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

0,3

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

Prog

(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

Crossrefs

Cf. A252867, A253050.

Sequence in context: A029440 A061337 A352767 * A103684 A105103 A086669

Adjacent sequences: A253048 A253049 A253050 * A253052 A253053 A253054

Keyword

nonn

Author

N. J. A. Sloane, Jan 01 2015

Status

approved