Mathematicians find a new class of digitally delicate primes
quantamagazine.org
quantamagazine.org
Whether the thumb and big toes are fingers is something that depends on definitions, but in the strictest sense they are excluded from fingers and toes, and thus humans generally have 20 digits, eight fingers, and eight toes by this definition.
111 = 3
11 = 2
1 = 1
= 0
-1 = -1
-11 = -2
-111 = -3
Using that logic, base 1 would only contain 0, hence it should not be able to expresss more values that 0.
Is your example actually a valid base 1?
[1]: https://en.wikipedia.org/wiki/Decimal
Another prime might NOT be delicate in base 10 but be delicate in base 8.
In all seriousness though, it's fascinating that we know this number exists but have no idea what it is. What a treasure hunt..
I would probably look into using some sort of large integer sieve if I were trying to search for an example. Thinking about it though, I am even unsure whether it is decidable that N is widely digitally delicate.
Your number is broken up into 32-bit coefficients to 2^32 as a compounding power (sum over a_i * 2^32i) -- and then you can do it as tensor(ish) operations, if you fiddle with carry/mod steps in between to keep the coefficients the right size and partial results properly aligned.
(By "any" I mean any arithmetic progression a (mod m) where a and m don't share any common prime factor.)
So, for example, there is a string of a billion consecutive primes, each of whose decimal expansions ends in a billion ones.
In principle you could use the method to find it, possibly the first example would be around 10^(10^10) or so.
The gateway to all of this is the quantitative form of Dirichlet's theorem on prime numbers in arithmetic progressions: given coprime a and m, there are infinitely many primes which are a (mod m). (So, in other words, Shiu's result with "just one prime".) In this case, we know asymptotically how many such primes there are, and we can get upper bounds on the error. The proof is highly non-constructive, and uses complex analysis applied to the so-called "L-functions".
Using a variety of trickery ("sieve methods", "the circle method", etc.) this can then be leveraged into other interesting results.
1. We first prove there are infinitely many primes.
2. Assume the number of primes is finite.
3. Thus, there exists a number that is the product of all primes.
4. Thus, there exists a number that is one more than the product of all primes, let us call this number A.
5. A is larger than any prime, thus different from any prime, thus not a prime.
6. Thus, there is at least one prime that A can be divided by.
7. But by it's very definition as the product of all primes + 1, the remainder of dividing this number by any prime is 1.
8. Thus, A is divisible by no prime, and prime.
9. Contradiction follows from assumption 2, thus assumption is false.
10. There are thus infinitely many primes.
11. There are finitely many number below 10^10^10^10^10
12. Thus, there must at exist infinite primes above 10^10^10^10^10
I don't know any of these numbers, but showing that a contradiction can be derived from assuming that such a number not exist is the common way to do it.
That doesn't sound very normal, probably no numbers you usually think about have these properties, why do mathematicians call them "normal"? Well, the mathematicians have proved that almost all real numbers are normal. It just turns out that we haven't the faintest idea how to find them and we never needed them for anything so we didn't notice how abundant they apparently are.
We aren't even sure if some weird numbers we do know about are normal, Pi looks pretty normal, but we can't prove it is.
This is equivalent to proving that certain slices of some arithmetic successions contain no primes. Not a professional mathematician and I didn't read the paper, but I suspect this is related to prime gaps works by Tao, who is cited in the article.
A prime with N digits has 9N ‘neighbors’.
There are about 10^N/ln(10^N) - 10^(N-1)/ln(10^(N-1)) primes with N digits, so the probability of a N-digit number being prime is about 1/ln(10^N), of it not being prime 1-1/ln(10^N).
So, the question is about the behavior of (1-1/ln(10^N))^9N as N ⇒ ∞. Google gives me a plot that’s essentially zero, so yes, most primes should be digitally delicate.
> Despite finding no specific examples,
Call me stupid, but no, I don’t “recognize” that. :P
Isn't the first digitally delicate prime simply 2? Or are single-digit primes excluded? Or am I missing something terribly fundamental?
edit Or is it that, because we can replace 2 by 3 and still get a prime, it doesn't count? Which is to say, replacing any digit in the number by any other digit, must always result in a non-prime?
I'm considering trivial to be something like all primes less than 100 or all primes divisible by some prime p, which itself is just a much fancier generalization of the claim that 2 is a noteworthy prime for being the only even prime.
I think this is the only set of non-trivial primes proven to be finite, and even then it is technically an infinite set of finites sets of primes.
https://math.stackexchange.com/questions/2289089/has-any-non...
Take the variation seen in https://en.wikipedia.org/wiki/Ulam_spiral#/media/File:Sacks_..., isn't there a Fibonacci spiral also contained?
If that is the case, then finding such a number immediately trivialises the search for the 'largest known prime' - take the largest known prime, tack this number on the back of it, and you now have a new largest prime. I guess you'd get bonus points if somehow it's also Mersenne.
Adding this prime to the end of a number guarantees a composite.
I wonder if that is a property the belongs to primes, or if there are non-trivial composite examples (any number ending in 2 has this property).
Similarly we can say there is a class of primes that have this property, what other classes of numbers also have this property?
This makes no sense. Let X be a number that satisfies your condition and consider the number XX formed prepending X to itself. You're asserting that XX is prime, but it is obviously composite, because it is divisible by X (just like say 2929 is 29*101).
For instance, 23 is semi-sturdy as you can replace 2 by 1 or 3 by 9 and both 13 and 29 are prime.
The interesting question then becomes: how many?
If there existed infinitely many 'semi-sturdy' primes, this would imply that several prime 2-tuples match infinitely many primes, but results in that area are limited AFAIK.
>“The story of mathematical research is that you don’t know beforehand if you can solve a challenging problem or whether it will lead to something important,” Pomerance said. “You can’t decide in advance: Today I’m going to do something valuable. Though it’s great, of course, when things turn out that way.”
In other words: 'Maybe someone, someday, will figure out if this matters. It'll be neat if it does'
It would have to be a pair ending in 9 and 1 so that at least one more digit is different between them (though potentially more if it’s ...999).