2, 1, 1, 2, 3, 1, 1, 0, 2, 1, 2, 2, 2, 2, 3, 1, 2, 2, 1, 2, 2, 2, 2, 1, 1, 1, 0, 1, 3, 1, 3, 3, 2, 1, 2, 1, 3, 2, 1, 1, 3, 2, 1, 1, 2, 2, 4, 0

Offset: 1

Author

Antti Karttunen, Nov 28 2017

Keywords

• nonn
• easy

Comments

In the prime factorization of n = p1^e1 * ... pk^ek, add together the number of trailing 1-bits in each exponent e when they are written in binary.

Links

• Antti Karttunen, Table of n, a(n) for n = 1..65537
• Index entries for sequences computed from exponents in factorization of n

Crossrefs

Cf. A000005, A007814, A028234, A056169, A067029, A077761, A162642, A295663.

Cf. A000290 (positions of zeros).

Programs

Mathematica

Table[IntegerExponent[DivisorSigma[0, n], 2], {n, 120}] (* Michael De Vlieger, Nov 28 2017 *)

PARI

a(n) = valuation(numdiv(n), 2); \\ Michel Marcus, Nov 30 2017

Python

from sympy import divisor_count
def A295664(n): return (~(m:=int(divisor_count(n))) & m-1).bit_length() # Chai Wah Wu, Jul 05 2022

Formula

Additive with a(p^e) = A007814(1+e).

a(1) = 0; for n > 1, a(n) = A007814(1+A067029(n)) + a(A028234(n)).

a(n) = A007814(A000005(n)).

a(n) >= A162642(n) >= A056169(n).

Sum_{k=1..n} a(k) ~ n * (log(log(n)) + B + C), where B is Mertens's constant (A077761) and C = Sum_{p prime} f(1/p) =-0.223720656976344505701..., where f(x) = -x + (1-x) * Sum_{k>=1} x^(2^k-1)/(1-x^(2^k)). - Amiram Eldar, Sep 28 2023