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A342511 Irregular triangle read by rows: T(n,k) is the number of substrings of n read in binary that are instances of the Zimin word Z_k. 1 <= n <= A342510(n). 4
1, 1, 3, 3, 6, 6, 1, 6, 6, 1, 10, 1, 10, 1, 10, 2, 10, 2, 10, 10, 2, 10, 1, 10, 3, 15, 3, 15, 2, 15, 4, 15, 2, 15, 3, 15, 4, 15, 4, 15, 4, 15, 1, 15, 2, 15, 3, 15, 4, 15, 1, 15, 4, 15, 3, 15, 6, 21, 6, 21, 4, 21, 6, 21, 3, 21, 6, 21, 6, 21, 5, 21, 4, 21, 5 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,3
LINKS
Peter Kagey, Table of n, a(n) for n = 0..10170 (first 2^12 rows)
Peter Kagey, Matching ABACABA-type patterns, Code Golf Stack Exchange.
Danny Rorabaugh, Toward the Combinatorial Limit Theory of Free Words, arXiv preprint arXiv:1509.04372 [math.CO], 2015.
Wikipedia, Sesquipower.
FORMULA
T(n,1) = A000217(A070939(n)).
EXAMPLE
n | binary |k=1, 2, 3
-----+---------+----------
0 | 0 | 1
1 | 1 | 1
2 | 10 | 3
3 | 11 | 3
4 | 100 | 6
5 | 101 | 6, 1
6 | 110 | 6
7 | 111 | 6, 1
8 | 1000 | 10, 1
9 | 1001 | 10, 1
10 | 1010 | 10, 2
11 | 1011 | 10, 2
12 | 1100 | 10
13 | 1101 | 10, 2
14 | 1110 | 10, 1
15 | 1111 | 10, 3
16 | 10000 | 15, 3
...
85 | 1010101 | 28, 11, 1
For n = 121, the binary expansion is "1111001", which has 28 nonempty substrings.
For k = 1, there are T(121,1) = 28 substrings that are instances of Z_1 = A.
For k = 2, there are T(121,2) = 7 substrings that are instances of Z_2 = ABA are:
(111)1001 with A = 1 and B = 1,
1(111)001 with A = 1 and B = 1,
(1111)001 with A = 1 and B = 11,
111(1001) with A = 1 and B = 00,
11(11001) with A = 1 and B = 100,
1(111001) with A = 1 and B = 1100, and
(1111001) with A = 1 and B = 11100.
CROSSREFS
Sequence in context: A307967 A131077 A357897 * A175520 A271668 A370291
KEYWORD
nonn,base,tabf,look
AUTHOR
Peter Kagey, Mar 14 2021
STATUS
approved

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Last modified March 28 16:34 EDT 2024. Contains 371254 sequences. (Running on oeis4.)