A092246 Odd "odious" numbers (A000069).

1, 7, 11, 13, 19, 21, 25, 31, 35, 37, 41, 47, 49, 55, 59, 61, 67, 69, 73, 79, 81, 87, 91, 93, 97, 103, 107, 109, 115, 117, 121, 127, 131, 133, 137, 143, 145, 151, 155, 157, 161, 167, 171, 173, 179, 181, 185, 191, 193, 199, 203, 205, 211, 213, 217, 223, 227, 229, 233

Offset: 1

Author

Benoit Cloitre, Feb 23 2004

Keywords

• nonn
• easy

Comments

In other words, numbers having a binary representation ending in 1, and an odd number of 1's overall. It follows that by decrementing an odd odious number, one gets an even evil number (A125592). - Ralf Stephan, Aug 27 2013

The members of the sequence may be called primitive odious numbers because every odious number is a power of 2 times one of these numbers. Note that the difference between consecutive terms is either 2, 4, or 6. - T. D. Noe, Jun 06 2007

From Gary W. Adamson, Apr 06 2010: (Start)

a(n) = A026147(n)-th odd number, where A026147 = (1, 4, 6, 7, 10, 11, ...); e.g.,

n: 1 2 3 4 5 6 7 8 9 10 11

n-th odd: 1 3 5 7 9 11 13 15 17 19 21

a(n): 1 7 11 13 19 21

etc. (End)

Numbers m, such that when merge-sorting lists of length m, the maximal number of comparisons is even: A003071(a(n)) = A230720(n). - Reinhard Zumkeller, Oct 28 2013

Fixed points of permutation pair A268717/A268718. - Antti Karttunen, Feb 29 2016

Links

• T. D. Noe, Table of n, a(n) for n = 1..1000
• Index entries for sequences related to binary expansion of n
• Index entries for sequences related to sorting

Crossrefs

Cf. A129771, A026147.

Cf. A230709 (complement).

Cf. A268717, A268718, A268673.

Programs

Haskell

a092246 n = a092246_list !! (n - 1)
a092246_list = filter odd a000069_list
-- Reinhard Zumkeller, Oct 28 2013

Mathematica

Table[If[n < 1, 0, 2 n - 1 - Mod[First@ DigitCount[n - 1, 2], 2]], {n, 120}] /. n_ /; EvenQ@ n -> Nothing (* Michael De Vlieger, Feb 29 2016 *)
Select[Range[1, 1001, 2], OddQ[Total[IntegerDigits[#, 2]]]&] (* Jean-François Alcover, Mar 15 2016 *)

PARI

is(n)=n%2&&hammingweight(n)%2 \\ Charles R Greathouse IV, Mar 21 2013

PARI

a(n)=4*n-if(hammingweight(n-1)%2,1,3) \\ Charles R Greathouse IV, Mar 22 2013

Python

def A092246(n): return (n<<2)-(1 if (n-1).bit_count()&1 else 3) # Chai Wah Wu, Mar 03 2023

Formula

a(n) = 4*n + 2*A010060(n-1) - 3;

a(n) = 2*A001969(n-1) + 1.