A371440 Count of iteration steps for the iterates in A267371 that end in 1.

0, 2, 5, 9, 12, 14, 17, 20, 25, 28, 32, 36, 39, 43, 46, 49, 52, 56, 59, 61, 64, 67, 71, 74, 76, 80, 83, 87, 90, 92, 95, 99, 102, 106, 109, 111, 114, 117, 119, 123, 126, 130, 133, 138, 140, 143, 147, 150, 156, 159, 163, 166, 168, 171, 174, 179, 183, 186, 190, 193, 195, 198

Offset

0,2

Comments

Conjecture: If iterate m terminates with prefix s ending in 1 then the subsequent iterate of A267371 is terminated by prefix s0.

Links

Michael S. Branicky, Table of n, a(n) for n = 0..4000

Example

a(0) = 0 since the initial string for the iteration process is 01.
a(2) = 5 since the first 5 prefixes are 0, 01, 010, 0100 and 0100101 determining the iterate 0100101001000100101 of length 19. The next prefix in the iteration process is 01001010.

Mathematica

(* function preLo[ ] is defined in A267371 *)
lastChar[n_] := Map[StringTake[Last[#], -1]&, NestList[{preLo[#], StringJoin[#[[2]], StringTake[#[[2]], preLo[#]]]}&, {1, "01"}, n]]a371440[n_] := Flatten[Position[lastChar[n], "1"]-1]a371440[200]

Prog

(Python)
from itertools import count, islice
def agen(): # generator of terms
    astr, k, mink = "01", 2, 1
    for i in count(0):
        if astr[-1] == "1": yield i
        for k in range(mink, len(astr)+1):
            if astr[1:].count(astr[:k]) == 0:
                break
        mink = max(mink, k)
        astr += astr[:k]
print(list(islice(agen(), 62))) # Michael S. Branicky, Mar 23 2024

Crossrefs

Cf. A267371.

Sequence in context: A099434 A191104 A287387 * A287409 A279171 A297465

Adjacent sequences: A371437 A371438 A371439 * A371441 A371442 A371443

Keyword

nonn

Author

Hartmut F. W. Hoft, Mar 23 2024

Status

approved