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

Author

Peter Kagey, Mar 31 2021

Keywords

• nonn
• base

Examples

			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.
		

Links

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

Crossrefs

Cf. A308092, A342937.

Programs

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