1,2

This sequence and A000069 give the unique solution to the problem of splitting the nonnegative integers into two classes in such a way that sums of pairs of distinct elements from either class occur with the same multiplicities [Lambek and Moser]. Cf. A000028, A000379.

En francais: les nombres paiens.

a(n)-A001285(n) = 2n-1 has been verified for n=0,1,2,...,400 - John W. Layman (layman(AT)math.vt.edu)

First differences give A036585. Observed by Franklin T. Adams-Watters, proved by Max Alekseyev, Aug 30 2006. This is equivalent to a(n) = 2*n + A010060(n). Proof: If the number of bits in n is odd then the last bit of a(n) is 1, and if the number of bits in n is even then the last bit of a(n) is 0. Hence the sequence of last bits is A010060. Therefore a(n) = 2*n + A010060(n).

Allouche, Jean-Paul and Cohen, Henri, Dirichlet series and curious infinite products, Bull. London Math. Soc. 17 (1985), 531-538.

J.-P. Allouche and J. Shallit, The ring of k-regular sequences, Theoretical Computer Sci., 98 (1992), 163-197.

J. Lambek and L. Moser, On some two way classifications of integers, Canad. Math. Bull. 2 (1959), 85-89.

M. D. McIlroy, The number of 1's in binary integers: bounds and extremal properties, SIAM J. Comput., 3 (1974), 255-261.

H. L. Montgomery, Ten Lectures on the Interface Between Analytic Number Theory and Harmonic Analysis, Amer. Math. Soc., 1996, p. 208.

D. J. Newman, A Problem Seminar, Springer; see Problem #89.

Shallit, J. O., On infinite products associated with sums of digits, J. Number Theory 21 (1985), 128-134.

N. J. A. Sloane, Table of n, a(n) for n = 1..10000

J.-P. Allouche and J. Shallit, The ring of k-regular sequences, Theoretical Computer Sci., 98 (1992), 163-197.

J. N. Cooper, D. Eichhorn and K. O'Bryant, Reciprocals of binary power series

Eric Weisstein's World of Mathematics, Evil Number

Index entries for sequences related to binary expansion of n

Index entries for "core" sequences

Note that 2n+1 is in the sequence iff 2n is not, and so this sequence has asymptotic density 1/2. - Franklin T. Adams-Watters, Aug 23 2006

a(n) = (1/2) * (4n + 1 - (-1)^A000120(n)). - Ralf Stephan (ralf(AT)ark.in-berlin.de), Sep 14 2003

G.f.: sum[k>=0, t(3+2t+3t^2)/(1-t^2)^2 * prod(l=0, k-1, 1-x^(2^l)), t=x^2^k]. - Ralf Stephan (ralf(AT)ark.in-berlin.de), Mar 25 2004

n such that A010060(n)=0 - Benoit Cloitre (benoit7848c(AT)orange.fr), Nov 15 2003

a(2*n+1) + a(2*n) = A017101(n) = 8*n+3. a(2*n+1) - a(2*n) gives the Thue-Morse sequence (3, 1 version): 3, 1, 1, 3, 1, 3, 3, 1, 1, 3, .... A001969(n) + A000069(n) = A016813(n) = 4*n+1. - DELEHAM Philippe (kolotoko(AT)wanadoo.fr), Feb 04 2004

a(0) = 0, a(2n) = a(n) + 2n, a(2n+1) = -a(n) + 6n + 3.

Let b(n) = 1 if sum of digits of n is even, -1 if it is odd; then Shallit (1985) showed that Prod_{n>=0} ((2n+1)/(2n+2))^b(n) = 1/sqrt(2).

a(n) = 2n + A010060(n). - Frank Adams-Watters (FrankTAW(AT)Netscape.net), Aug 28 2006

s := proc(n) local i, j, ans; ans := [ ]; j := 0; for i from 0 while j<n do if add(k, k=convert(i, base, 2)) mod 2=0 then ans := [ op(ans), i ]; j := j+1; fi; od; RETURN(ans); end; t1 := s(100); A001969 := n->t1[n]; # s(k) gives first k terms.

Select[Range[0, 300], EvenQ[DigitCount[ #, 2][[1]]] &]

(PARI) a(n)=2*n+subst(Pol(binary(n)), x, 1)%2

(PARI) a(n)=if(n<1, 0, if(n%2==0, a(n/2)+n, -a((n-1)/2)+3*n))

(MAGMA) [ n : n in [0..129] | IsEven(&+Intseq(n, 2)) ]; - from Sergei Haller (sergei(AT)sergei-haller.de), Dec 21 2006

Complement of A000069 (the odious numbers). Cf. A133009.

a(n)=2*n+A010060(n)=A000069(n)-(-1)^A010060(n). Cf. A018900.

The basic sequences concerning the binary expansion of n are A000120, A000788, A000069, A001969, A023416, A059015.

Cf. A036585 (differences), A010060.

Sequence in context: A128291 A081121 A080307 this_sequence A075311 A032786 A080309

Adjacent sequences: A001966 A001967 A001968 this_sequence A001970 A001971 A001972

easy,core,nonn,nice

njas

More terms from Robin Trew (trew(AT)hcs.harvard.edu).