A383670 Limiting word, as a sequence, obtained by prefixing with 0 the limiting sequence of s(n) defined by s(0) = 0, s(1) = 12, s(n) = the concatenation of s(n - 1) and s(n - 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, 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

Offset: 1

Author

Clark Kimberling, May 15 2025

Keywords

• nonn
• changed

Comments

The length of the n-th initial subword is A000045(n), for n>=1.

Examples

			s(0) = 1, s(1) = 12, s(2) = 120, s(3) = 12012, etc., so that the limiting word with 0 prefixed is 0120120120...
		

Crossrefs

Cf. A000045, A003849, A276885 (positions of 0), A001950 (positions of 1), A026352 (positions of 2), A383671.

Programs

Mathematica

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

Python

from math import isqrt
def A276885(n): return n+(n-1+isqrt(5*(n-1)**2)&-2)
def A001950(n): return (n+isqrt(5*n**2)>>1)+n
def A383670(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
    for i, f in enumerate((A276885, A001950)):
        if f(bsearch(f, n))==n: return i
    return 2 # Chai Wah Wu, May 21 2025