A077459 Numbers k such that k and 3*k have the same digital binary sum.

0, 3, 6, 7, 12, 14, 15, 24, 28, 30, 31, 45, 48, 51, 56, 60, 62, 63, 75, 89, 90, 93, 96, 99, 102, 103, 109, 112, 115, 120, 124, 126, 127, 139, 150, 151, 155, 177, 178, 180, 185, 186, 189, 192, 195, 198, 199, 204, 206, 207, 217, 218, 221, 224, 227, 230, 231, 237

Offset: 1

Author

Benoit Cloitre, Dec 01 2002

Keywords

• base
• nonn
• easy

Comments

Numbers k such that valuation(C(3*k,k),2) = valuation(C(2*k,k),2). - Benoit Cloitre, Mar 20 2004

From Robert Israel, Dec 13 2018: (Start)

If n is in the sequence, then so is 2*n.

If m and n are in the sequence and 2^k>3*n then 2^k*m+n is in the sequence. (End)

Contains all terms of A000225 except 1. - David A. Corneth, Dec 13 2018

Let's call "primitive" those a(n) not of the form 2^k*a(i) + a(j) with 2^k > 3*a(j), i > 1. The primitive terms not of the form 2^n-1 are (45, 89, 93, 109, 139, 151, 177, 217, 221, 237, ...) = (101101, 1011001, 1011101, 1101101, 10001011, 10010111, 10110001, 11011001, 11011101, 11101101, ...) in binary. Are all of them of the form x*2^k + 1 or 2^k + x where x = 2^m - 2^n - 1, m > n+1 (base-2 repunit with one "interior" digit 0, so that 3x has one bit 1 less than x)? - M. F. Hasler, Dec 13 2018

Examples

			51 binary representation is [1, 1, 0, 0, 1, 1] and 3*51=153 binary representation is [1, 0, 0, 1, 1, 0, 0, 1], both binary sum = 4, hence 51 is in the sequence.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000225.

Programs

Magma

[n: n in [0..300] | Valuation(Binomial(3*n, n), 2) eq Valuation(Binomial(2*n, n), 2)]; // Vincenzo Librandi, Dec 14 2018

Maple

filter:= n -> convert(convert(n,base,2),`+`)=convert(convert(3*n,base,2),`+`):
select(filter, [$0..1000]); # Robert Israel, Dec 13 2018

Mathematica

digitSum[n_] := Total@IntegerDigits[n, 2]; Select[Range[0, 250],  digitSum[#] == digitSum[3#] &] (* Amiram Eldar, Dec 13 2018 *)

PARI

is(n)=hammingweight(n)==hammingweight(3*n) \\ Charles R Greathouse IV, Mar 27 2013