A267115 Bitwise-AND of the exponents of primes in the prime factorization of n, a(1) = 0.

0, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 0, 1, 1, 1, 4, 1, 0, 1, 0, 1, 1, 1, 1, 2, 1, 3, 0, 1, 1, 1, 5, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 2, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 6, 1, 1, 1, 0, 1, 1, 1, 2, 1, 1, 0, 0, 1, 1, 1, 0, 4, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1

Offset

1,4

Comments

The sums of the first 10^k terms, for k = 1, 2, ..., are 13, 105, 826, 7440, 71558, 707625, 7053959, 70473172, 704531711, 7044701307, 70445097231, ... . Apparently, the asymptotic mean of this sequence is limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 0.7044... . - Amiram Eldar, Sep 09 2022

Links

Antti Karttunen, Table of n, a(n) for n = 1..10000

Index entries for sequences computed from exponents in factorization of n

Formula

If A028234(n) = 1 [when n is a power of prime, in A000961], a(n) = A067029(n), otherwise a(n) = A067029(n) AND a(A028234(n)). [Here AND stands for bitwise-and, A004198.]

Example

For n = 24 = 2^3 * 3^1, bitwise-and of 3 and 1 ("11" and "01" in binary) gives 1, thus a(24) = 1.
For n = 210 = 2^1 * 3^1 * 5^1 * 7^1, bitwise-and of 1, 1, 1 and 1 gives 1, thus a(210) = 1.
For n = 720 = 2^4 * 3^2 * 5^1, bitwise-and of 4, 2 and 1 ("100", "10" and "1" in binary) gives zero, thus a(720) = 0.

Mathematica

{0}~Join~Table[BitAnd @@ Map[Last, FactorInteger@ n], {n, 2, 120}] (* Michael De Vlieger, Feb 07 2016 *)

Prog

(Scheme, two variants)
(define (A267115 n) (let loop ((n (A028234 n)) (z (A067029 n))) (cond ((= 1 n) z) (else (loop (A028234 n) (A004198bi z (A067029 n))))))) ;; A004198bi implements bitwise-and (see A004198).
;; A recursive version using memoizing definec-macro:
(definec (A267115 n) (if (= 1 (A028234 n)) (A067029 n) (A004198bi (A067029 n) (A267115 (A028234 n)))))
(PARI) a(n)=my(f = factor(n)[, 2]); if (#f == 0, return (0)); my(b = f[1]); for (k=2, #f, b = bitand(b, f[k]); ); b; \\ Michel Marcus, Feb 07 2016
(PARI) a(n)=if(n>1, fold(bitand, factor(n)[, 2]), 0) \\ Charles R Greathouse IV, Aug 04 2016
(Python)
from functools import reduce
from operator import and_
from sympy import factorint
def A267115(n): return reduce(and_, factorint(n).values()) if n > 1 else 0 # Chai Wah Wu, Aug 31 2022

Crossrefs

Cf. A000961, A004198, A028234, A067029, A267114, A267116, A268387.

Cf. A002035 (indices of odd numbers), A072587 (indices of even numbers that occur after a(1)).

Cf. A267117 (indices of zeros).

Sequence in context: A074761 A037861 A145037 * A328919 A277647 A296134

Adjacent sequences: A267112 A267113 A267114 * A267116 A267117 A267118

Keyword

nonn

Author

Antti Karttunen, Feb 03 2016

Status

approved