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Number of factors in Lyndon factorization of binary expansion of n.
+10
48
1, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 2, 4, 3, 4, 4, 5, 2, 3, 2, 4, 3, 3, 2, 5, 3, 4, 3, 5, 4, 5, 5, 6, 2, 3, 2, 4, 2, 3, 2, 5, 3, 4, 2, 4, 3, 3, 2, 6, 3, 4, 3, 5, 4, 4, 3, 6, 4, 5, 4, 6, 5, 6, 6, 7, 2, 3, 2, 4, 2, 3, 2, 5, 3, 3, 2, 4, 2, 3, 2, 6, 3, 4, 3, 5, 4, 3, 2, 5, 3, 4, 3, 4, 3, 3, 2, 7, 3, 4, 3, 5, 3, 4, 3, 6, 4, 5, 3, 5, 4, 4, 3, 7, 4, 5, 4, 6, 5, 5, 4, 7
COMMENTS
Any binary word has a unique factorization as a product of nonincreasing Lyndon words (see Lothaire). a(n) = number of factors in Lyndon factorization of binary expansion of n.
It appears that a(n) = k for the first time when n = 2^(k-1)+1.
We define the Lyndon product of two or more finite sequences to be the lexicographically maximal sequence obtainable by shuffling the sequences together. For example, the Lyndon product of (231) with (213) is (232131), the product of (221) with (213) is (222131), and the product of (122) with (2121) is (2122121). A Lyndon word is a finite sequence that is prime with respect to the Lyndon product. Equivalently, a Lyndon word is a finite sequence that is lexicographically strictly less than all of its cyclic rotations. Every finite sequence has a unique (orderless) factorization into Lyndon words, and if these factors are arranged in lexicographically decreasing order, their concatenation is equal to their Lyndon product. - Gus Wiseman, Nov 12 2019
REFERENCES
M. Lothaire, Combinatorics on Words, Addison-Wesley, Reading, MA, 1983. See Theorem 5.1.5, p. 67.
G. Melançon, Factorizing infinite words using Maple, MapleTech Journal, vol. 4, no. 1, 1997, pp. 34-42
EXAMPLE
n=25 has binary expansion 11001, which has Lyndon factorization (1)(1)(001) with three factors, so a(25) = 3.
Here are the Lyndon factorizations for small values of n:
.0.
.1.
.1.0.
.1.1.
.1.0.0.
.1.01.
.1.1.0.
.1.1.1.
.1.0.0.0.
.1.001.
.1.01.0.
.1.011.
.1.1.0.0.
...
MATHEMATICA
lynQ[q_]:=Array[Union[{q, RotateRight[q, #]}]=={q, RotateRight[q, #]}&, Length[q]-1, 1, And];
lynfac[q_]:=If[Length[q]==0, {}, Function[i, Prepend[lynfac[Drop[q, i]], Take[q, i]]][Last[Select[Range[Length[q]], lynQ[Take[q, #]]&]]]];
Table[Length[lynfac[IntegerDigits[n, 2]]], {n, 0, 30}] (* Gus Wiseman, Nov 12 2019 *)
CROSSREFS
Cf. A001037 (number of Lyndon words of length m); A102659 (list thereof). Ignoring the first digit gives A211097.
Irregular triangle read by rows where row n gives the lengths of the components in the Lyndon factorization of the binary expansion of n.
+10
14
1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 2, 1, 1, 3, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 3, 1, 1, 4, 1, 2, 1, 1, 1, 2, 2, 1, 3, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 2, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1
COMMENTS
We define the Lyndon product of two or more finite sequences to be the lexicographically maximal sequence obtainable by shuffling the sequences together. For example, the Lyndon product of (231) with (213) is (232131), the product of (221) with (213) is (222131), and the product of (122) with (2121) is (2122121). A Lyndon word is a finite sequence that is prime with respect to the Lyndon product. Equivalently, a Lyndon word is a finite sequence that is lexicographically strictly less than all of its cyclic rotations. Every finite sequence has a unique (orderless) factorization into Lyndon words, and if these factors are arranged in lexicographically decreasing order, their concatenation is equal to their Lyndon product. For example, (1001) has sorted Lyndon factorization (001)(1).
EXAMPLE
Triangle begins:
0: () 20: (1211) 40: (12111) 60: (111111)
1: (1) 21: (122) 41: (123) 61: (11112)
2: (11) 22: (131) 42: (1221) 62: (111111)
3: (11) 23: (14) 43: (15) 63: (111111)
4: (111) 24: (11111) 44: (1311) 64: (1111111)
5: (12) 25: (113) 45: (132) 65: (16)
6: (111) 26: (1121) 46: (141) 66: (151)
7: (111) 27: (113) 47: (15) 67: (16)
8: (1111) 28: (11111) 48: (111111) 68: (1411)
9: (13) 29: (1112) 49: (114) 69: (16)
10: (121) 30: (11111) 50: (1131) 70: (151)
11: (13) 31: (11111) 51: (114) 71: (16)
12: (1111) 32: (111111) 52: (11211) 72: (13111)
13: (112) 33: (15) 53: (1122) 73: (133)
14: (1111) 34: (141) 54: (1131) 74: (151)
15: (1111) 35: (15) 55: (114) 75: (16)
16: (11111) 36: (1311) 56: (111111) 76: (1411)
17: (14) 37: (15) 57: (1113) 77: (16)
18: (131) 38: (141) 58: (11121) 78: (151)
19: (14) 39: (15) 59: (1113) 79: (16)
MATHEMATICA
lynQ[q_]:=Array[Union[{q, RotateRight[q, #]}]=={q, RotateRight[q, #]}&, Length[q]-1, 1, And];
lynfac[q_]:=If[Length[q]==0, {}, Function[i, Prepend[lynfac[Drop[q, i]], Take[q, i]]][Last[Select[Range[Length[q]], lynQ[Take[q, #1]]&]]]];
Table[Length/@lynfac[If[n==0, {}, IntegerDigits[n, 2]]], {n, 0, 50}]
CROSSREFS
Ignoring the first digit gives A329325. Positions of rows of length 2 are A329327. Numbers whose reversed binary expansion is a Lyndon word are A328596. Length of the co-Lyndon factorization of the binary expansion is A329312.
Irregular triangle read by rows where row n gives the lengths of the components in the Lyndon factorization of the binary expansion of n with first digit removed.
+10
13
1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 3, 2, 1, 3, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 3, 1, 4, 2, 1, 1, 2, 2, 3, 1, 4, 1, 1, 1, 1, 1, 3, 1, 2, 1, 1, 3, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 4, 1, 5, 3, 1, 1, 5, 4, 1, 5, 2, 1
COMMENTS
We define the Lyndon product of two or more finite sequences to be the lexicographically maximal sequence obtainable by shuffling the sequences together. For example, the Lyndon product of (231) with (213) is (232131), the product of (221) with (213) is (222131), and the product of (122) with (2121) is (2122121). A Lyndon word is a finite sequence that is prime with respect to the Lyndon product. Every finite sequence has a unique (orderless) factorization into Lyndon words, and if these factors are arranged in lexicographically decreasing order, their concatenation is equal to their Lyndon product. For example, (1001) has sorted Lyndon factorization (001)(1).
EXAMPLE
Triangle begins:
1: () 21: (22) 41: (23) 61: (1112)
2: (1) 22: (31) 42: (221) 62: (11111)
3: (1) 23: (4) 43: (5) 63: (11111)
4: (11) 24: (1111) 44: (311) 64: (111111)
5: (2) 25: (13) 45: (32) 65: (6)
6: (11) 26: (121) 46: (41) 66: (51)
7: (11) 27: (13) 47: (5) 67: (6)
8: (111) 28: (1111) 48: (11111) 68: (411)
9: (3) 29: (112) 49: (14) 69: (6)
10: (21) 30: (1111) 50: (131) 70: (51)
11: (3) 31: (1111) 51: (14) 71: (6)
12: (111) 32: (11111) 52: (1211) 72: (3111)
13: (12) 33: (5) 53: (122) 73: (33)
14: (111) 34: (41) 54: (131) 74: (51)
15: (111) 35: (5) 55: (14) 75: (6)
16: (1111) 36: (311) 56: (11111) 76: (411)
17: (4) 37: (5) 57: (113) 77: (6)
18: (31) 38: (41) 58: (1121) 78: (51)
19: (4) 39: (5) 59: (113) 79: (6)
20: (211) 40: (2111) 60: (11111) 80: (21111)
For example, the trimmed binary expansion of 41 is (01001), with Lyndon factorization (01)(001), so row 41 is {2,3}.
MATHEMATICA
lynQ[q_]:=Array[Union[{q, RotateRight[q, #]}]=={q, RotateRight[q, #]}&, Length[q]-1, 1, And];
lynfac[q_]:=If[Length[q]==0, {}, Function[i, Prepend[lynfac[Drop[q, i]], Take[q, i]]][Last[Select[Range[Length[q]], lynQ[Take[q, #1]]&]]]];
Table[Length/@lynfac[Rest[IntegerDigits[n, 2]]], {n, 100}]
CROSSREFS
Keeping the first digit gives A329314. Positions of singleton rows are A329327. Numbers whose reversed binary expansion is a Lyndon word are A328596. Length of the co-Lyndon factorization of the binary expansion is A329312.
The binary expansion of a(n) is the second through n-th terms of A000002 - 1.
+10
3
0, 1, 3, 6, 12, 25, 50, 101, 203, 406, 813, 1627, 3254, 6508, 13017, 26034, 52068, 104137, 208275, 416550, 833101, 1666202, 3332404, 6664809, 13329618, 26659237, 53318475, 106636950, 213273900, 426547801, 853095602, 1706191204, 3412382409, 6824764818
EXAMPLE
a(11) = 813 has binary expansion q = {1, 1, 0, 0, 1, 0, 1, 1, 0, 1}, and q + 1 is {2, 2, 1, 1, 2, 1, 2, 2, 1, 2}, which is the second through 11th terms of A000002.
MATHEMATICA
kolagrow[q_]:=If[Length[q]<2, Take[{1, 2}, Length[q]+1], Append[q, Switch[{q[[Length[Split[q]]]], q[[-2]], Last[q]}, {1, 1, 1}, 0, {1, 1, 2}, 1, {1, 2, 1}, 2, {1, 2, 2}, 0, {2, 1, 1}, 2, {2, 1, 2}, 2, {2, 2, 1}, 1, {2, 2, 2}, 1]]]
kol[n_Integer]:=If[n==0, {}, Nest[kolagrow, {1}, n-1]];
Table[FromDigits[kol[n]-1, 2], {n, 30}]
Numbers whose reversed binary expansion has co-Lyndon factorization of length 2.
+10
3
2, 3, 5, 9, 11, 17, 19, 23, 33, 35, 37, 39, 43, 47, 65, 67, 69, 71, 75, 79, 83, 87, 95, 129, 131, 133, 135, 137, 139, 143, 147, 149, 151, 155, 159, 163, 167, 171, 175, 183, 191, 257, 259, 261, 263, 265, 267, 271, 275, 277, 279, 283, 287, 291, 293, 295, 299
COMMENTS
First differs from A329327 in lacking 77 and having 83. The co-Lyndon product of two or more finite sequences is defined to be the lexicographically minimal sequence obtainable by shuffling the sequences together. For example, the co-Lyndon product of (231) and (213) is (212313), the product of (221) and (213) is (212213), and the product of (122) and (2121) is (1212122). A co-Lyndon word is a finite sequence that is prime with respect to the co-Lyndon product. Equivalently, a co-Lyndon word is a finite sequence that is lexicographically strictly greater than all of its cyclic rotations. Every finite sequence has a unique (orderless) factorization into co-Lyndon words, and if these factors are arranged in a certain order, their concatenation is equal to their co-Lyndon product. For example, (1001) has sorted co-Lyndon factorization (1)(100).
EXAMPLE
The reversed binary expansion of each term together with their co-Lyndon factorizations:
2: (01) = (0)(1)
3: (11) = (1)(1)
5: (101) = (10)(1)
9: (1001) = (100)(1)
11: (1101) = (110)(1)
17: (10001) = (1000)(1)
19: (11001) = (1100)(1)
23: (11101) = (1110)(1)
33: (100001) = (10000)(1)
35: (110001) = (11000)(1)
37: (101001) = (10100)(1)
39: (111001) = (11100)(1)
43: (110101) = (11010)(1)
47: (111101) = (11110)(1)
65: (1000001) = (100000)(1)
67: (1100001) = (110000)(1)
69: (1010001) = (101000)(1)
71: (1110001) = (111000)(1)
75: (1101001) = (110100)(1)
79: (1111001) = (111100)(1)
MATHEMATICA
colynQ[q_]:=Array[Union[{RotateRight[q, #], q}]=={RotateRight[q, #], q}&, Length[q]-1, 1, And];
colynfac[q_]:=If[Length[q]==0, {}, Function[i, Prepend[colynfac[Drop[q, i]], Take[q, i]]]@Last[Select[Range[Length[q]], colynQ[Take[q, #]]&]]];
Select[Range[100], Length[colynfac[Reverse[IntegerDigits[#, 2]]]]==2&]
CROSSREFS
Length of the co-Lyndon factorization of the binary expansion is A329312. Cf. A059966, A060223, A102659, A211097, A211100, A275692, A328594, A328595, A328596, A329131, A329313, A329314, A329325, A329359.
Numbers whose bijective base-2 representation is a Lyndon word.
+10
2
1, 2, 4, 8, 10, 16, 18, 22, 32, 34, 36, 38, 42, 46, 64, 66, 68, 70, 74, 76, 78, 86, 94, 128, 130, 132, 134, 136, 138, 140, 142, 146, 148, 150, 154, 156, 158, 170, 174, 182, 190, 256, 258, 260, 262, 264, 266, 268, 270, 274, 276, 278, 280, 282, 284, 286, 292, 294, 298, 300, 302, 308
COMMENTS
A Lyndon word is a word which is lexicographically smaller than all its nontrivial rotations.
From the Chen-Fox-Lyndon theorem, every word can be written in a unique way as a concatenation of a nonincreasing sequence of Lyndon words. Since each natural number has a unique string representation in bijective bases, it can also be written exactly one way as a concatenation of these numbers in nonincreasing lexicographic order, in bijective base-2.
FORMULA
Observation: a(n) = 2* A326774(n-1), n >= 2. (At least for the terms from the Data section). - Omar E. Pol, Dec 09 2020
EXAMPLE
1 and 2 are in this sequence, since their bijective base-2 representations are also just "1" and "2", and words of just one letter have no nontrivial rotations.
3 is not in this sequence, since written in bijective base-2 it becomes "11", which is equal to its single nontrivial rotation.
108 is not in this sequence, since in bijective base-2 it becomes "212212", which is larger than two of its nontrivial rotations (both equal to "122122"). However, "212212" can be uniquely split into the lexicographically nonincreasing sequence of Lyndon words "2", "122" and "12", corresponding to 2, 10 and 4 in this sequence.
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