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A090706
Number of numbers having in binary representation the same number of zeros and ones as n has.
7
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
OFFSET
0,6
COMMENTS
a(n) = binomial(A070939(n)-1, A000120(n)-1).
LINKS
Eric Weisstein's World of Mathematics, Binary
Eric Weisstein's World of Mathematics, Digit Count
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}] (* Jean-François 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
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