A265917 a(n) = floor(A070939(n)/A000120(n)) where A070939(n) is the binary length of n and A000120(n) is the binary weight of n.

1, 2, 1, 3, 1, 1, 1, 4, 2, 2, 1, 2, 1, 1, 1, 5, 2, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 6, 3, 3, 2, 3, 2, 2, 1, 3, 2, 2, 1, 2, 1, 1, 1, 3, 2, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 7, 3, 3, 2, 3, 2, 2, 1, 3, 2, 2, 1, 2, 1, 1, 1, 3, 2, 2, 1, 2, 1, 1

Offset: 1

Author

Alex Ratushnyak, Dec 18 2015

Keywords

• nonn
• base

Comments

1/a(n) gives a very rough approximation of the density of 1-bits in the binary representation (A007088) of n. This is 1 if more than half of the bits of n are 1. - Antti Karttunen, Dec 19 2015

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000120, A007088, A070939, A135941, A199238, A265918.

Programs

Mathematica

Table[Floor[IntegerLength[n, 2]/Total@ IntegerDigits[n, 2]], {n, 120}] (* Michael De Vlieger, Dec 21 2015 *)

PARI

a(n) = #binary(n)\hammingweight(n); \\ Michel Marcus, Dec 19 2015

Python

for n in range(1, 88):
    print(str((len(bin(n))-2) // bin(n).count('1')), end=',')