A383671 The limiting word that starts with 0, as a sequence, generated by s(0) = 0, s(1) = 12, s(n) = concatenation of s(n - 2) and s(n - 1).

0, 1, 2, 1, 2, 0, 1, 2, 1, 2, 0, 1, 2, 0, 1, 2, 1, 2, 0, 1, 2, 1, 2, 0, 1, 2, 0, 1, 2, 1, 2, 0, 1, 2, 0, 1, 2, 1, 2, 0, 1, 2, 1, 2, 0, 1, 2, 0, 1, 2, 1, 2, 0, 1, 2, 1, 2, 0, 1, 2, 0, 1, 2, 1, 2, 0, 1, 2, 0, 1, 2, 1, 2, 0, 1, 2, 1, 2, 0, 1, 2, 0, 1, 2, 1, 2

Offset: 0

Author

Clark Kimberling, May 15 2025

Keywords

• nonn

Comments

There are two distinct limiting words generated by s(0) = 0, s(1) = 12, s(n) = s(n - 2)s(n - 1). This one is given by s(2n) for n>=0; the other, given by s(2n-1) for n>=0, is 1201201212012... In both limiting words, the length of the n-th initial subword is A000045(n+1), for n>=1.

Examples

			Initial subwords: s(0)=0, s(1)=12, s(2)=012, s(3)=12012, s(4)= 01212012, of lengths 1, 2, 3, 5, 8 (Fibonacci numbers).
		

Crossrefs

Cf. A000045, A003849, A047924 (positions of 0), A026356 (positions of 1), A022413 (positions of 2), A383670.

Programs

Mathematica

s[0] = "0"; s[1] = "12"; s[n_] := StringJoin[s[n - 2], s[n - 1]];
Join[{0}, IntegerDigits[FromDigits[s[10]]]]

Python

from math import isqrt
def A047924(n): return ((m:=(n+isqrt(5*n**2)>>1)+1)+isqrt(5*m**2)>>1)+m+1
def A026356(n): return (n+1+isqrt(5*(n-1)**2)>>1)+n
def A383671(n):
    def bsearch(f, n):
        kmin, kmax = 0, 1
        while f(kmax) <= n:
            kmax <<= 1
        kmin = kmax>>1
        while True:
            kmid = kmax+kmin>>1
            if f(kmid) > n:
                kmax = kmid
            else:
                kmin = kmid
            if kmax-kmin <= 1:
                break
        return kmin
    if n<3: return n
    for i, f in enumerate((A047924, A026356)):
        if f(bsearch(f,n+1))==n+1: return i
    return 2 # Chai Wah Wu, May 21 2025