
What is the largest bi-truncatable prime? - headalgorithm
https://www.primepuzzles.net/puzzles/puzz_950.htm
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Someone
So, is there a truly largest one?

Let’s assume we have B(n) bitruncatable primes of length (n)

For B(n+2), we have at most 45 times that number (digit added at the front
can’t be a zero; digit added at the back must be odd)

The fraction of them that’s prime is about 2/log(10^(n+1)) (factor 2 added
because we already dropped all even numbers; n+1 in the denominator as being
halfway between n and n+2 digits; neither factors affect the conclusion)

That gives us the recurrence relation

    
    
        B(n+2) ~= 90B(n)/log(10^(n+1))
                = 90B(n)/((n+1)log(10))
                = (90/log(10)) B(n)/(n+1)
    

That number gets smaller and smaller once n > 90/log(10) (about 39)

So, I think it’s very likely there is a largest one.

This line of thought applies to all integer bases > 2 (there isn’t a largest
one in base 2 because there isn’t any in base 2)

~~~
sotsoguk
I think B(n+2) is at most B(n)*36. you can’t append a 5 to get a prime number.

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misja
Am I overlooking something, or is it obvious that if this is the single
largest bi-truncatable prime known so far, that any larger one has to contain
this prime in its middle?

So finding a new largest bi-truncatable prime is just a matter of trying to
add each single digit at both sides of this prime; if any of them is again a
prime, we have a new largest and if none of them is (which I guess is the case
because the author must have tried this out already), then there exists no
larger bi-truncatable prime.

~~~
dmurray
If by "this" you mean the 83-digit prime in the question, then no. There's no
indication that there are no other 83-digit bitruncatable primes - that's just
the only one the author had found.

If you mean the sole 97-digit solution given in the comments, the implication
that all the bitruncatable primes have been found and enumerated by your
method, and the assertion that there are no such 99-digit primes - then yes,
that looks pretty legit, though I'd want to see an independent verification
before accepting it as a proof, since computer programs have bugs.

~~~
jwilk
> I'd want to see an independent verification

I wrote a solver for this problem: [https://github.com/jwilk/bitruncatable-
primes](https://github.com/jwilk/bitruncatable-primes)

I haven't run it yet, because I don't have a powerful-enough machine at hand.
It needs ~6 GB of RAM and ~20 CPU core-hours.

~~~
jwilk
After fixing exorbitant memory use, I have now independenty verified that

7228828176786792552781668926755667258635743361825711373791931117197999133917737137399993737111177
(97 digits)

is indeed the biggest bi-truncatable prime.

This was done under assumption that zero digits are not allowed. If they are
allowed, bigger bi-truncatable primes exists, such as:

90072457733413689120801410250233316614403998951220231333193991731791997911317971131797197339199333933
(101 digits)

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tomxor
I find it difficult be interested in these, primes are primes; bi-truncatable
are base-10 specific. It feels more like numerology in the same way that we
assign meaning to powers of 10 and the natural world does not.

~~~
tedunangst
"What's the smallest power of pi that spells your name when encoded in
base64?"

~~~
enriquto
There's probably an infinite set of such powers inside any neighborhood of
zero.

~~~
hervature
I was going to say, that irrationality would mean that any power of pi would
contain your name given enough digits.

~~~
Someone
That logic is incorrect. The base 2 Liouville number
([https://en.wikipedia.org/wiki/Liouville_number](https://en.wikipedia.org/wiki/Liouville_number))
is transcendental, so surely irrational, but mostly contains two repeated
characters.

