A342162 a(n) = difference between the starting positions of the first two occurrences of n in the Champernowne string (cf. A033307), where the first n is not required to be at its natural position.

11, 9, 13, 14, 15, 16, 17, 18, 19, 20, 180, 1, 13, 37, 55, 73, 91, 109, 127, 145, 221, 17, 1, 34, 37, 55, 73, 91, 109, 127, 231, 35, 17, 1, 55, 37, 55, 73, 91, 109, 241, 53, 35, 17, 1, 76, 37, 55, 73, 91, 251, 71, 53, 35, 17, 1, 97, 37, 55, 73, 261, 89, 71, 53, 35, 17, 1, 118, 37, 55, 271, 107

Offset: 0

Author

Scott R. Shannon, Mar 03 2021

Keywords

• nonn
• base

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, which is not required to be at its natural position in the string, 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).

This sequence is similar to A337227 but here the first appearance of n can be anywhere in string. The first example where n appears before its natural position in the string is n = 12. See the examples.

Examples

			a(0) = 11 as the string '0' first appears at position 1 and again at position 12, giving a difference of 11.
a(10) = 180 as the string '10' first appears at position 11 and again at position 191 (where it forms the start of the number 100), giving a difference of 180.
a(12) = 13 as the string '12' first appears at position 2 (the first number to appear before its natural position) and again at position 15 (its natural position), giving a difference of 13. This is the first term to differ from A337227.
		

Links

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

Crossrefs

Cf A033307, A337227, A007376, A331162.

Programs

Python

from itertools import count
def A342162(n):
    s1, s2, m = tuple(int(d) for d in str(n)), tuple(), -1
    l = len(s1)
    for i, s in enumerate(int(d) for k in count(0) for d in str(k)):
        s2 = (s2+(s,))[-l:]
        if s2 == s1:
            if m >= 0:
                return i-m
            m = i # Chai Wah Wu, Feb 18 2022