A341406 Run lengths of the bits in A308092, read in binary.

2, 1, 8, 1, 3, 2, 3, 3, 3, 4, 3, 5, 3, 6, 3, 7, 2, 1, 10, 1, 11, 1, 9, 1, 2, 1, 9, 2, 2, 1, 8, 1, 5, 1, 8, 1, 3, 1, 2, 1, 8, 1, 3, 1, 3, 1, 8, 1, 3, 1, 1, 1, 2, 1, 8, 1, 3, 1, 1, 2, 2, 1, 8, 1, 3, 2, 5, 1, 8, 1, 3, 2, 3, 1, 2, 1, 8, 1, 3, 2, 3, 2, 2, 1, 8, 1

Offset

1,1

Links

Table of n, a(n) for n=1..86.

Peter Kagey, Run lengths of bits and run lengths of an auxiliary sequence, Mathematics Stack Exchange.

Example

In binary, A308092 is 1, 10, 11, 111, 1110, 11100, 111000, 1110000, .... The sequence begins with a(1) = 2 ones followed by a(2) = 1 zeros, a(3) = 8 ones, a(4) = 1 zeros, a(5) = 3 ones, a(6) = 2 zeros, and so on.
This sequence first disagrees with A342937 at n = 51, where a(51) = 1 and
A342937(51) = 2.

Prog

(Python)
from itertools import groupby
def aupton(terms):
  A308092, bstr, rl_lst, rl_idx, n = [1, 2], "110", [2], 2, 3
  while len(rl_lst) < terms:
    an = int(bstr[:n], 2) - int(bstr[:n-1], 2)
    A308092, bstr, n = A308092 + [an], bstr + bin(an)[2:], n+1
    new_runs = [len(list(g)) for k, g in groupby(bstr[rl_idx:])]
    if len(new_runs) > 1:
      rl_idx += sum(new_runs[:-1])
      rl_lst.extend(new_runs[:-1]) # don't take last one in case mid-run
  return rl_lst[:terms]
print(aupton(86)) # Michael S. Branicky, Apr 03 2021

Crossrefs

Cf. A308092, A342937.

Sequence in context: A318318 A318314 A230369 * A342937 A353571 A254027

Adjacent sequences: A341403 A341404 A341405 * A341407 A341408 A341409

Keyword

nonn,base

Author

Peter Kagey, Mar 31 2021

Status

approved