A045671 Extension of Beatty sequence; complement of A045672.

0, 1, 2, 3, 5, 6, 7, 9, 10, 11, 13, 14, 15, 16, 17, 19, 20, 21, 23, 24, 25, 27, 28, 29, 30, 31, 33, 34, 35, 37, 38, 39, 41, 42, 43, 44, 45, 47, 48, 49, 51, 52, 53, 55, 56, 57, 59, 60, 61, 63, 64, 65, 66, 67, 69, 70

Offset: 0

Author

Aviezri S. Fraenkel

Keywords

• nonn

Comments

Sequence can also be characterized by a special numeration system-see above reference.

(s,t)-sequences; the case s=2, t=2.

For n>=1, these are the positions of 1 in the fixed point of the morphism 0->11, 1->1110; see A285671. Conjecture: -1 < n*r - a(n) < 2 for n>=0, where r = (1 + sqrt(17))/4. - Clark Kimberling, May 02 2017

Links

• Shiri Artstein-avidan, Aviezri S. Fraenkel and Vera T. Sós, A two-parameter family of an extension of Beatty sequences, Discrete Math., 308 (2008), 4578-4588. doi:10.1016/j.disc.2007.08.070
• A. S. Fraenkel, Heap games, numeration systems and sequences, Annals of Combinatorics, 2 (1998), 197-210.
• A. S. Fraenkel, Recent results and questions in combinatorial game complexities, Theoretical Computer Science, vol. 249, no. 2 (2000), 265-288.
• A. S. Fraenkel, New games related to old and new sequences, INTEGERS, Electronic J. of Combinatorial Number Theory, Vol. 4, Paper G6, 2004.
• Clark Kimberling, Complementary Equations, Journal of Integer Sequences, Vol. 10 (2007), Article 07.1.4.
• Index entries for sequences related to Beatty sequences

Crossrefs

Cf. A026366, A045672, A285671.

Programs

Mathematica

s=2; t=2;
mex:=First[Complement[Range[1,Max[#1]+1],#1]]&;
a[0]=0; b[n_]:=b[n]=s*a[n]+t*n;
a[n_]:=a[n]=mex[Flatten[Table[{a[i],b[i]},{i,0,n-1}]]];
Table[a[n],{n,200}] (* A045671 *)
Table[b[n],{n,200}] (* A045672 *)
(* Clark Kimberling, Apr 02 2011 *)
s = Nest[Flatten[# /. {0 -> {1, 1}, 1 -> {1, 1, 1, 0}}] &, {0}, 10]; (* A285671 *)
Flatten[Position[s, 0]];  (* A045672 *)
Flatten[Position[s, 1]];  (* A045671 *)
(* - Clark Kimberling, May 02 2017 *)

Formula

a(n) = mex{a(i), b(i):0 <= iA045672, mex S=least integer >= 0 not in sequence S.

a(n) = (1+sqrt(17))/4*n+O(1). - Benoit Cloitre, Apr 23 2008