Why does the continued fraction of Champerowne's number have so large entries?

If we calculate the continued fraction of Champerowne's number $$0.1234567891011121314151617181920\cdots$$ it turns out that it contains very large entries

How can I understand this intuitively ?

The decimal representation of Champerowne's number must have parts which make the number look like a rational because of appearing periods, but it is not obvious where these (large) periods should be.

Many zeros (making the number look like to have a terminating decimal expansion) occur later, so they do not explain the large entries.

Any ideas ?

• How many continuants have you sampled? Are you asking about the first few terms, or the longer term behavior? – hardmath Jan 14 '17 at 18:05
• Very large numbers occur quite soon. The sequence of the number of digits of the entries starts with : $$0 , 1 , 1 , 1 , 6 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 2 , 166 , 1 , 1 , 1 , 2 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 2 , 1 , 3 , 1 , 2 , 1 , 2 , 2504 , 1 , 1 , 1 , 1 , 1 , 1 , 2 , 1 , 1$$ according to PARI/GP, which obviously gives the value $0$ for the number $0$ – Peter Jan 14 '17 at 18:13
• Looking at the Wikipedia article, it seems to me the large entries occur when the decimal expansion is of the form $...9899100...$. – Rahul Jan 14 '17 at 18:17
• @hardmath I would be content to understant the $166$-digit entry and the $2504$-digit-entry – Peter Jan 14 '17 at 18:28
• Also see this paper, which proves that Champernowne's number has irrationality measure $10$. I don't totally understand it, but skimming it suggests that @RosieF's rational approximations are in fact the key ones (they appear in identity 3.1 of the paper). – Micah Jan 14 '17 at 18:31

As a first-order approximation, it is very close to

$10/81 = 0.1 + 0.02 + 0.003 + \dots = 0.123456790123\dots$

As a second,

$1/9801 = 0.0001 + 0.000002 + \dots = 0.0001020304050607080910111213\dots$

so it should not come as a surprise that Champernowne's [sic] number is close to the sum of $10/81$ and another rational which is a multiple of $1/(9801\cdot10^8)$, the factor $10^8$ being needed to shift the $10111213\dots$ part 8 decimal places to the right, to leave the initial $0.12345678$ alone (the $9$ is already in the correct place w.r.t. the following $10$). In fact

$10/81-1002/980100000000 \approx 0.12345678910111213\dots96979900010203040506070809\dots$

• I knew the first approximtaion-fraction, but the second is very impressive indeed! – Peter Jan 14 '17 at 18:30