

A303065


Number of numbers < n whose binary representation has the same difference between the numbers of 0's and 1's as n does.


0



0, 0, 0, 0, 1, 1, 2, 0, 0, 1, 2, 1, 3, 2, 3, 0, 0, 2, 3, 3, 4, 4, 5, 1, 5, 6, 7, 2, 8, 3, 4, 0, 0, 1, 2, 4, 3, 5, 6, 4, 4, 7, 8, 5, 9, 6, 7, 1, 5, 10, 11, 8, 12, 9, 10, 2, 13, 11, 12, 3, 13, 4, 5, 0, 0, 1, 2, 6, 3, 7, 8, 9, 4, 9, 10, 10, 11, 11, 12, 5, 5, 12, 13
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OFFSET

1,7


COMMENTS

First occurrence of k, k=0,1,2,...: 0, 4, 6, 12, 20, 22, 25, 26, 28, 44, 49, ..., .  Robert G. Wilson v, Feb 08 2018


LINKS

Table of n, a(n) for n=1..83.


FORMULA

a(n) = 0 iff n belongs to A097110.  Rémy Sigrist, May 16 2018


EXAMPLE

There are two numbers below 6 with number of 1's in the binary representation minus number of 0's equal to 1, namely 1 and 5, therefore a(6)=2.
There are 3 numbers below 12 with number of 1's in the binary representation minus number of 0's equal to 0, namely 2, 9, 10, therefore a(12)=3.


MATHEMATICA

d[n_] := DigitCount[n, 2, 1]  DigitCount[n, 2, 0]; f[n_] := Block[{fd = d[n], c = k = 0}, While[k < n, If[d@ k == fd, c++]; k++]; c]; Array[f, 83, 0] (* Robert G. Wilson v, Feb 08 2018 *)


PROG

(Python)
d=[0]*200
for n in range(1024):
b = bin(n)[2:]
c0 = b.count('0')
c1 = len(b)  c0
diff = c0  c1
print str(d[100+diff])+', ',
d[100+diff] += 1


CROSSREFS

Cf. A037861, A097110.
Sequence in context: A079113 A144874 A333365 * A325406 A257900 A039971
Adjacent sequences: A303062 A303063 A303064 * A303066 A303067 A303068


KEYWORD

nonn,base


AUTHOR

Alex Ratushnyak, Apr 17 2018


STATUS

approved



