A273059 -id:A273059 - cp's OEIS frontend

A274640 Counterclockwise square spiral constructed by greedy algorithm, so that each row, column, and diagonal contains distinct numbers.

1, 2, 3, 4, 2, 3, 4, 5, 6, 1, 4, 6, 2, 1, 6, 5, 3, 1, 5, 2, 6, 1, 2, 4, 5, 3, 7, 8, 5, 4, 9, 7, 8, 3, 10, 11, 4, 7, 8, 6, 3, 9, 5, 7, 8, 9, 10, 11, 12, 6, 8, 9, 11, 10, 12, 13, 7, 6, 10, 9, 12, 13, 14, 15, 8, 2, 9, 12, 7, 10, 11, 13, 14, 10, 9, 6, 13, 5, 3, 15, 16, 7, 1, 10, 13, 12, 14, 11, 15, 3, 8, 5, 1, 12, 11, 14, 7, 4, 2, 16, 9, 17, 1, 8, 11

Offset: 0

Zak Seidov and Kerry Mitchell, Jun 30 2016

Presumably every row, column, and diagonal is a permutation of the natural numbers, but is there a proof? - N. J. A. Sloane, Jul 10 2016
The n-th cell in the spiral has coordinates x = A174344(n+1), y = A274923(n+1). - N. J. A. Sloane, Jul 11 2016
From Robert G. Wilson v, Dec 25 2016: (Start) [Memo: all these numbers need to decreased by 1, since the offset here is 0. See A324481. - N. J. A. Sloane, Jul 23 2017. Furthermore, the numbers don't seem correct, even after subtracting 1. - N. J. A. Sloane, Jul 04 2019]
Index of first appearance of k = 1,2,3,...: 1, 2, 3, 7, 8, 15, 17, 25, 35, 41, 47, 61, 62, 89, 98, 99, 121, 129, 130, 143, 197, 208, 225, 239, 271, ..., .
1 appears at: 1, 4, 12, 19, 22, 33, 42, 68, 79, 120, 179, 194, 302, 311, 445, 489, 511, 558, 630, 708, 847, 877, 907, ..., .
2 appears at: 2, 5, 9, 16, 48, 52, 70, 73, 88, 95, 110, 146, 280, 291, 309, 327, 488, 605, 656, 681, 735, 778, 1000, ..., .
3 appears at: 3, 6, 10, 23, 29, 36, 56, 76, 97, 105, 153, 168, 184, 252, 338, 437, 457, 670, 818, 906, 953, 967, ..., . (End).

			The spiral begins:
.
   9--16---2---4---7--14--11--12---1---5---8
   |                                       |
  17   8--15--14--13--12---9--10---6---7   3
   |   |                               |   |
   1   2   4--11--10---3---8---7---9  13  15
   |   |   |                       |   |   |
   8   9   7   3---5---6---1---2   4  12  11
   |   |   |   |               |   |   |   |
  11  12   8   1   2---4---3   6   5  10  14
   |   |   |   |   |       |   |   |   |   |
  15   7   6   5   3   1---2   4   8  11  12
   |   |   |   |   |           |   |   |   |
  14  10   3   2   4---5---6---1   7   9  13
   |   |   |   |                   |   |   |
   7  11   9   6---1---2---4---5---3   8  10
   |   |   |                           |   |
   4  13   5---7---8---9--10--11--12---6   1
   |   |                                   |
  12  14--10---9---6--13---5---3--15--16---7
   |
  10--15---1--12--16---8--14--13--11--18--17
.
The 8 spokes (A274924-A274931) begin:
  E:  1, 2, 4,  8, 11, 12, 16,  9, 19, 24, 22, ...
  NE: 1, 3, 2,  9,  7,  8, 12, 15, 13, 17, 20, ...
  N:  1, 4, 6,  3, 12, 14, 15, 18, 20, 26, 25, ...
  NW: 1, 2, 3,  4,  8,  9,  7, 11, 14, 10, 22, ...
  W:  1, 3, 5,  6,  7, 15, 10, 17, 13, 25, 14, ...
  SW: 1, 4, 6,  5, 14, 10, 11, 23, 16, 18, 21, ...
  S:  1, 5, 2,  9, 13,  8,  7, 11, 10, 17, 19, ...
  SE: 1, 6, 5, 12, 16, 17, 21, 24, 27, 13, 15, ...

Alois P. Heinz, Table of n, a(n) for n = 0..20000
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.
Alois P. Heinz, Distribution of a(n) for n <= 4010000
Kerry Mitchell, Color-coded version of spiral, (1): the colors represent the values, from black (small) to white (large) (jpg file, low resolution)
Kerry Mitchell, Color-coded version of spiral, (1a): the colors represent the values, from black (small) to white (large) (tiff file, much higher resolution)
Kerry Mitchell, Color-coded version of spiral, (2): values <= 100 are black and those > 100 are white.
Zak Seidov, Distribution of a(n) for first 20001 terms

Cf. A274641 (the same spiral, but starting with 0 not 1), A174344, A274923.
The 8 spokes are A274924-A274931.
The East-West axis is A275877 (see also A324680), the North-South axis is A276036.
Positions of 1's and 2's give A273059 and A275116.
In the same spirit as the infinite Sudoku array A269526.
Cf. A324481 (position of first n).
Cf. A274821 (the same construction on a hexagonal tiling).

#  Maple program from Alois P. Heinz, Jul 12 2016:
fx:= proc(n) option remember; `if`(n=1, 0, (k->
       fx(n-1)+sin(k*Pi/2))(floor(sqrt(4*(n-2)+1)) mod 4))
     end:
fy:= proc(n) option remember; `if`(n=1, 0, (k->
       fy(n-1)-cos(k*Pi/2))(floor(sqrt(4*(n-2)+1)) mod 4))
     end:
b:= proc() 0 end:
a:= proc(n) local x,y,s,i,t,m;
      x, y:= fx(n+1), fy(n+1);
      if b(x, y) > 0 then b(x, y)
    else s:={};
    for i do t:=b(x+i,y+i); if t>0 then s:=s union {t} else break fi od;
    for i do t:=b(x-i,y-i); if t>0 then s:=s union {t} else break fi od;
    for i do t:=b(x+i,y-i); if t>0 then s:=s union {t} else break fi od;
    for i do t:=b(x-i,y+i); if t>0 then s:=s union {t} else break fi od;
    for i do t:=b(x+i,y  ); if t>0 then s:=s union {t} else break fi od;
    for i do t:=b(x-i,y  ); if t>0 then s:=s union {t} else break fi od;
    for i do t:=b(x  ,y+i); if t>0 then s:=s union {t} else break fi od;
    for i do t:=b(x  ,y-i); if t>0 then s:=s union {t} else break fi od;
         for m while m in s do od;
         b(x,y):= m
      fi
    end:
seq(a(n), n=0..1000);

fx[n_] := fx[n] = If[n == 1, 0, Function[k, fx[n-1] + Sin[k*Pi/2]][Mod[Floor[Sqrt[4*(n-2)+1]], 4]]]; fy[n_] := fy[n] = If[n == 1, 0, Function[k, fy[n-1] - Cos[k*Pi/2]][Mod[Floor[Sqrt[4*(n-2)+1]], 4]]]; Clear[b]; b[, ] = 0; a[n_] := Module[{x, y, s, i, t, m}, {x, y} = {fx[n+1], fy[n+1]}; If[b[x, y] > 0, b[x, y], s = {};
For[i=1, True, i++, t=b[x+i, y+i]; If[t>0, s=Union[s,{t}], Break[]]];
For[i=1, True, i++, t=b[x-i, y-i]; If[t>0, s=Union[s,{t}], Break[]]];
For[i=1, True, i++, t=b[x+i, y-i]; If[t>0, s=Union[s,{t}], Break[]]];
For[i=1, True, i++, t=b[x-i, y+i]; If[t>0, s=Union[s,{t}], Break[]]];
For[i=1, True, i++, t=b[x+i, y  ]; If[t>0, s=Union[s,{t}], Break[]]];
For[i=1, True, i++, t=b[x-i, y  ]; If[t>0, s=Union[s,{t}], Break[]]];
For[i=1, True, i++, t=b[x  , y+i]; If[t>0, s=Union[s,{t}], Break[]]];
For[i=1, True, i++, t=b[x  , y-i]; If[t>0, s=Union[s,{t}], Break[]]];
m = 1; While[MemberQ[s, m], m++]; b[x, y] = m]]; Table[a[n], {n, 0, 1000}] (* Jean-François Alcover, Nov 14 2016, after Alois P. Heinz *)

class Lines: # manage lines in direction d = dx + dy*1j
    def _init_(self, d):
        self.lines={}; self.t = d.real/d.imag if d.imag else None
    def _call_(self, pos): # Return the line through pos in direction d
        index = pos.imag if self.t is None else pos.real - pos.imag*self.t
        if index not in self.lines: self.lines[index] = Values()
        return self.lines[index]
class Values(set): # the set of used numbers on a given line
    def next(self, n): # return least k >= n not on this line
        return min(m+1 for m in self if m+1 >= n and m+1 not in self
                   ) if n in self else n
def A274640(): # generator of the sequence, see below for possible usage
    lines = [Lines(d) for d in (1, 1+1j, 1j, 1-1j)]; pos = 0
    for side in range(9**9):
        for _ in range(side//2 + 1):
            n = 1; lines_here = [L(pos) for L in lines]
            while any(n < (n := L.next(n)) for L in lines_here): pass
            yield n; any(L.add(n) for L in lines_here); pos += 1j**side
[a for a,A274640(),range(99))%5D%20%23%20_M.%20F.%20Hasler"> in zip(A274640(),range(99))] # _M. F. Hasler, Feb 01 2025

Corrected and extended by Alois P. Heinz, Jul 12 2016

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).

Original entry on oeis.org

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

Paul D. Hanna, Jun 04 2008

nonn

Sequence A140100 = {X(n), n >= 1} is the complement of the current sequence, while the sequence of differences, A140102 = {Y(n)-X(n), n >= 1}, forms the complement of the sequence of sums, A140103 = {Y(n)+X(n), n >= 1}.
Compare with A140099(n) = [n*(1+t)], a Beatty sequence involving the tribonacci constant t = t^3 - t^2 - 1 = 1.83928675521416113255...
Theorem: A140099(n) - A140101(n) is always in {-1,0,1} (see A275926). (See also A276385.)
Comments from N. J. A. Sloane, Aug 30 2016: (Start) This is the same problem as the "Greedy Queens in a spiral" problem described in A273059. In A273059 the queens come in sets of 4, each set of 4 being on the same shell around the central square.
a(n) specifies the shell number for the successive sets of 4 (taking the central square to be shell 0, the 8 squares around the center to be shell 1, etc.).
For example, the queens at squares 9, 13, 17, 21 in the spiral (terms A273059(2)-A273059(5)) are all on shell a(1) = 2. The next four queens, at squares 82, 92, 102, 112, are on shell a(2) = 5.
The four "spokes" in A273059 are given in A275916-A275919. The precise connection with the current sequence is that a(n) = nearest integer to (1 + sqrt(A275917(n-1)+1))/2.
This sequence links together the spokes A275916-A275919 in the sense that A275918(n) = A275917(n)+2*a(n+1), A275919(n) = A275917(n)+4*a(n+1), and A275916(n+1) = A275917(n)+6*a(n+1).
(End)
Conjecture: a(n) = A003144(n) + n. (This is from my notebook Lattices 115 page 20 from Oct 25 2016.  It is now a theorem - see the Dekking et al. paper.) - N. J. A. Sloane, Jul 22 2019
The sequence is "tribonacci-synchronized"; this means there is a finite automaton recognizing the tribonacci representation of (n,a(n)) input in parallel, where a shorter input is padded with leading zeros.   This finite automaton has 23 states and was verified with Walnut.  In particular this finite automaton and a similar one for A140101 was used to verify that (conjecture of J. Cassaigne) either a(b(n)) = a(n)+b(n) or b(a(n)) = a(n)+b(n) for all n>=1, where b(n) = A140100(n). - Jeffrey Shallit, Oct 04 2022

			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.

Robbert Fokkink, Gerard Francis Ortega, Dan Rust, Corner the Empress, arXiv:2204.11805. See Table 3.

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

Cf. A140100 (complement); A140102, A140103, A275926 (deviation from A140099).
Cf. related Beatty sequences: A140098, A140099; A000201.
Cf. A058265 (tribonacci constant).
Cf. Greedy Queens in a spiral, A273059, A275916, A275917, A275918, A275919, A275925.
See also A276385.
For first differences of A140100, A140101, A140102, A140103 see A305392, A305374, A305393, A305394.
The indicator function of this sequence is A305386.

Maple
```
See link.
```

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 *)

/* 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))))))}

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)

Terms computed independently by Reinhard Zumkeller and Joshua Zucker

Edited and a(0)=0 added by N. J. A. Sloane, Aug 30 2016

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).

Original entry on oeis.org

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

Paul D. Hanna, Jun 04 2008

nonn

Sequence A140101 = {Y(n), n >= 1} is the complement of the current sequence, while the sequence of differences, A140102 = {Y(n) - X(n), n >= 1}, forms the complement of the sequence of sums, A140103 = {Y(n) + X(n), n >= 1}.
Compare with A140098(n) = floor(n*(1+1/t)), a Beatty sequence involving the tribonacci constant t = t^3 - t^2 - 1 = 1.83928675521416113255...
Conjecture: A140100(n) - A140098(n) = A276404(n) is always 0 or 1; see A276406 for the positions where a difference of 1 occurs.
This is the same problem as the "Greedy Queens in a spiral" problem described in A273059. See the Dekking et al. paper and comments in A140101. - N. J. A. Sloane, Aug 30 2016
The sequence is "tribonacci-synchronized"; this means there is a finite automaton recognizing the tribonacci representation of (n,a(n)) input in parallel, where a shorter input is padded with leading zeros.   This finite automaton has 22 states and was verified with Walnut.  In particular this finite automaton and a similar one for A140101 was used to verify that (conjecture of J. Cassaigne) either a(b(n)) = a(n)+b(n) or b(a(n)) = a(n)+b(n) for all n>=1, where b(n) = A140101(n). - Jeffrey Shallit, Oct 04 2022

			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,...].

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

Cf. A140101 (complement); A140102, A140103, A276404, A276405, A276406, A275926.
Cf. related Beatty sequences: A140098, A140099; A000201.
Cf. A058265 (tribonacci constant).
Cf. Greedy Queens in a spiral, A273059.
For first difference of A140100, A140101, A140102, A140103 see A305392, A305374, A305393, A305394.

Maple
```
See link.
```

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 *)

/* 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))))))}

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)

Terms computed independently by Reinhard Zumkeller and Joshua Zucker

Edited by N. J. A. Sloane, Aug 30 2016

A275925 Trajectory of 3 under repeated application of the morphism sigma: 3 -> 3656, 5 -> 365656, 6 -> 3656656.

Original entry on oeis.org

3, 6, 5, 6, 3, 6, 5, 6, 6, 5, 6, 3, 6, 5, 6, 5, 6, 3, 6, 5, 6, 6, 5, 6, 3, 6, 5, 6, 3, 6, 5, 6, 6, 5, 6, 3, 6, 5, 6, 5, 6, 3, 6, 5, 6, 6, 5, 6, 3, 6, 5, 6, 6, 5, 6, 3, 6, 5, 6, 5, 6, 3, 6, 5, 6, 6, 5, 6, 3, 6, 5, 6, 3, 6, 5, 6, 6, 5, 6, 3, 6, 5, 6, 5, 6, 3, 6, 5, 6, 6, 5, 6, 3, 6, 5, 6, 5, 6, 3, 6

Offset: 1

N. J. A. Sloane, Aug 29 2016

nonn

Versions of this sequence arises in so many different ways in the analysis of the Lonely Queens problem described in A140100-A140103 that it is convenient to define THETA(a,b,c) to be the result of replacing {6,5,3} here by {a,b,c} respectively. - N. J. A. Sloane, Mar 19 2019
Conjecture 1: This sequence is a compressed version of A140101 (see that entry for details). [This was formerly stated as a theorem, but I am no longer sure I have a proof. - N. J. A. Sloane, Sep 29 2018. It is true: see the Dekking et al. paper. - N. J. A. Sloane, Jul 22 2019]
From Michel Dekking, Dec 12 2018: (Start)
Let tau be the tribonacci morphism from A092782, but on the alphabet {6,5,3}, i.e., tau(3)=6, tau(5)=63, tau(6)=65. Then tau^3 is given by
      3 -> 6563, 5 -> 656365, 6 -> 6563656.
Let sigma be the morphism generating (a(n)). Then sigma is conjugate to tau^3 with conjugating word u = 656:
    (656)^{-1} tau^3(3) 656 = 3656 = sigma(3)
    (656)^{-1} tau^3(5) 656 = 365656 = sigma(5)
    (656)^{-1} tau^3(6) 656 = 3656656 = sigma(6).
It follows that tau and sigma generate the same language, in particular the frequencies of corresponding letters are equal.
Added Mar 03 2019: Since tau and sigma are irreducible morphisms (which means that their incidence matrices are irreducible), all of their fixed points have the same collection of subwords, this is what is called the language of tau, respectively sigma. See Lemma 3 of Allouche et al. (2003) for background.
(End)
From N. J. A. Sloane, Mar 03 2019: (Start)
The tribonacci word A092782 is the limit S_oo where f is the morphism 1 -> 12, 2 -> 13, 3 -> 1; S_0 = 1, and S_n = f(S_{n-1}).
The present sequence is the limit T_oo where
sigma: 3 -> 3656, 5 -> 365656, 6 -> 3656656; T_0 = 3, and T_n = sigma(T_{n-1}).
Conjecture 2: For all k=0,1,2,..., the following two finite words are identical:
S_{3k+2} with 1,2 mapped to 6,5 respectively, and 3 fixed,
T_{k+1} with its initial 3 moved to the end.
Example for k=1:
S_5 =    1, 2, 1, 3, 1, 2, 1, 1, 2, 1, 3, 1, 2, 1, 2, 1, 3, 1, 2, 1, 1, 2, 1, 3,
T_2 = 3, 6, 5, 6, 3, 6, 5, 6, 6, 5, 6, 3, 6, 5, 6, 5, 6, 3, 6, 5, 6, 6, 5, 6,
Note that S_{3k+2} has length A000073(3k+5) and always ends with a 3.
The conjecture would imply that if we omit the initial 3 here, and change 6 to 1, 5 to 2, and leave 3 fixed, we get A092782. Alternatively, if we omit the initial 3 here, and change 6 to 0, 5 to 1, and 3 to 2, we get A080843.
(End)
From Michel Dekking, Mar 11 2019: (Start)
Proof of Conjecture 2.
It is convenient to apply the letter to letter map 1->6, 2->5, 3->3 from the start, which changes f^3 to tau^3. Let alpha := tau^3.
We prove by induction that 3 alpha^n(3) = sigma^n(3) 3.
This is true for n=1: 3 alpha(3) = 3 6563 = sigma(3) 3.
The conjugation observation in my comment from December 12 implies that for all words w from the language of tau:
   alpha(w) 656 = 656 sigma(w).
Applying this with the word w = alpha^n(3) yields
      3 alpha^{n+1}(3) 656 = 3 656 sigma(alpha^n(3)) =
      sigma(3 alpha^n(3)) = sigma(sigma^n(3) 3) =
      sigma^{n+1}(3) 3656,
where we used the induction hypothesis in the second line.  Removing the 656's at the end completes the induction step.
(End)
Lengths of runs of 2's in A276788. - John Keith, May 15 2022

			The first few generations of the iteration are:
3
3656
365636566563656563656656
3656365665636565636566563656365665636565636566563656656365656365665636563656656\
   3656563656656365656365665636563656656365656365665636566563656563656656
...

N. J. A. Sloane, Table of n, a(n) for n = 1..35890
J.-P. Allouche, M. Baake, J. Cassaigns, and D. Damanik, Palindrome complexity, arXiv:math/0106121 [math.CO], 2001; Theoretical Computer Science, 292 (2003), 9-31.
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.
Index entries for sequences that are fixed points of mappings

Cf. A140100, A140101, A273059, A000073.
See A276790 and A277745 for other versions. See also A276788 and A080843, A092782.
For partial sums see A305373, also A276796, A276797, A276798.

SubstitutionSystem[{3 -> {3, 6, 5, 6}, 5 -> {3, 6, 5, 6, 5, 6}, 6 -> {3, 6, 5, 6, 6, 5, 6}}, {3}, 3] // Last (* Jean-François Alcover, Jan 21 2018 *)

Theorem: The partial sums of the generalized version THETA(r,s,t) (see Comments) are given by the following formula: Sum_{i=1..n} THETA(r,s,t)(i) = r*A276796(n-1) + s*A276797(n-1) + t*A276798(n-1). - N. J. A. Sloane, Mar 23 2019

cp's OEIS Frontend

A275915 First differences of A273059.

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A275916 a(n) = A273059(4n).

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A275919 a(n) = A273059(4n+3).

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A275917 a(n) = A273059(4n+1).

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A275918 a(n) = A273059(4n+2).

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A274640 Counterclockwise square spiral constructed by greedy algorithm, so that each row, column, and diagonal contains distinct numbers.

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A275925 Trajectory of 3 under repeated application of the morphism sigma: 3 -> 3656, 5 -> 365656, 6 -> 3656656.

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