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Consider the number c made from the first digits of $2^n$. To be more precise, the $n$-th decimal digit of $c$ is the first digit of $2^n$. The first digits from $c$ are :

$0.24813612512481361251248136125124813612512481371251249137125124913712512491371361 24913713612491371361$

At first sight, the number appears to be rational because apparent patterns appear showing periods. In fact, the continued fraction of $c$ has very large convergents. I calculated the first $20 000$ digits from $c$ with PARI and found a convergent with amazing $5817$ digits! The terms afterwards are totally normal. This leads to the conjecture that $c$ is transcendental. Has anyone an idea how this can be proven ?

A similar situation is observed in champernowne's constant constant $0.12345678910111213...$ I read in the internet that this number also has extreme convergents without having obvious periods. Does anyone know why the large convergents occur ?

Jyrki Lahtonen
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Peter
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    Interesting question (+1). I'd guess it is almost surely transcendental although it does not seem we have enough tools currently to prove or disprove this. – Balarka Sen Dec 24 '13 at 11:38
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    FYI : http://oeis.org/search?q=2%2C4%2C8%2C1%2C3%2C6%2C1%2C2%2C5%2C1%2C2%2C4&language=japanese&go=%E6%A4%9C%E7%B4%A2 – mathlove Dec 24 '13 at 11:39
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    I am surprised that this sequence is already in OEIS. – Peter Dec 24 '13 at 11:41
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    It is well known (and follows easily from Kronecker density theorem) that the decimal expansion of a power of two can begin with any finite sequence of digits. In particular any fixed length sequence of digits occurs infinitely many times. It is not clear at all what that implies about this question. For this number that result implies that any sequence that can occur (within the obvious but not easy to describe constraints), occurs infinitely often, but does that help at all? – Jyrki Lahtonen Nov 27 '24 at 09:05
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    What's a "convergent", in this context? – Toph Nov 27 '24 at 09:25
  • Convergent in the context of continued fractions and irrational numbers is rational approximation for the irrational number that comes from the continued fraction, for example $\frac{22}{7}$ is convergent for $\pi$, another convergent of $\pi$ is $\frac{355}{113}$ – Math Admiral Nov 27 '24 at 13:25

1 Answers1

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This not a complete answer.

One thing that can be said is that this number is not a normal one, as $1$ appears more often that $2$, $2$ more often that $3$, etc.

Also, it is definitely not rational, as the probability of a digit to be equal to $1$ is $\log_{10} 2$, which is an irrational number.

Also, it can be expressed as $$ \alpha=\sum_{n=1}^\infty \frac{\left\lfloor 10^{n\log_{10}2-\lfloor n\log_{10}2\rfloor}\right\rfloor}{10^n}. $$

Yiorgos S. Smyrlis
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