

A090706


Number of numbers having in binary representation the same number of zeros and ones as n has.


6



1, 1, 1, 1, 1, 2, 2, 1, 1, 3, 3, 3, 3, 3, 3, 1, 1, 4, 4, 6, 4, 6, 6, 4, 4, 6, 6, 4, 6, 4, 4, 1, 1, 5, 5, 10, 5, 10, 10, 10, 5, 10, 10, 10, 10, 10, 10, 5, 5, 10, 10, 10, 10, 10, 10, 5, 10, 10, 10, 5, 10, 5, 5, 1, 1, 6, 6, 15, 6, 15, 15, 20, 6, 15, 15, 20, 15, 20, 20, 15, 6, 15, 15, 20, 15, 20
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OFFSET

0,6


COMMENTS

a(n) = binomial(A070939(n)1, A000120(n)1).


LINKS

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Eric Weisstein's World of Mathematics, Binary
Eric Weisstein's World of Mathematics, Digit Count
Index entries for sequences related to binary expansion of n


FORMULA

a(n) = binomial(A070939(n)1, A023416(n)).


EXAMPLE

n=25>'11001': a(25) = #{'10011'>19, '10101'>21, '10110'>22, '11001'>25, '11010'>26, '11100'>28} = 6.
n=23>'1_0111' has 5 bits, and the lower 4 bits can be shuffled. There are 1 zero and 3 ones, so the number of combinations is C(4,1) = 4 (the zero can be in 4 positions).
n=31>'1_1111': C(4,4) = 1.
n=33>'1_00001': C(5,1) = 5 (the one can be in 5 positions).
n=35>'1_00011': C(5,2) = 10. Ruud H.G. van Tol, Apr 17 2014


MATHEMATICA

a[n_] := Binomial[Length[b = IntegerDigits[n, 2]]1, Count[b, 0]]; a[0] = 1; Table[a[n], {n, 0, 100}] (* JeanFrançois Alcover, Apr 25 2014 *)


PROG

(PARI) A090706 = n>binomial(#binary(n)1, hammingweight(n)(n>0)) \\ About 20% faster than the alternative "...1)+!n".  M. F. Hasler, Jan 04 2014


CROSSREFS

Cf. A007088, A007318, A014312.
Sequence in context: A140408 A047080 A036064 * A284548 A228371 A176971
Adjacent sequences: A090703 A090704 A090705 * A090707 A090708 A090709


KEYWORD

nonn,base


AUTHOR

Reinhard Zumkeller, Jan 15 2004


EXTENSIONS

Missing a(0)=1 added and offset adjusted by Reinhard Zumkeller, Dec 19 2012


STATUS

approved



