A006364 Numbers k with an even number of 1's in binary, ignoring last bit.

0, 1, 6, 7, 10, 11, 12, 13, 18, 19, 20, 21, 24, 25, 30, 31, 34, 35, 36, 37, 40, 41, 46, 47, 48, 49, 54, 55, 58, 59, 60, 61, 66, 67, 68, 69, 72, 73, 78, 79, 80, 81, 86, 87, 90, 91, 92, 93, 96, 97, 102, 103, 106, 107, 108, 109, 114, 115, 116, 117, 120, 121, 126, 127, 130, 131, 132

Offset: 1

Author

N. J. A. Sloane

Keywords

• base
• nonn
• nice
• easy

Comments

Equivalently, numbers k such that k has an odd number of 1's in binary if and only if k is odd. - Aaron Weiner, Jun 19 2013

Numbers k with an even number of trailing zeros in the binary representation of k!, A011371(k). - Amiram Eldar, Sep 05 2024

Examples

			G.f. = x + 6*x^2 + 7*x^3 + 10*x^4 + 11*x^5 + 12*x^6 + 13*x^7 + 18*x^8 + ...
		

References

• E. R. Berlekamp, J. H. Conway and R. K. Guy, Winning Ways, Academic Press, NY, 2 vols., 1982, see p. 111.
• R. K. Guy, Impartial games, pp. 35-55 of Combinatorial Games, ed. R. K. Guy, Proc. Sympos. Appl. Math., 43, Amer. Math. Soc., 1991.
• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
• J.-P. Allouche and J. Shallit, The Ring of k-regular Sequences, II, Theoret. Computer Sci., 307 (2003), 3-29.
• Index entries for sequences related to binary expansion of n.

Crossrefs

Cf. A000120, A001969, A011371.

Programs

Haskell

a006364 n = a006364_list
a006364_list = filter (even . a000120. (`div` 2)) [0..]
-- Reinhard Zumkeller, Oct 03 2011

Mathematica

Select[Range[0,150],EvenQ[Count[Most[IntegerDigits[#,2]],1]]&] (* Harvey P. Dale, Nov 03 2011 *)
a[ n_] := Which[ n < 1, 0, Mod[n, 2] > 0, a[n - 1] + 1, Mod[n, 4] > 0, 3 n - a[n/2 - 1], True, n + a[n/2]]; (* Michael Somos, Dec 21 2016 *)

PARI

a(n)=if(n<1,0,if(n%2==0,if(n%4==0,a(n/2)+n,-a((n-2)/2)+3*n),a(n-1)+1)) \\ Ralf Stephan

PARI

is(n)=hammingweight(n>>1)%2==0 \\ Charles R Greathouse IV, Jun 19 2013

Formula

Union of 2*A001969 and 2*A001969+1. With initial index 0: a(2n+1) = a(2n)+1, a(4n) = a(2n)+4n, a(4n+2) = -a(2n)+12n+6. - Ralf Stephan, Oct 17 2003

Conjecture: a(n) = 2*n + (-1)^(n-A000120(n-1)) - (3+(-1)^n)/2. - Velin Yanev, Dec 21 2016