OFFSET
1,1
COMMENTS
Here an overlap means some initial part of the binary word matches exactly the end part of the word. More precisely if B = b_1,b_2,...,b_n is the word, and k is the largest value for which b_i=b_n-k+i for 1 <= i <= k, k < n, then B is said to have a maximum overlap of k. The smallest possible overlap is 0 and largest possible overlap is n-1.
The trivial overlap n=k is ignored.
All terms are even, because a word and its bitwise complement have the same maximum overlap.
LINKS
Sean A. Irvine, Rows n=1..38 flattened
H. Harborth, Endliche 0-1-Folgen mit gleichen Teilblöcken, J. für Reine Angewandte Math. 271 (1974), 139-154.
Sean A. Irvine, Java program (github)
EXAMPLE
For n=3, the maximum overlaps are as follows:
000 2,
001 0,
010 1,
011 0,
100 0,
101 1,
110 0,
111 2;
thus row 3 of the triangle is 4, 2, 2 (4 with overlap 0, 2 with overlap 1, 2 with overlap 2).
The triangle begins:
2;
2, 2;
4, 2, 2;
6, 6, 2, 2;
12, 10, 6, 2, 2;
20, 22, 12, 6, 2, 2;
...
PROG
(Python)
def maxoverlap(n):
b = bin(n)[2:]
for k in range(len(b)-1, -1, -1):
if b.startswith(b[-k:]): return k
def T(n, k): return 2*sum(maxoverlap(i) == k for i in range(2**(n-1), 2**n))
print([T(n, k) for n in range(1, 12) for k in range(n)]) # Michael S. Branicky, Jun 24 2021
(Python) # faster version, using maxoverlap above
from collections import Counter
def row(n):
c = Counter(maxoverlap(i) for i in range(2**(n-1), 2**n))
return [2*c[k] for k in range(n)]
def table(r): return [i for n in range(1, r+1) for i in row(n)]
print(table(11)) # Michael S. Branicky, Jun 24 2021
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Sean A. Irvine, Jun 20 2021
STATUS
approved