

A068076


Number of positive integers < n with the same number of 1's in their binary expansions as n.


6



0, 1, 0, 2, 1, 2, 0, 3, 3, 4, 1, 5, 2, 3, 0, 4, 6, 7, 4, 8, 5, 6, 1, 9, 7, 8, 2, 9, 3, 4, 0, 5, 10, 11, 10, 12, 11, 12, 5, 13, 13, 14, 6, 15, 7, 8, 1, 14, 16, 17, 9, 18, 10, 11, 2, 19, 12, 13, 3, 14, 4, 5, 0, 6, 15, 16, 20, 17, 21, 22, 15, 18, 23, 24, 16, 25, 17, 18, 6, 19, 26, 27, 19
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OFFSET

1,4


COMMENTS

From Rémy Sigrist, Dec 23 2018: (Start)
This sequence is related to the combinatorial number system:
 if n = Sum_{k=1..h} 2^c_k with 0 <= c_1 < c_2 < ... < c_h,
 then a(n) = Sum_{k=1..h} binomial(c_k, k) (with binomial(n, r) = 0 if n < r).
(End)


LINKS

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
Wikipedia, Combinatorial number system


FORMULA

a(n) = A263017(n)  1.  Antti Karttunen, May 22 2017


EXAMPLE

The binary expansion of 22 (10110) has 3 1's, as do those of the 6 smaller numbers 7, 11, 13, 14, 19 and 21, so a(22)=6.


MATHEMATICA

w[n_] := Plus@@IntegerDigits[n, 2]; a[n_] := Plus@@MapThread[Binomial, {Flatten[Position[Reverse[IntegerDigits[n, 2]], 1]]1, Range[w[n]]}]


PROG

(PARI) a(n)=my(k=hammingweight(n)); sum(i=1, n1, hammingweight(i)==k) \\ Charles R Greathouse IV, Sep 24 2012
(PARI) a(n) = my (v=0, k=0); for (c=0, oo, if (n==0, return (v), n%2, k++; if (c>=k, v+=c!/k!/(ck)!)); n\=2) \\ Rémy Sigrist, Dec 23 2018
(Python)
def a(n):
x=bin(n)[2:].count("1")
return sum(1 for i in range(n) if bin(i)[2:].count("1")==x) # Indranil Ghosh, May 24 2017


CROSSREFS

One less than A263017.
Cf. A067587, also A000120 for numerous references.
Sequence in context: A290537 A272569 A344788 * A138498 A276669 A307596
Adjacent sequences: A068073 A068074 A068075 * A068077 A068078 A068079


KEYWORD

nonn


AUTHOR

Dean Hickerson, Feb 16 2002


EXTENSIONS

Edited by John W. Layman, Feb 20 2002


STATUS

approved



