A140101 Start with Y(0)=0, X(1)=1, Y(1)=2. For n > 1, choose least positive integers Y(n) > X(n) such that neither Y(n) nor X(n) appear in {Y(k), 1 <= k < n} or {X(k), 1 <= k < n} and such that Y(n)-X(n) does not appear in {Y(k)-X(k), 1 <= k < n} or {Y(k)+X(k), 1 <= k < n}; sequence gives Y(n) (for X(n) see A140100).
0, 2, 5, 8, 11, 13, 16, 19, 22, 25, 28, 31, 33, 36, 39, 42, 45, 48, 50, 53, 56, 59, 62, 65, 68, 70, 73, 76, 79, 81, 84, 87, 90, 93, 96, 99, 101, 104, 107, 110, 113, 116, 118, 121, 124, 127, 130, 133, 136, 138, 141, 144, 147, 150, 153, 156, 158, 161, 164, 167, 170, 173
Offset: 0
Keywords
Examples
Start with Y(0)=0, X(1)=1, Y(1)=2 ; Y(1)-X(1)=1, Y(1)+X(1)=3. Next choose X(2)=3 and Y(2)=5; Y(2)-X(2)=2, Y(2)+X(2)=8. Next choose X(3)=4 and Y(3)=8; Y(3)-X(3)=4, Y(3)+X(3)=12. Next choose X(4)=6 and Y(4)=11; Y(4)-X(4)=5, Y(4)+X(4)=17. Continue to choose the least positive X and Y > X not appearing earlier such that Y-X and Y+X do not appear earlier as a difference or sum. This sequence gives the y-coordinates of the positive quadrant in the construction given in the examples for A140100.
References
- Robbert Fokkink, Gerard Francis Ortega, Dan Rust, Corner the Empress, arXiv:2204.11805. See Table 3.
Links
- N. J. A. Sloane, Table of n, a(n) for n = 0..50000, Sep 13 2016 (First 1001 terms from Reinhard Zumkeller)
- F. Michel Dekking, Jeffrey Shallit, and N. J. A. Sloane, Queens in exile: non-attacking queens on infinite chess boards, Electronic J. Combin., 27:1 (2020), #P1.52.
- Eric Duchêne and Michel Rigo, A morphic approach to combinatorial games: the Tribonacci case. RAIRO - Theoretical Informatics and Applications, 42, 2008, pp 375-393. doi:10.1051/ita:2007039. [Also available here]
- Robbert Fokkink and Dan Rust, A modification of Wythoff's Nim, arXiv:1904.08339 [math.CO], 2019.
- Jeffrey Shallit, Some Tribonacci conjectures, arXiv:2210.03996 [math.CO], 2022.
- Jeffrey Shallit, Using Walnut to Prove Results About Sequences in the OEIS, Seminar, Oct 18 2
- 022
- Jeffrey Shallit, Automaton to be called yaut.txt in Walnut format, recognizing (n, Y(n)) in parallel, in Tribonacci representation.
- N. J. A. Sloane, Maple program for A140100, A140101, A140102, A140103
Crossrefs
Programs
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Maple
See link.
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Mathematica
y[0] = 0; x[1] = 1; y[1] = 2; y[n_] := y[n] = For[yn = y[n - 1] + 1, True, yn++, For[xn = x[n - 1] + 1, xn < yn, xn++, xx = Array[x, n - 1]; yy = Array[y, n - 1]; If[FreeQ[xx, xn] && FreeQ[xx, yn] && FreeQ[yy, xn] && FreeQ[yy, yn] && FreeQ[yy - xx, yn - xn] && FreeQ[yy + xx, yn - xn], x[n] = xn; Return[yn]]]]; Table[y[n], {n, 0, 100}] (* Jean-François Alcover, Jun 17 2018 *) -
PARI
/* Print (x,y) coordinates of the positive quadrant */ {X=[1];Y=[2];D=[1];S=[3];print1("["X[1]","Y[1]"],"); for(n=1,100,for(j=2,2*n,if(setsearch(Set(concat(X,Y)),j)==0,Xt=concat(X,j); for(k=j+1,3*n,if(setsearch(Set(concat(Xt,Y)),k)==0, if(setsearch(Set(concat(D,S)),k-j)==0,if(setsearch(Set(concat(D,S)),k+j)==0, X=Xt;Y=concat(Y,k);D=concat(D,k-j);S=concat(S,k+j); print1("["X[ #X]","Y[ #Y]"],");break);break))))))}
Formula
CONJECTURE: the limit of a(n)/n = 1+t and limit of X(n)/n = 1+1/t so that limit of a(n)/X(n) = t = tribonacci constant (A058265), and thus the limit of [a(n) + X(n)]/[a(n) - X(n)] = t^2 and the limit of [a(n)^2 + X(n)^2]/[a(n)^2 - X(n)^2] = t.
Conjectured recursion: Take first differences: 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 2, ... (appears to consist of only 3's and 2's); list the run lengths: 3, 1, 6, 1, 5, 1, 6, 1, 3, 1, 6, 1, 5, 1, 6, 1, ... (it appears that every second term is 1 and the other terms are 3, 5, and 6); and bisect, getting 3, 6, 5, 6, 3, 6, 5, 6, 6, 5, 6, 3, 6, ... This is (although I do not have a proof) the recursively defined A275925. Thanks to Alois P. Heinz for providing enough terms of A273059 to enable a (morally) convincing check of this conjecture. - N. J. A. Sloane, Aug 30 2016
From Michel Dekking, Mar 17 2019: (Start)
This conjecture can be reformulated as follows (cf. A140100).
The first differences of (a(n)) = (Y(n)) as a word are given by
3 delta(x),
where x is the tribonacci word x = A092782, and delta is the morphism
1 -> 3333332,
2 -> 333332,
3 -> 3332.
This conjecture implies the frequency conjecture above: let N(i,n) be the number of letters i in a(1)a(2)...a(n). Then simple counting gives
a(7*N(1,n)+6*N(2,n)+4*N(3,n)) = 20*N(1,n)+17*N(2,n)+11*N(3,n), where we neglected the first symbol of a = Y.
It is well known (see, e.g., A092782) that the frequencies of 1, 2 and 3 in x are respectively 1/t, 1/t^2 and 1/t^3. Dividing all the N(i,n) by n, and letting n tend to infinity, we then have to see that
20/t + 17/t^2 + 11/t^3 = (1+t)*(7/t + 6/t^2 + 4/t^3).
This is a simple verification, using t^3 = t^2 + t + 1.
(End)
Extensions
Terms computed independently by Reinhard Zumkeller and Joshua Zucker
Edited and a(0)=0 added by N. J. A. Sloane, Aug 30 2016
Comments