A140100 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 X(n) (for Y(n) see A140101).
1, 3, 4, 6, 7, 9, 10, 12, 14, 15, 17, 18, 20, 21, 23, 24, 26, 27, 29, 30, 32, 34, 35, 37, 38, 40, 41, 43, 44, 46, 47, 49, 51, 52, 54, 55, 57, 58, 60, 61, 63, 64, 66, 67, 69, 71, 72, 74, 75, 77, 78, 80, 82, 83, 85, 86, 88, 89, 91, 92, 94, 95, 97, 98, 100, 102, 103, 105, 106
Offset: 1
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. CONSTRUCTION: PLOT OF (A140100(n), A140101(n)). This sequence gives the x-coordinates of the following construction. Start with an x-y coordinate system and place an 'o' at the origin. Define an open position as a point not lying in the same row, column, or diagonal (slope +1/-1) as any point previously given an 'o' marker. From then on, place an 'o' marker at the first open position with integer coordinates that is nearest the origin and the y-axis in the positive quadrant, while simultaneously placing markers at rotationally symmetric positions in the remaining three quadrants. Example: after the origin, begin placing markers at x-y coordinates: n=1: (1,2), (2,-1), (-1,-2), (-2,1); n=2: (3,5), (5,-3), (-3,-5), (-5,3); n=3: (4,8), (8,-4), (-4,-8), (-8,4); n=4: (6,11), (11,-6), (-6,-11), (-11,6); n=5: (7,13), (13,-7), (-7,-13), (-13,7); ... The result of this process is illustrated in the following diagram (see A273059 for an equivalent picture - _N. J. A. Sloane_, Aug 30 2016). ----------------+---o------------ --o-------------+---------------- ----o-----------+---------------- ----------------+--o------------- --------o-------+---------------- -----------o----+---------------- ----------------+o--------------- --------------o-+---------------- ++++++++++++++++o++++++++++++++++ ----------------+-o-------------- ---------------o+---------------- ----------------+----o----------- ----------------+-------o-------- -------------o--+---------------- ----------------+------------o--- ----------------+--------------o- ------------o---+---------------- Graph: no two points lie in the same row, column, diagonal, or antidiagonal. Points in the positive quadrant are at (A140100(n), A140101(n)). A140101 begins: [2,5,8,11,13,16,19,22,25,28,31,33,36,39,42,...].
Links
- N. J. A. Sloane, Table of n, a(n) for n = 1..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, Automaton to be called xaut.txt in Walnut format, recognizing (n, X(n)) in parallel, in Tribonacci representation
- N. J. A. Sloane, Maple program for A140100, A140101, A140102, A140103
Programs
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Maple
See link.
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Mathematica
y[0] = 0; x[1] = 1; y[1] = 2; x[n_] := x[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], y[n] = yn; Return[xn]]]]; Table[x[n], {n, 1, 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 X(n)/n = 1+1/t and limit of Y(n)/n = 1+t where the limit of Y(n)/X(n) = t = tribonacci constant (A058265), and thus the limit of (Y(n) + X(n))/(Y(n) - X(n)) = t^2 and the limit of (Y(n)^2 + X(n)^2)/(Y(n)^2 - X(n)^2) = t.
From Michel Dekking, Mar 16 2019: (Start)
It is conjectured in A305392 that the first differences of (X(n)) as a word are given by 212121 delta(x), where x is the tribonacci word x = A092782, and delta is the morphism
1 -> 2212121212121,
2 -> 22121212121,
3 -> 2212121.
This conjecture implies the frequency conjectures above: let N(i,n) be the number of letters i in x(1)x(2)...x(n). Then simple counting gives
X(13*N(1,n)+11*N(2,n)+7*N(3,n)) = 20*N(1,n)+17*N(2,n)+11*N(3,n), where we neglected the first 6 symbols of X.
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*1/t + 17*1/t^2 + 11*1/t^3 = (1+1/t)*(13*1/t + 11*1/t^2 + 7*1/t^3).
This is a simple verification. (End)
Extensions
Terms computed independently by Reinhard Zumkeller and Joshua Zucker
Edited by N. J. A. Sloane, Aug 30 2016
Comments