A253051 A compressed version of A253050 (A252867 mod 2): replace every substring 01 in A253050 with 1, every 001 with 2, every 0001 with 3, every 00001 with 4, etc.

1, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 3, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 2, 1, 2, 2, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 3, 1, 1, 2, 2, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1

Offset: 0

Author

N. J. A. Sloane, Jan 01 2015

Keywords

• nonn

Comments

The replacements are to be done in reverse order, of course, starting with the longest string of 0 and working backwards.

The first 4 appears at about term 3879. When does the first 5 appear?

Equivalent to run lengths of 0's. - Chai Wah Wu, Jan 01 2015

a(n) < 5 for n <= 10^6. - Chai Wah Wu, Jan 14 2015

Links

• Chai Wah Wu, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A252867, A253050.

Programs

Python

A253051_list, c, l1, l2, s, b = [1], 1, 2, 1, 3, set()
for _ in range(10**6):
    i = s
    while True:
        if not (i in b or i & l1) and i & l2:
            if i & 1:
                A253051_list.append(c)
                c = 0
            else:
                c += 1
            l2, l1 = l1, i
            b.add(i)
            while s in b:
                b.remove(s)
                s += 1
            break
        i += 1 # Chai Wah Wu, Jan 01 2015