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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).
14
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
1,6
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.
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
MATHEMATICA
Table[IntegerExponent[DivisorSigma[0, n], 2], {n, 120}] (* Michael De Vlieger, Nov 28 2017 *)
PROG
(Scheme) ;; With memoization-macro definec.
(definec (A295664 n) (if (= 1 n) 0 (+ (A007814 (+ 1 (A067029 n))) (A295664 (A028234 n)))))
(define (A295664 n) (A007814 (A000005 n)))
(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
CROSSREFS
Cf. A000290 (positions of zeros).
Sequence in context: A071460 A369862 A218867 * A250213 A033794 A218856
KEYWORD
nonn,easy
AUTHOR
Antti Karttunen, Nov 28 2017
STATUS
approved