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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.
8
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, 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
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