A175356 Those positive integers n where, when written in binary, there are exactly k number of runs (of either 0's or 1's) each of exactly k length, for all k where 1<=k<=m, for some positive integer m.

1, 19, 25, 27, 8984, 8988, 9016, 9100, 9112, 9116, 9784, 10008, 10012, 10040, 12568, 12572, 12600, 12680, 12686, 12728, 12740, 12742, 12744, 12750, 12760, 12764, 12856, 13192, 13198, 13240, 13880, 14104, 14108, 14136, 14476, 14488, 14492, 14532, 14534, 14536

Offset: 1

Author

Leroy Quet, Apr 22 2010

Keywords

• base
• nonn

Comments

A "run" of 0's is not immediately bounded by any 0's, and a "run" of 1's is not immediately bounded by any 1's.

There are exactly (m*(m+1)/2)! / Product_{k=1 to m} k! numbers in the sequence each of m^3/3 + m^2/2 + m/6 binary digits, for all m >= 1, and none of any other number of binary digits.

A005811(a(n)) is triangular, i.e., in A000217. - Michael S. Branicky, Jan 19 2021

Examples

			9016 in binary is 10001100111000. There is exactly one run of one binary digit, two runs of two binary digits, and three runs of three binary digits. (Note that it doesn't matter if the runs are of 0's or of 1's.) So, 9016 is in the sequence.
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 1..12664
• Rémy Sigrist, PARI program for A175356

Crossrefs

Cf. A000330, A022915, A175061, A175357.

Cf. A005811, A000217.

Programs

PARI

\\ See Links section.

Python

from itertools import groupby
def ok(n):
  runlens = [len(list(g)) for k, g in groupby(bin(n)[2:])]
  return all(runlens.count(k) == k for k in range(1, max(runlens)+1))
def aupto(limit): return [m for m in range(1, limit+1) if ok(m)]
print(aupto(14536)) # Michael S. Branicky, Jan 19 2021

Extensions

More terms from Rémy Sigrist, Feb 06 2019