A342937 Run lengths of A324608.

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, 2, 2, 1, 8, 1, 3, 1, 3, 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, 3, 2

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

A324608 begins 1, 1, 2, 3, 3, 3, 3, 3, 3, 3, 3, 10, 11, 11, 11, 13, ..., with a(1) = 2 ones, a(2) = 1 two, a(3) = 8 threes, a(4) = 1 ten, a(5) = 3 elevens, and so on.
This sequence first disagrees with A341406 at n = 51, where a(51) = 2 and
A341406(51) = 1.

Mathematica

a[1]=1; a[2]=2; a[n_]:=a[n]=FromDigits[Flatten[IntegerDigits[#, 2]&/@Table[a[k], {k, n-1}]][[;; n]], 2]-Total@Table[a[m], {m, n-1}]
Length/@Split@Table[Count[IntegerDigits[a[l], 2], 1], {l, 300}] (* Giorgos Kalogeropoulos, Mar 30 2021 *)

Prog

(Python)
from itertools import groupby
def aupton(terms):
  A324608, bstr, rl_lst, rl_idx, n = [1, 1], "110", [], 0, 3
  while len(rl_lst) < terms:
    an = int(bstr[:n], 2) - int(bstr[:n-1], 2)
    binan = bin(an)[2:]
    A324608, bstr, n = A324608 + [binan.count('1')], bstr + binan, n+1
    new_runs = [len(list(g)) for k, g in groupby(A324608[rl_idx:])]
    if len(new_runs) > 0:
      rl_lst.extend(new_runs[:-1]) # don't take last one in case mid-run
      rl_idx += sum(new_runs[:-1])
  return rl_lst[:terms]
print(aupton(86)) # Michael S. Branicky, Mar 30 2021

Crossrefs

Cf. A308092, A324608, A341406.

Sequence in context: A318314 A230369 A341406 * A353571 A372775 A254027

Adjacent sequences: A342934 A342935 A342936 * A342938 A342939 A342940

Keyword

nonn,base

Author

Peter Kagey, Mar 29 2021

Status

approved