Prime Generating Irrational Number

Question

Does there exist an irrational number such that every time it is multiplied by 100 its integer part gives a prime number?

$$ \phi= 0,a_0a_1a_2a_3\cdots$$ $$ \lfloor 10^{2n}\phi \rfloor \in \mathbb P,\quad \forall n \in \mathbb N$$

Or in a more general way multiply by $10^{p.n}$, where p is a fixed prime number.

For example, let $\phi = 0,1163\cdots$ $$\lfloor 10^2 \phi \rfloor = 11$$ And $$\lfloor 10^4 \phi \rfloor = 1163$$ While 11 and 1163 are primes, 63 by itself is not. So, $a_{2i}a_{2i+1}$ is not necessarily prime for $i \in \mathbb N$

@RoddyMacPhee 25 but i think it is irrelevant. Can you be more spesific about your question? — okstrok, Sep 28 '19 at 13:19
you have to find a non repeating string of them. how many strings of them are there ? — , Sep 28 '19 at 13:20
@RoddyMacPhee not the string of primes <100. For example phi can be something like 0,1163... when you multiply by 100 you get 11 and when you do that again you get 1163 both of which are primes while 63 is not — okstrok, Sep 28 '19 at 13:23
@okstrok, I suggest editing your question to include that great example to help readers understand what you mean. — Joe, Sep 28 '19 at 13:30
@Joe Done. Thanks for feedback. I hope it is more clear now. — okstrok, Sep 28 '19 at 14:03
The "Related" column on the right of my screen has a very closely related question. — Lee Mosher, Sep 28 '19 at 14:14
Here is a number generating $11$ primes: $0.0307032101271101875723\ldots$. — Christian Blatter, Sep 28 '19 at 15:42
@ChristianBlatter I found something similar with brute force up to 28 digits but my question is not about finitely many digits — okstrok, Sep 28 '19 at 21:47
still somewjat relates to what I sa8d but coprimes counted by $\varphi(100)$ — , Sep 29 '19 at 14:16

score 1 · Accepted Answer · answered Oct 02 '19 at 15:30

This is similar to right-truncatable primes, but removing two decimal digits at a time.

Equivalently, we are looking for right-truncatable primes in base $100$. http://oeis.org/A076586 tells us that there are exactly $9823399067$ such primes. Hence there is no infinite example and no such irrational number.

For the more general question of multiplying by $10^{p\cdot n}$, we can note that the natural density of primes is $0$, and that arbitrarily large prime gaps exists, to conclude that the number of right-truncatable primes in any given base is almost certainly finite. However, I don't believe a proof exists. See for example Proof for finite number of truncatable primes .

Prime Generating Irrational Number

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