A276194 Odd numbers whose binary representation contains an even number of 1's and at least one 0.

5, 9, 17, 23, 27, 29, 33, 39, 43, 45, 51, 53, 57, 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, 237

Offset: 1

Author

Lei Zhou, Oct 20 2016

Keywords

• nonn
• base

Examples

			Binary expansions of odd integers in decimal and binary forms are as follows:
   1 ->     1, no;
   3 ->    11, no;
   5 ->   101, yes, so a(1)=5;
   7 ->   111, no;
   9 ->  1001, yes so a(2)=9;
  11 ->  1011, no;
  13 ->  1101, no;
  15 ->  1111, no;
  17 -> 10001, yes so a(3)=17.
		

Links

• Lei Zhou, Table of n, a(n) for n = 1..10000
• Index entries for sequences related to binary expansion of n

Crossrefs

Cf. A005408.

Intersection of A129771 and A062289.

Programs

Mathematica

BNDigits[m_Integer] :=
  Module[{n = m, d, t = {}},
   While[n > 0, d = Mod[n, 2]; PrependTo[t, d]; n = (n - d)/2]; t];
c = 1;
Table[While[c = c + 2; d = BNDigits[c]; ld = Length[d];
  c1 = Total[d]; ! (EvenQ[c1] && (c1 < ld))]; c, {n, 1, 57}]

PARI

isok(n) = my(b=binary(n)); (n % 2) && (vecmin(b)==0) && !(vecsum(b) % 2); \\ Michel Marcus, Oct 21 2016

PARI

seq(N) = {
  my(bag = List(), cnt = 0, n = 1);
  while(cnt < N,
        if (hammingweight(n)%2 == 0 && hammingweight(n+1) > 1,
            listput(bag, n); cnt++);
        n += 2);
  return(Vec(bag));
};
seq(57)  \\ Gheorghe Coserea, Oct 25 2016

Formula

a(2^n - floor(n/2)) = 4*2^n + 1, for all n >= 0. - Gheorghe Coserea, Oct 24 2016