A129771 Evil odd numbers.

3, 5, 9, 15, 17, 23, 27, 29, 33, 39, 43, 45, 51, 53, 57, 63, 65, 71, 75, 77, 83, 85, 89, 95, 99, 101, 105, 111, 113, 119, 123, 125, 129, 135, 139, 141, 147, 149, 153, 159, 163, 165, 169, 175, 177, 183, 187, 189, 195, 197, 201, 207, 209, 215, 219, 221, 225, 231, 235

Offset: 1

Author

Tanya Khovanova, May 16 2007

Keywords

• nonn
• easy
• base

Comments

A heuristic argument suggests that, as n tends to infinity, a(n)/n converges to 4. - Stefan Steinerberger, May 17 2007

These numbers may be called primitive evil numbers because every evil number is a power of 2 multiplied by one of these numbers. Note that the difference between consecutive terms is either 2, 4, or 6. - T. D. Noe, Jun 06 2007

If m is in the sequence, then so is 2m-1 because in binary, m is x1 and 2m-1 is x01. Presumably the numbers that generate the whole sequence by application of n -> 2n-1 are the evil numbers times 4 plus 3. - Ralf Stephan, May 25 2013

Links

• Francisco J. Muñoz, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)
• Francisco J. Muñoz and Juan Carlos Nuño, Rule-based Generation of de Bruijn Sequences: Memory and Learning, arXiv:2507.09764 [cs.FL], 2025. See p. 9.

Crossrefs

Intersection of A001969 and A005408.

Supersequence of A093688.

Cf. A092246 (odd odious numbers).

Column 2 of A277880, positions of 1's in A277808 (2's in A277822).

Cf. A000069, A128309, A277823, A277902.

Programs

Mathematica

Select[Range[300], OddQ[ # ] && EvenQ[DigitCount[ #, 2, 1]] &] (* Stefan Steinerberger, May 17 2007 *)
Select[Range[300], EvenQ[Plus @@ IntegerDigits[ #, 2]] && OddQ[ # ] &]

PARI

is(n)=n%2 && hammingweight(n)%2==0 \\ Charles R Greathouse IV, Mar 21 2013

PARI

a(n)=4*n-if(hammingweight(n-1)%2,3,1) \\ Charles R Greathouse IV, Mar 21 2013

Python

def A129771(n): return (((m:=n-1)<<1)+(m.bit_count()&1^1)<<1)+1 # Chai Wah Wu, Mar 09 2023

Formula

a(n) = 2*A000069(n) + 1. a(n) is 1 plus twice odious numbers.

a(n) = A128309(n) + 1. a(n) is 1 plus odious even numbers.

A132680(a(n)) = A132680((a(n)-1)/2) + 2. - Reinhard Zumkeller, Aug 26 2007

a(n) = 4n + O(1). - Charles R Greathouse IV, Mar 21 2013

a(n) = A001969(1+A000069(n)) = A277902(A277823(n)). - Antti Karttunen, Nov 05 2016

Extensions

More terms from Stefan Steinerberger, May 17 2007