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

1,1

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.

Links

Rémy Sigrist, Table of n, a(n) for n = 1..43

Rémy Sigrist, C++ program

Example

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.

Prog

(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.

Crossrefs

Cf. A030302, A030303, A033307, A337227.

Sequence in context: A101961 A285055 A344712 * A268530 A090847 A318780

Adjacent sequences: A351750 A351751 A351752 * A351754 A351755 A351756

Keyword

nonn,base

Author

Scott R. Shannon, Feb 18 2022

Extensions

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

Status

approved