the question as posed is easily seen to be equivalent to asking how to show that a certain number is irrational.

the number referred to as $\Phi$ is defined briefly beneath the question. it is a characteristic number encoding an infinite sequence of binary bits. in this case the information concerns the parity of the odd primes modulo $4$.

question is there a quick demonstration of the irrationality of $\Phi$?

i have a few other curiosities about the distribution of the binary digits of this number (or of the coefficients of the corresponding 2-adic integer), but the simple number theoretic question is all i wish to ask here.

in fact since the deep properties of the primes seem to be encrypted beyond the polynomial domain, one might judge it very likely that $\Phi$ is transcendental, but the assertion of its irrationality is a much more modest claim.

let $P \subset \mathbb{N}$ be the odd primes, numbered in ascending order so $p_0=3$, $p_1=5$ and so on. define the parity function on primes $\phi \in 2^{\mathbb{N}}$ by $$ \phi(n) = \phi_n = \frac12 (1 + i^{p_n-1}) $$ this gives rise to the sequence $\Phi_m$ where: $$ \Phi_m = \sum_{k=0}^m \frac{\phi_k}{2^k} $$ with $$ \Phi = \lim_{m \to \infty} \Phi_m = 1.010110011... = 1.34960... $$

  • $\begingroup$ @mixedmath sorry if my exposition was confusing. Dirichlet's theorem is a beautiful and profound result. i hope one day to really get to grips with the proof, but i need a bit more preparation to develop the required attention span $\endgroup$ – David Holden Aug 5 '14 at 13:38

It is known by a result of Daniel Shiu that there are arbitrarily long runs of consecutive primes congruent to $1$ modulo $4$. (It also would work for $3$ modulo $4$ or in fact any admissible congruence class.)

Thus, the binary expansion of your number will contain arbitrarily long stings of the same digit and thus cannot be eventually periodic, whence the number is irrational.

There is some discussion of this and other methods to see this in the MO question The prime numbers modulo $k$, are not periodic

Moreover, for what could be called a decimal version of your problem see Number made from ending digits of primes

In particular, note that the answer to such a question will be the same for any base and minor modifications, as the eventual nonperiodicity is known for any modulus as mentioned above.

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    $\begingroup$ thank you quid. very good answer and useful link. $\endgroup$ – David Holden Aug 5 '14 at 13:47
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    $\begingroup$ Very nice and simple, and a nice result of Daniel Shiu's. $\endgroup$ – davidlowryduda Aug 5 '14 at 13:50

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