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A337227 a(n) = difference between the starting positions of the first two occurrences of n in the Champernowne string (starting at 0) 01234567891011121314151617181920... (cf. A033307). 10
11, 9, 13, 14, 15, 16, 17, 18, 19, 20, 180, 1, 19, 37, 55, 73, 91, 109, 127, 145, 221, 166, 1, 19, 37, 55, 73, 91, 109, 127, 231, 149, 233, 1, 19, 37, 55, 73, 91, 109, 241, 132, 243, 244, 1, 19, 37, 55, 73, 91, 251, 115, 253, 254, 255, 1, 19, 37, 55, 73, 261, 98, 263, 264, 265, 266, 1, 19 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,1
COMMENTS
Consider the infinite string
01234567891011121314151617181920... (cf. A033307)
formed by the concatenation of all decimal digits of all nonnegative numbers. From the position of the first digit of the first occurrence of the number n find the number of digits one has to move forward to get to the start of the second occurrence of n. This is a(n).
LINKS
EXAMPLE
The infinite string corresponding to the concatenation of all decimal digits >=0 starts "012345678910111213141516171819202122232425....".
a(0) = 11 because '0' appears at positions 1 and 12.
a(1) = 9 because '1' appears at positions 2 and 11.
a(10) = 180 because '10' appears starting at positions 11 and 191.
a(11) = 1 because '11' appears starting at positions 13 and 14.
PROG
(Python)
from itertools import count
def A337227(n):
s1 = tuple(int(d) for d in str(n))
s2 = s1
for i, s in enumerate(int(d) for k in count(n+1) for d in str(k)):
s2 = s2[1:]+(s, )
if s2 == s1:
return i+1 # Chai Wah Wu, Feb 18 2022
CROSSREFS
Cf. A342162.
Sequence in context: A090075 A004500 A342162 * A254716 A240757 A206422
KEYWORD
nonn,base,look
AUTHOR
STATUS
approved

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Last modified March 28 22:04 EDT 2024. Contains 371254 sequences. (Running on oeis4.)