login
A385111
Triangle read by rows: T(n,k) is the least cardinality of a set of binary strings of length k such that any binary string of length n contains one of the strings in the set as a substring; 0 <= k <= n.
0
1, 1, 2, 1, 2, 4, 1, 2, 3, 8, 1, 2, 3, 6, 16, 1, 2, 3, 5, 10, 32, 1, 2, 3, 4, 9, 20, 64, 1, 2, 3, 4, 7, 15
OFFSET
0,3
COMMENTS
For any binary string s of length k, there is a string of length n such that all its length-k substrings are cyclic shifts of s. (Simply repeat s until the length is n.) This implies that T(n,k) >= A000031(k).
It is clear that T(n,k) is nonincreasing in n for any fixed k, so T(n,k) reaches a limiting value as n increases. Apparently, this limiting value is A000031(k). For 0 <= k <= 4, the smallest values of n for which T(n,k) = A000031(k) are 0, 1, 3, 6, 9, respectively.
Equivalently, for k >= 1, T(n,k) is the least number of edges that must be removed from the binary (k-1)-dimensional de Bruijn graph such that the remaining graph does not contain any walk of length n-k+1. For a fixed k and sufficiently large n, this is the least number of edges that must be removed in order to make the remaining graph acyclic.
FORMULA
T(n,n) = 2^n.
T(n,k) <= T(n-1,k).
T(n,k) >= A000031(k).
EXAMPLE
Triangle begins:
n\k| 0 1 2 3 4 5 6 7 8 9
---+--------------------------------
0 | 1
1 | 1 2
2 | 1 2 4
3 | 1 2 3 8
4 | 1 2 3 6 16
5 | 1 2 3 5 10 32
6 | 1 2 3 4 9 20 64
7 | 1 2 3 4 7 15 ? 128
8 | 1 2 3 4 7 14 ? ? 256
9 | 1 2 3 4 6 12 ? ? ? 512
Any binary string of length n >= 3 contains one of the substrings 00, 01, 11 of length k = 2. The length-n strings 00...0, 11...1, and 0101... show that 00, 11, and either 01 or 10 are necessary, so T(n,2) = 3 for n >= 3.
CROSSREFS
Cf. A000031.
Main diagonal gives A000079.
Sequence in context: A174843 A253572 A141539 * A340547 A376033 A327844
KEYWORD
nonn,tabl,more
AUTHOR
STATUS
approved