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Search: a305392 -id:a305392
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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
32
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
0,2
COMMENTS
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
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.
022
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)
EXAMPLE
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.
MAPLE
See link.
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 *)
PROG
(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))))))}
CROSSREFS
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.
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Jun 04 2008
EXTENSIONS
Terms computed independently by Reinhard Zumkeller and Joshua Zucker
Edited and a(0)=0 added by N. J. A. Sloane, Aug 30 2016
STATUS
approved
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).
+0
25
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
1,2
COMMENTS
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
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.
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)
EXAMPLE
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,...].
MAPLE
See link.
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 *)
PROG
(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))))))}
CROSSREFS
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.
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Jun 04 2008
EXTENSIONS
Terms computed independently by Reinhard Zumkeller and Joshua Zucker
Edited by N. J. A. Sloane, Aug 30 2016
STATUS
approved
Term-by-term differences of A140101 and A140100; also, equals the complement of A140103, which is the term-by-term sums of A140101 and A140100, where A140101 is the complement of A140100.
+0
15
0, 1, 2, 4, 5, 6, 7, 9, 10, 11, 13, 14, 15, 16, 18, 19, 21, 22, 23, 24, 26, 27, 28, 30, 31, 32, 33, 35, 36, 37, 38, 40, 41, 42, 44, 45, 46, 47, 49, 50, 52, 53, 54, 55, 57, 58, 59, 61, 62, 63, 64, 66, 67, 68, 70, 71, 72, 73, 75, 76, 78, 79, 80, 81, 83, 84, 85, 87, 88, 89, 90, 92, 93
OFFSET
0,3
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.
FORMULA
a(n) = A140101(n) - A140100(n).
Theorem: the limit of A140103(n)/A140102(n) = t^2 = 3.38297576...
where the limit of A140101(n)/A140100(n) = t = 1.839286755...
and t = tribonacci constant satisfies: t^3 = 1 + t + t^2.
MAPLE
See link.
MATHEMATICA
nmax = 100; 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 | yn] && FreeQ[yy, xn | yn] && FreeQ[yy - xx, yn - xn] && FreeQ[yy + xx, yn - xn], y[n] = yn; Return[xn]]]];
Do[x[n], {n, 1, nmax}];
Join[{0}, yy - xx] (* Jean-François Alcover, Aug 01 2018 *)
PROG
(PARI) {X=[1]; Y=[2]; D=[1]; S=[3]; print1(Y[1]-X[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(Y[ #X]-X[ #Y]", "); break); break))))))}
CROSSREFS
Cf. A140103 (complement); A140100, A140101; A058265.
For first differences of A140100, A140101, A140102, A140103 see A305392, A305374, A305393, A305394.
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Jun 04 2008
EXTENSIONS
Terms computed by Reinhard Zumkeller.
Offset and initial term changed by N. J. A. Sloane, Oct 10 2016
STATUS
approved
Term-by-term sums of A140101 and A140100; also, equals the complement of A140102, which is the term-by-term differences of A140101 and A140100, where A140101 is the complement of A140100.
+0
15
3, 8, 12, 17, 20, 25, 29, 34, 39, 43, 48, 51, 56, 60, 65, 69, 74, 77, 82, 86, 91, 96, 100, 105, 108, 113, 117, 122, 125, 130, 134, 139, 144, 148, 153, 156, 161, 165, 170, 174, 179, 182, 187, 191, 196, 201, 205, 210, 213, 218, 222, 227, 232, 236, 241, 244, 249
OFFSET
1,1
COMMENTS
Conjecture: a(n) = A003145(n) + n. This is the most direct connection between the Greedy Queens sequence and the tribonacci word that I know. - Michel Dekking, Mar 19 2019. [My notes show that I made this conjecture on Jul 20 2018. There are many similar conjectures relating the two problems. For example A140100 = A003145(n)-A003144(n), A140101(n) = A003146(n)-A003145(n), a(n) = A003146(n)-A003144(n). - N. J. A. Sloane, Mar 19 2019] All these conjectures are now theorems - see the Dekking et al. paper. - N. J. A. Sloane, Jul 22 2019
REFERENCES
Robbert Fokkink, Gerard Francis Ortega, Dan Rust, Corner the Empress, arXiv:2204.11805. See Table 2.
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.
FORMULA
a(n) = A140100(n) + A140101(n).
Conjecture: the limit of A140103(n)/A140102(n) = t^2 = 3.38297576... (cf. A276800) where the limit of A140101(n)/A140100(n) = t = 1.839286755.. and t = tribonacci constant satisfies: t^3 = 1 + t + t^2.
MAPLE
See link.
MATHEMATICA
nmax = 100; 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 | yn] && FreeQ[yy, xn | yn] && FreeQ[yy - xx, yn - xn] && FreeQ[yy + xx, yn - xn], y[n] = yn; Return[xn]]]];
Do[x[n], {n, 1, nmax}];
yy + xx (* Jean-François Alcover, Aug 01 2018 *)
PROG
(PARI) {X=[1]; Y=[2]; D=[1]; S=[3]; print1(Y[1]-X[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(Y[ #X]-X[ #Y]", "); break); break))))))}
CROSSREFS
Cf. A140102 (complement); A140100, A140101; A058265, A276800.
For first differences of A140100, A140101, A140102, A140103 see A305392, A305374, A305393, A305394.
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Jun 04 2008
EXTENSIONS
Terms computed by Reinhard Zumkeller
STATUS
approved
First differences of A140101.
+0
9
2, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 2, 3, 3
OFFSET
0,1
COMMENTS
Or, prefix A276788 with a 1 and then add 1 to every term.
This relation between A003144 and A140101 is a conjecture (Daniel Forgues remarks would trivially follow from this relation). - Michel Dekking, Mar 18 2019
The lengths of the successive runs of 3's are given by A275925.
a(n) seems to take only the values 2 or 3, where {a(n), a(n+1)} may be {3, 2} or {2, 3} or {3, 3}, but not {2, 2}. The second differences of A140101 (first differences of this sequence) thus seem to take only the values -1 or 0 or 1. - Daniel Forgues, Aug 19 2018
Conjecture: This sequence is 2.TTW(3,3,2) where TTW is the ternary tribonacci word defined in A080843, or equally it is THETA(3,3,2), where THETA is defined in A275925. - N. J. A. Sloane, Mar 19 2019
All these conjectures are now theorems - see the Dekking et al. paper. - N. J. A. Sloane, Jul 22 2019
LINKS
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.
FORMULA
a(n) = A140101(n+1)-A140101(n).
CROSSREFS
For first differences of A140100, A140101, A140102, A140103 see A305392, A305374, A305393, A305394.
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Jun 09 2018
STATUS
approved
First differences of A140102.
+0
9
1, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1
OFFSET
1,2
COMMENTS
Although initially this agrees with A293630, the sequences are distinct.
From Michel Dekking, Mar 18 2019: (Start)
Let x be the tribonacci word x = A092782 = 1,2,1,3,1,2,1,1,...
Consider the morphism delta:
1 -> 1112,
2 -> 112,
3 -> 12.
Conjecture: (a(n)) = 12 delta(x).
(End)
Conjecture: This sequence (prefixed by 1 since A140102 should really begin with 0) is 1.TTW(1,2,1) where TTW is the ternary tribonacci word defined in A080843, or equally it is THETA(1,2,1), where THETA is defined in A275925. - N. J. A. Sloane, Mar 19 2019
All these conjectures are now theorems - see the Dekking et al. paper. - N. J. A. Sloane, Jul 22 2019
LINKS
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.
FORMULA
a(n) = A140102(n+1)-A140102(n), n >= 1.
CROSSREFS
For first differences of A140100, A140101, A140102, A140103 see A305392, A305374, A305393, A305394.
Cf. A293630.
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Jun 23 2018
STATUS
approved
First differences of A140103.
+0
8
5, 4, 5, 3, 5, 4, 5, 5, 4, 5, 3, 5, 4, 5, 4, 5, 3, 5, 4, 5, 5, 4, 5, 3, 5, 4, 5, 3, 5, 4, 5, 5, 4, 5, 3, 5, 4, 5, 4, 5, 3, 5, 4, 5, 5, 4, 5, 3, 5, 4, 5, 5, 4, 5, 3, 5, 4, 5, 4, 5, 3, 5, 4, 5, 5, 4, 5, 3, 5, 4, 5, 3, 5, 4, 5, 5, 4, 5, 3, 5, 4, 5, 4, 5, 3, 5, 4, 5, 5, 4, 5, 3, 5, 4, 5, 4
OFFSET
1,1
COMMENTS
Conjecture: this sequence is the ternary tribonacci word on the alphabet {5,4,3}, i.e., (a(n)) is the unique fixed point of the morphism 5 -> 54, 4 -> 53, 3 -> 5; see A092782. - Michel Dekking, Mar 13 2019
An equivalent conjecture: This sequence (prefixed by 3 since A140103 should really begin with 0) is 3.TTW(5,4,3) where TTW is the ternary tribonacci word defined in A080843, or equally it is THETA(5,4,3), where THETA is defined in A275925. There are similar conjectures for the first differences of A140100, A140101, A140102. - N. J. A. Sloane, Mar 14 2019 and Mar 19 2019
All these conjectures are now theorems - see the Dekking et al. paper. - N. J. A. Sloane, Jul 22 2019
LINKS
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.
FORMULA
a(n) = A140103(n+1) - A140103(n).
CROSSREFS
For first differences of A140100, A140101, A140102, A140103 see A305392, A305374, A305393, A305394.
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Jun 23 2018
STATUS
approved

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