Displaying 1-10 of 62 results found.
Partial sums of the infinite self-similar tribonacci word, written in the form A080843.
+20
7
0, 1, 1, 3, 3, 4, 4, 4, 5, 5, 7, 7, 8, 8, 9, 9, 11, 11, 12, 12, 12, 13, 13, 15, 15, 16, 16, 18, 18, 19, 19, 19, 20, 20, 22, 22, 23, 23, 24, 24, 26, 26, 27, 27, 27, 28, 28, 30, 30, 31, 31, 31, 32, 32, 34, 34, 35, 35, 36, 36, 38, 38, 39, 39, 39, 40, 40, 42, 42, 43, 43
COMMENTS
This sequence produces a formula for the A-numbers A278040, specifying the positions (or indices) of 1's in A080843, namely A(n) = 4*n+1 - a(n-1), with a(-1) = 0.
FORMULA
a(n) = Sum_{j=0..n} A080843(n), n >= 0. a(n) = z_A(n) + 2*z_C(n) = A276797(n+1) + 2*( A276798(n+1) - 1), where z_A(n) gives the number of A-numbers from A278040 not exceeding n, similarly for z_C(n) with the C-numbers from A278041. - Wolfdieter Lang, Dec 13 2018
First differences of ternary tribonacci word A080843.
+20
3
1, -1, 2, -2, 1, -1, 0, 1, -1, 2, -2, 1, -1, 1, -1, 2, -2, 1, -1, 0, 1, -1, 2, -2, 1, -1, 2, -2, 1, -1, 0, 1, -1, 2, -2, 1, -1, 1, -1, 2, -2, 1, -1, 0, 1, -1, 2, -2, 1, -1, 0, 1, -1, 2, -2, 1, -1, 1, -1, 2, -2, 1, -1, 0, 1, -1, 2, -2, 1, -1, 2, -2, 1, -1, 0, 1, -1, 2, -2, 1, -1, 1, -1, 2, -2, 1, -1, 0, 1, -1, 2, -2, 1
Abelian complexity function of tribonacci word ( A080843).
+20
2
3, 3, 4, 3, 4, 4, 4, 3, 4, 4, 4, 4, 4, 4, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 3, 4, 5, 5, 4, 4, 4, 4, 4, 5, 5, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 3, 4, 4, 5, 4, 5, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 4, 5, 4, 4, 4, 5, 5, 4, 4, 4, 4, 4, 5, 5, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 3, 4, 4, 4
COMMENTS
For all n, a(n) equals 3,4,5,6, or 7.
The values 3,4,5,6, and 7 are all obtained infinitely often.
The first 6 occurs when n=342. The first 7 occurs when n=3914.
REFERENCES
G. Richomme, K. Saari, L. Q. Zamboni, Balance and Abelian Complexity of the Tribonacci word, Adv. Appl. Math. 45 (2010) 212-231.
Orders of abelian cubes in the tribonacci word A080843.
+20
1
4, 6, 7, 11, 13, 17, 18, 20, 24, 26, 27, 30, 31, 33, 37, 38, 40, 41, 42, 43, 44, 48, 50, 51, 55, 57, 61, 62, 63, 64, 68, 70, 74, 75, 77, 79, 81, 85, 86, 87, 88, 92, 94, 95, 98, 99, 101, 105, 107, 108, 111, 112, 114, 116, 118, 119, 122, 123, 125, 129, 131, 132
COMMENTS
An abelian cube is a word of the form x x' x'', where x' and x'' are permutations of x, like the English word "deeded". The order of an abelian cube is the length of x.
FORMULA
There is a deterministic finite automaton of 1169 states that takes n in its tribonacci representation as input and accepts if and only if there is an abelian cube of order n. It can be obtained with the Walnut theorem-prover.
EXAMPLE
Here are the earliest-appearing abelian cubes of the first few orders:
n = 4: 2010.0102.0102
n = 6: 102010.010201.010201
n = 7: 0102010.0102010.1020100
n = 11: 02010010201.01020100102.01020100102
Orders of additive cubes in the tribonacci word A080843.
+20
1
3, 4, 6, 7, 10, 11, 13, 14, 16, 17, 18, 20, 21, 23, 24, 26, 27, 30, 31, 33, 34, 37, 38, 40, 41, 42, 43, 44, 45, 47, 48, 50, 51, 54, 55, 57, 58, 60, 61, 62, 63, 64, 65, 67, 68, 70, 71, 74, 75, 77, 78, 79, 81, 82, 84, 85, 86, 87, 88, 89, 91, 92, 94, 95, 97, 98, 99
COMMENTS
An additive cube is three consecutive blocks of the same length and same sum.
There is a tribonacci automaton of 4927 states recognizing the set of these orders (in tribonacci representation).
EXAMPLE
The first few examples of additive cubes of different lengths in the tribonacci word are 020.101.020 (order 3), 2010.0102.0102 (order 4), and 102010.010201.010201 (order 6)
Decimal expansion of constant related to complexity of the tribonacci word ( A080843).
+20
0
1, 0, 6, 0, 5, 2, 3, 8, 2, 9, 1, 0, 2, 6, 6, 3, 6, 1, 2, 0, 7, 9, 7, 2, 6, 9, 6, 3, 7, 5, 6, 3, 3, 5, 5, 7, 7, 4, 3, 2, 2, 9, 4, 2, 8, 3, 3, 5, 9, 4, 7, 4, 4, 6, 1, 0, 8, 1, 7, 8, 8, 3, 9, 9, 3, 8, 7, 4, 9, 4, 7, 0, 1, 4, 1, 0, 1, 8, 4, 7, 0, 1, 0, 1, 8, 5, 6, 2, 2, 0, 8, 7, 4, 3, 7, 0, 3, 9, 2, 9, 3, 3, 5, 0, 1, 8, 1
COMMENTS
The minimal polynomial of this constant is x^3 - 13*x^2 + 27*x - 17, and it is its unique real root. - Amiram Eldar, May 27 2023
FORMULA
Equals 9+(6*t-4)/(t^2+1), where t is the tribonacci constant A058265. Equals (13 + (847 - 33*sqrt(33))^(1/3) + (11 * (77 + 3*sqrt(33)))^(1/3))/3. - Amiram Eldar, May 27 2023
EXAMPLE
10.6052382910266361207972696375...
MATHEMATICA
RealDigits[x /. FindRoot[x^3 - 13*x^2 + 27*x - 17, {x, 10}, WorkingPrecision -> 120]][[1]] (* Amiram Eldar, May 27 2023 *)
Quasiperiods of the ternary tribonacci sequence ( A080843).
+20
0
7, 13, 14, 24, 25, 26, 27, 44, 45, 46, 47, 48, 49, 50, 51, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 149, 150, 151, 152, 153, 154, 155, 156, 157, 158, 159, 160, 161, 162, 163, 164, 165, 166, 167, 168, 169, 170, 171, 172, 173, 174, 175, 176
COMMENTS
A word x is a quasiperiod of another (possibly infinite) word w if you can cover all of the symbols of w by translates of x.
Numbers k such that the ternary tribonacci sequence ( A080843) has a Lyndon factor of length k.
+20
0
1, 2, 3, 4, 5, 6, 7, 9, 11, 13, 16, 18, 20, 22, 24, 29, 31, 35, 37, 40, 42, 44, 53, 55, 57, 64, 66, 68, 77, 79, 81, 97, 99, 101, 110, 112, 121, 123, 125, 134, 136, 145, 147, 149, 178, 180, 189, 191, 193, 215, 217, 226, 228, 230, 246, 248, 250, 259, 261, 270
COMMENTS
A "factor" is a contiguous subblock. A factor is "Lyndon" if it is lexicographically least among all its cyclic shifts.
Numbers k such that there are two distinct abelian squares of order k in the tribonacci word A080843.
+20
0
15, 34, 59, 90, 96, 97, 102, 134, 137, 170, 171, 172, 178, 183, 215, 240, 252, 259, 262, 289, 321, 333, 364, 370, 371, 387, 389, 391, 402, 408, 411, 445, 457, 470, 482, 489, 516, 519, 538, 556, 557, 563, 594, 600, 601, 606, 638, 665, 674, 675, 676, 682, 687
COMMENTS
An abelian square is a word of the form x x' where x' is a permutation of x, like the English word "reappear". The order of an abelian square x x' is the length of x.
The tribonacci word has abelian squares of all orders. If we consider two abelian squares x x' and y y' to be the same if y is a permutation of x, then some orders have only 1 abelian square (up to this equivalence), while others have 2, and these are the only possibilities. There is a 463-state automaton that recognizes the tribonacci representation of those terms k in this sequence. All this can be proved with the Walnut theorem prover.
EXAMPLE
For k = 15, the two distinct abelian squares are 100102010102010.010201001020101 and 020102010010201.010201001020102.
a(n) is the largest k such that the tribonacci word A080843 contains a block of length k*n that is an abelian k-th power.
+20
0
2, 2, 2, 3, 2, 4, 5, 2, 2, 2, 4, 2, 8, 2, 2, 2, 4, 3, 2, 6, 2, 2, 2, 8, 2, 4, 3, 2, 2, 3, 6, 2, 3, 2, 2, 2, 10, 3, 2, 3, 3, 3, 3, 13, 2, 2, 2, 4, 2, 5, 4, 2, 2, 2, 5, 2, 6, 2, 2, 2, 6, 3, 3, 4, 2, 2, 2, 12, 2, 3, 2, 2, 2, 5, 5, 2, 3, 2, 3, 2, 21, 2, 2, 2, 4, 3
COMMENTS
An abelian k-th power consists of k blocks x_1 x_2 ... x_k such that each x_i is a permutation of x_1. For example, the English word "deeded" is an abelian 3rd power.
EXAMPLE
For n=4, A080843 contains the block 2010.0102.0102, which is an abelian 3rd power. But no blocks of size 16 in A080843 are abelian 4th powers.
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