A099244 Greatest common divisor of length of n in binary representation and its number of ones.

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

Offset

1,3

Comments

For k >= 2, n in the range [2^(k-1)..2^k - 2] have binary length k but fewer than k 1's, thus a(n) is a proper divisor of k, and if k is a prime then a(n) = 1. - Ctibor O. Zizka, Jun 19 2021

Links

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Index entries for sequences related to binary expansion of n.

Formula

a(n) = gcd(A070939(n), A000120(n)).

a(A000225(n)) = n and a(m) < n for m < A000225(n).

Mathematica

a[n_] := GCD[BitLength[n], DigitCount[n, 2, 1]]; Array[a, 100] (* Amiram Eldar, Jul 16 2023 *)

Prog

(Haskell)
a099244 n = gcd (a070939 n) (a000120 n)
-- Reinhard Zumkeller, Oct 10 2013
(Python)
from math import gcd
def a(n): b = bin(n)[2:]; return gcd(len(b), b.count('1'))
print([a(n) for n in range(1, 106)]) # Michael S. Branicky, Jun 17 2021

Crossrefs

Cf. A000120, A000225, A007088, A070939, A099245, A099246, A099247, A099248, A099249.

Sequence in context: A280688 A285728 A157925 * A260221 A014671 A029365

Adjacent sequences: A099241 A099242 A099243 * A099245 A099246 A099247

Keyword

base,easy,nonn

Author

Reinhard Zumkeller, Oct 08 2004

Status

approved