Is a number whose infinite decimal part is the sequence of even numbers, transcendental? How about a number whose infinite decimal part is the odd numbers? Would the odds be more difficult to prove since they contain almost the entire sequence of primes?
Is 0.24681012141618202224… a Transcendental Number?
nt.number-theorytranscendental-number-theory
Related Solutions
Yes. It's known to be transcendental. The sequence of coefficients of your number is a variant of a Sturmian sequence. It has very low complexity. The definition of this: let the digit sequence be $a_1,a_2,a_3\ldots$ taking values in $ \lbrace 0,1\ldots,d-1 \rbrace ^{\mathbb N}$. A subword of length $k$ is a string $a_ia_{i+1}a_{i+2}\ldots a_{i+k-1}$. The complexity, $p(k)$, is a function from $\mathbb N$ to $\mathbb N$ taking $k$ to the number of subwords of the sequence of length $k$.
In a 2007 paper in the Annals of Mathematics (vol 165, p547--565), Adamczewski and Bugeaud (On the complexity of algebraic numbers. I. Expansions in integer bases) showed that if a number is algebraic, then its digit sequence in base $b$ has complexity satisfying $p(k)/k\to\infty$.
In your case, the complexity of the sequence of base 2 digits satisfies $p(k)=2k$. How to see this? Define a map $f$ from $[0,1)$ to $\lbrace 0,1\rbrace$ by $f(x)=1$ if $x\in [0,1/2)$ and 0 otherwise. The $n$th term of your sequence is $f(\alpha n\bmod 1)$, where $\alpha=1/(2\pi)$. Write $T$ for the transformation from $[0,1)$ to itself given by $T(x)=x+\alpha\bmod 1$. Then the $n$th term is just $f(T^n0)$. The sub-block of the digit sequence of length $k$ starting at the $j$th term is $f(T^j0)\ldots f(T^{j+k-1}0)$. Since the $T^j0$ are dense in $[0,1)$, we need to ask how many blocks $f(x)\ldots f(T^{k-1}x)$ are possible.
Consider taking $x=0$ and moving it around the circle (=$[0,1)$) once. As you move it, the $T^ix$ also each move around the circle one time. The sequence changes each time one of the $(T^ix)_{0\le i< k}$ crosses 0 or 1/2. This is a total of $2k$ changes. Hence the sequence takes on $2k$ values as $x$ moves around the circle, hence the estimate for the complexity.
This is not a complete answer, but I would say that the "standard" way to prove the transcendence of $\pi$ is as a corollary of the more general fact that $e^\alpha$ is transcendental for all nonzero algebraic $\alpha$. For general $\alpha$, one has to come up with a general method for dealing with those pesky algebraic numbers in the exponent. But for $\alpha=1$, clever ad hoc arguments are possible. For example, in the book Making Transcendence Transparent by Burger and Tubbs (which I highly recommend as a source for further details), they show how to write down explicitly a polynomial $\mathcal{P}_n(z)$ such that $\mathcal{P}_n(1), \mathcal{P}_n(2), \ldots, \mathcal{P}_n(d)$ provide exceptionally good rational approximations to $e, e^2, \ldots, e^d$ respectively. This proof does exploit special properties of the series $\sum_n z^n\!/n!$ so this perhaps vindicates your intuition that $e$ is easier because we have a nicer series for $e$ than for $\pi$. On the other hand, this argument is a bit circular, because isn't $${\pi \over 4} = 1 - {1\over 3} + {1\over 5} - {1\over 7} + \cdots$$ a "nice" formula for $\pi$? Well, maybe, but it's not "nice" in a way that lets us prove transcendence! Hmmm…
So I think that the answer is that we don't know of a way to prove the transcendence of $\pi$ that is significantly simpler than a proof of a more general result, whereas we do know some ad hoc tricks that work for $e$. In principle this could change in the future if, for example, someone finds an amazingly simple ad hoc proof for the transcendence of $\pi$, or (less likely) a dramatic simplification of the proof of the transcendence of $e^\alpha$ for all nonzero algebraic $\alpha$.
Best Answer
In point of fact, K. Mahler proved in this paper that, if $p(x)$ in a non-constant polynomial such that $p(n) \in \mathbb{N}$ for every $n\in \mathbb{N}$, then the number
$$0.p(1)p(2)p(3)p(4)\ldots,$$
which is formed concatenating after the decimal point the values of $p(1), p(2), p(3), \ldots$ (in that order), is a transcendental and non-Liouville number.