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

1,2

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

Links

Rémy Sigrist, Table of n, a(n) for n = 1..12664

Rémy Sigrist, PARI program for A175356

Example

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.

Prog

(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

Crossrefs

Cf. A000330, A022915, A175061, A175357.

Cf. A005811, A000217.

Sequence in context: A105504 A151900 A087951 * A363802 A134255 A061841

Adjacent sequences: A175353 A175354 A175355 * A175357 A175358 A175359

Keyword

base,nonn

Author

Leroy Quet, Apr 22 2010

Extensions

More terms from Rémy Sigrist, Feb 06 2019

Status

approved