A371719 Number of length-n strongly factor-symmetric binary words.

1, 2, 4, 6, 10, 14, 22, 26, 38, 42, 58, 62, 78, 82, 110, 106, 130, 134, 166, 162, 190, 190, 238, 226, 254, 250, 318, 294, 330, 330, 378, 366, 418, 406, 490, 430, 486, 486, 586, 538, 570, 574, 650, 618, 690, 646, 786, 714, 766, 742, 870, 818, 890, 858, 966, 874, 970

Offset

0,2

Comments

Suppose w is a word with p 0's and q 1's. Let d(i,j) denote the number of distinct blocks in w having exactly i 0's and j 1's, for 0<=i<=p and 0<=j<=q. Then w is said to be strongly factor symmetric if d(p-i,q-j) = d(i,j) for all i and j.

a(n) is even for n > 0 by symmetry. - Michael S. Branicky, Apr 04 2024

Links

Bert Dobbelaere, Table of n, a(n) for n = 0..120

Example

For example, for n = 5, the word 00101 is strongly-factor symmetric.

Prog

(Python)
from itertools import product
from collections import defaultdict
def sfs(w): # is strongly factor symmetric
    p, q, d = w.count('0'), w.count('1'), defaultdict(lambda: set())
    d[0, 0] = set([""])
    for b in range(len(w)):
        for e in range(b+1, len(w)+1):
            i, j = w[b:e].count('0'), w[b:e].count('1')
            d[i, j].add(w[b:e])
    return all(len(d[p-i, q-j])==len(d[i, j]) for i in range(p+1) for j in range(q+1))
def a(n):
    if n == 0: return 1
    return 2*sum(1 for w in product("01", repeat=n-1) if sfs("0"+"".join(w)))
print([a(n) for n in range(17)]) # Michael S. Branicky, Apr 04 2024

Crossrefs

Sequence in context: A103445 A001747 A048670 * A333315 A307889 A239951

Adjacent sequences: A371716 A371717 A371718 * A371720 A371721 A371722

Keyword

nonn

Author

Jeffrey Shallit, Apr 04 2024

Extensions

a(21)-a(27) from Michael S. Branicky, Apr 04 2024

a(28)-a(56) from Bert Dobbelaere, Apr 06 2024

Status

approved