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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.
3
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
1,2
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
MATHEMATICA
Table[Floor[IntegerLength[n, 2]/Total@ IntegerDigits[n, 2]], {n, 120}] (* Michael De Vlieger, Dec 21 2015 *)
PROG
(Python)
for n in range(1, 88):
print(str((len(bin(n))-2) // bin(n).count('1')), end=', ')
(PARI) a(n) = #binary(n)\hammingweight(n); \\ Michel Marcus, Dec 19 2015
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Alex Ratushnyak, Dec 18 2015
STATUS
approved