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).

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

Offset: 0

Author

Peter Kagey, Mar 14 2021

Keywords

• nonn
• base
• tabf
• look

Examples

			   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.
		

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.

Crossrefs

Cf. A000217, A070939, A342510, A342512.

Formula

T(n,1) = A000217(A070939(n)).