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A175061
A positive integer n is included if n, when written in binary, is made of run-lengths (lengths of runs of 0's as well as of runs of 1's) that form a permutation of some number of consecutive positive integers starting with 1.
5
1, 4, 6, 35, 39, 49, 55, 57, 59, 536, 540, 560, 572, 624, 632, 776, 782, 784, 798, 880, 888, 900, 902, 912, 926, 944, 956, 964, 966, 968, 974, 984, 988, 16775, 16783, 16835, 16847, 16867, 16871, 17159, 17183, 17283, 17311, 17379, 17383, 17935, 17951
OFFSET
1,2
COMMENTS
Think of binary n as a string S of 0's and 1's. By a "run" of 0's or 1's, it is meant either a substring all of contiguous 0's, each run bounded by 1's or the edge of S; or a substring all of contiguous 1's, each run bounded by 0's or the edge of S.
This sequence contains those terms of A161001 that each contain a run of length 1.
LINKS
EXAMPLE
536 in binary is 1000011000. This contains a run of one 1, followed by a run of four 0's, followed by a run of two 1's, followed finally by a run of three 0's. So the run lengths are (1,4,2,3). And since this is a permutation of (1,2,3,4), then 536 is in the sequence.
PROG
(Python)
from itertools import groupby
def ok(n):
runlengths = [len(list(g)) for k, g in groupby(bin(n)[2:])]
return sorted(runlengths) == list(range(1, len(runlengths)+1))
print([n for n in range(1, 17952) if ok(n)]) # Michael S. Branicky, Jan 04 2021
(Python) # alternate that directly generates terms
from itertools import permutations
def runlength(r): # all terms with runlengths a permutation of 1, ..., r
c = ['1', '0']
return sorted([int("".join([c[j%2]*p[j] for j in range(r)]), 2)
for p in permutations(range(1, r+1))])
def aupto(nn):
r, out = 1, []
while len(out) < nn:
out += runlength(r)
r += 1
return out[:nn]
print(aupto(47)) # Michael S. Branicky, Jan 04 2021
CROSSREFS
KEYWORD
base,nonn
AUTHOR
Leroy Quet, Dec 12 2009
EXTENSIONS
Extended by Ray Chandler, Dec 16 2009
STATUS
approved