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A337227
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a(n) = difference between the starting positions of the first two occurrences of n in the Champernowne string (starting at 0) 01234567891011121314151617181920... (cf. A033307).
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10
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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
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OFFSET
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0,1
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COMMENTS
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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).
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LINKS
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EXAMPLE
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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.
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PROG
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(Python)
from itertools import count
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:
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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