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%I #23 May 05 2021 16:54:29
%S 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,
%T 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,
%U 2,2,1,3,2,2,1,2,1,1,1,3,2,2,1,2,1,1
%N 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.
%C 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
%H Michael De Vlieger, <a href="/A265917/b265917.txt">Table of n, a(n) for n = 1..10000</a>
%t Table[Floor[IntegerLength[n, 2]/Total@ IntegerDigits[n, 2]], {n, 120}] (* _Michael De Vlieger_, Dec 21 2015 *)
%o (Python)
%o for n in range(1, 88):
%o print(str((len(bin(n))-2) // bin(n).count('1')), end=',')
%o (PARI) a(n) = #binary(n)\hammingweight(n); \\ _Michel Marcus_, Dec 19 2015
%Y Cf. A000120, A007088, A070939, A135941, A199238, A265918.
%K nonn,base
%O 1,2
%A _Alex Ratushnyak_, Dec 18 2015
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