A351753 Take the first n digits on the binary Champernowne string (cf. A030302); a(n) gives the starting index of the second occurrence of this n-digit string within the binary Champernowne string.

2, 4, 5, 12, 12, 12, 213, 517, 517, 517, 517, 517, 517, 517, 517, 517, 14457, 189569, 258049, 258049, 14144865, 14144865, 14144865, 131391133, 131391133, 199844657, 199844657, 199844657, 1196986333, 1196986333, 5176897753, 5176897753, 5176897753, 5176897753

Offset: 1

Author

Scott R. Shannon, Feb 18 2022

Keywords

• nonn
• base

Comments

The twenty-first n-digit string is '110111001011101111000' (1808238 decimal) which cannot be readily split into consecutive smaller values implying it is likely its next occurrence is in its natural position, i.e., a(21) = 35876058.

Examples

			The binary Champernowne string starts 110111001011101111000100110101011....
a(1) = 2 as the second occurrence of '1' within the string starts at index 2.
a(2) = 4 as the second occurrence of '11' within the string starts at index 4.
a(3) = 5 as the second occurrence of '110' within the string starts at index 5.
a(4) = 12 as the second occurrence of '1101' within the string starts at index 12.
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 1..43
• Rémy Sigrist, C++ program

Crossrefs

Cf. A030302, A030303, A033307, A337227.

Programs

Python

from itertools import count
def A351753(n):
    s1, s2 = tuple(), tuple()
    for i, s in enumerate(int(d) for n in count(1) for d in bin(n)[2:]):
        if i < n:
            s1 += (s,)
            s2 += (s,)
        else:
            s2 = s2[1:]+(s,)
            if s1 == s2:
                return i-n+2 # Chai Wah Wu, Feb 18 2022
(C++) // See Links section.

Extensions

a(18)-a(20) corrected and a(21)-a(34) added by Chai Wah Wu, Feb 18 2022