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).

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

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

Table of n, a(n) for n=0..67.

Scott R. Shannon, Image of the first 100000 terms.

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. A007376, A331162, A033307.

Cf. A342162.

Sequence in context: A090075 A004500 A342162 * A254716 A240757 A206422

Adjacent sequences: A337224 A337225 A337226 * A337228 A337229 A337230

Keyword

nonn,base,look

Author

Scott R. Shannon and N. J. A. Sloane, Aug 19 2020

Status

approved