Displaying 1-10 of 14 results found.
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
FORMULA
and t = tribonacci constant satisfies: t^3 = 1 + t + t^2.
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}];
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))))))}
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
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
0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0
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).
+10
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
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... 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. (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.
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 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.
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]]]];
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
The indicator function of this sequence is A305386.
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).
+10
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
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... 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
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. 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.
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.
A140101 begins: [2,5,8,11,13,16,19,22,25,28,31,33,36,39,42,...].
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]]]];
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. Greedy Queens in a spiral, A273059.
Trajectory of 3 under repeated application of the morphism sigma: 3 -> 3656, 5 -> 365656, 6 -> 3656656.
+10
14
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
COMMENTS
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] 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)
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)
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)
FORMULA
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
EXAMPLE
The first few generations of the iteration are:
3
3656
365636566563656563656656
3656365665636565636566563656365665636565636566563656656365656365665636563656656\
3656563656656365656365665636563656656365656365665636566563656563656656
...
MATHEMATICA
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 *)
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
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
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
COMMENTS
Although initially this agrees with A293630, the sequences are distinct. 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
2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 2, 1
COMMENTS
a(n) seems to take only the values 1 or 2, where {a(n), a(n+1)} may be {2, 1} or {1, 2} or {2, 2}, but not {1, 1}, and where {a(n), a(n+1), a(n+2), a(n+3)} may be {2, 1, 2, 1} or {1, 2, 1, 2} or {1, 2, 2, 1}, but not {2, 1, 1, 2}. The second differences of A140100 (first differences of this sequence) thus seem to take only the values -1 or 0 or 1. - Daniel Forgues, Aug 17 2018 Let x be the tribonacci word x = A092782 = 1,2,1,3,1,2,1,1,... Consider the morphism delta:
1 -> 2212121212121,
2 -> 22121212121,
3 -> 2212121.
Conjecture: (a(n)) = 212121 delta(x).
(End)
Conjecture: This sequence (prefixed by 1 since A140100 should really begin with 0) is 1.TTW(2,1,1) where TTW is the ternary tribonacci word defined in A080843, or equally it is THETA(2,1,1), 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
0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0
COMMENTS
This sequence is "Tribonacci-automatic"; there is a 14-state automaton that takes the Tribonacci representation of n as input and accepts if and only if a(n) = 1. - Jeffrey Shallit, Oct 05 2022 This sequence is given by deleting the 2s in the Tribonacci word A080843 - Daniel Rust, Feb 28 2023
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