For those who enjoy burning cpu cycles !
m1277 = 2 ^ 1277 - 1 is not prime.
It easy to check it with the Lucas-Lehmer test.
But we don't know any of its divisors, which is quite fascinating.
Up to 100 it is relatively simple, if you remove the even numbers, the multiples of 3 (summing up their digits recursively gives 3 6 or 9), and those that end at 5, you end up with 25 numbers out of which 22 (88%) are primes. If you further exclude 49 which you should remember is 7^2 from the multiplication table, and 77 that is obviously divisible by 7, you are left with 22/23 chance a number that passes these exclusion criteria is not 91 and hence it is a prime.
The fixed point here being that if you add up a list of 1 digits, you'll always reach the same number (`sum([1]) = 1`). The best known is probably the hailstone sequence.
The standard divisibility rule for 3, 6 and 9 in base 10 is to sum the digits until you only have one left and check if it's one of those. Here, 5+7=12, 1+2=3, so 57 is divisible by 3.
Math is not my strong suit at all, so I probably won't grok this, but that kind of blows my mind, so I'm curious... how?! That works for any arbitrarily large number?
Math is crazy!... still don't want to study it though!
It is basically because $10 mod 3 == 1$ (as 10 = 3*3 + 1). So if you are in the ring modulo 3, where every number is equal to the remainder of its division by 3, the sum of the digits of the number in its decimal representation equals the number itself (modulo 3), because in that ring 10 is actually 1, so the 10s in the decimal sum become 1s. Ie if n_k is the kth digit of n, you have
Hence, n is divisible by 3 iff $n mod 3 == 0$ iff $(n_0 + n_1 + n_2 + ...) mod 3 == 0$.
Of course, summing up the digits may not give you a 1-digit number, but it gives you a number that you know is divisible by 3 (if the original number is divisible by 3). So you can apply the same idea/process again, summing up the digits of that number, and get another number that is divisible by 3. Repeat until you end up with one digit (hence the recursion mentioned).
When checking whether it is a multiple of some k, you can add/subtract multiples of k without changing the result, and those 99...9 are multiples of both 3 and 9.
I wonder what the underlying human intuition is for 'prime-ness' and why it might break down with larger numbers. Odd numbers in the rightmost position? The 'shape' of the number (phonaesthesia, the bouba/kiki effect)? Maybe they just sort of feel scary?
For n < 100 to be composite you need a factor < sqrt(100) = 10. Rules for 2, 3, 5 are easy to try quickly. That leaves 7, but up to 7*9 you should remember it from multiplication tables. 77 is quite obviously divisible by 11 too, and then it's 7*13 = 91 as the last boss. But I feel that once you realize how special 91 is in that context, you won't forget it again.
The example given in the article is 2^127 - 1, which was historically proved to be prime without computers using a clever method now known as the Lucas-Lehmer test. Your algorithm is not practical for that number.
Or if we can expand quantum superposition algorithms from 2^N states, for quantum circuits with N control qubits, to 2^(T*N) superpositions over T time steps, via some kind of superposition tree recursion. The number of superpositions increasing exponentially for T steps (and then reducing for another T steps) on a single recursive physical circuit.
That is not supported by the physical laws we have, but it is an interesting idea.
The obvious and naive method described above is O(sqrt(N)). For N ~= 2 ^ 127, that is about 2 ^ 64. / The Lucas-Lehmer method described in the article is better (how much better is an exercise for the reader).
You are assuming division itself is an O(1) operation. However, it also scales with the size of the number. So more correct would be to say that this naive method is O(sqrt(N) log(N) log(log(N))).
The HN rule is that anything can be posted if it is accessible to everyone, including via a paywall workaround if it has a paywall that is porous.
The community has converged on this being the least-worst approach after wrestling with the issue for well over a decade, and it's sufficient to helpfully post the archive link without a snarky editorial comment :)
i found out that most articles behind paywalls dont even need the wayback machine to be viewed
if you are using firefox, just enter reading mode and you can read the entire article without popups in your preferred background, text color, font, etc
http://calcoastrails.com/
(factorial(170,141,183,460,469,231,731,687,303,715,884,105,726) + 1)%(170,141,183,460,469,231,731,687,303,715,884,105,727) == 0
My fun fact is that this type of operation (repeatedly applying a child operation until you reach a fixed point) is called persistence.
https://en.wikipedia.org/wiki/Persistence_of_a_number
The fixed point here being that if you add up a list of 1 digits, you'll always reach the same number (`sum([1]) = 1`). The best known is probably the hailstone sequence.
https://en.wikipedia.org/wiki/Collatz_conjecture
I'm partial to multiplicative persistence.
https://www.youtube.com/watch?v=Wim9WJeDTHQ [15m]
What does this part mean? For example 57.
The standard divisibility rule for 3, 6 and 9 in base 10 is to sum the digits until you only have one left and check if it's one of those. Here, 5+7=12, 1+2=3, so 57 is divisible by 3.
Math is crazy!... still don't want to study it though!
Of course, summing up the digits may not give you a 1-digit number, but it gives you a number that you know is divisible by 3 (if the original number is divisible by 3). So you can apply the same idea/process again, summing up the digits of that number, and get another number that is divisible by 3. Repeat until you end up with one digit (hence the recursion mentioned).
123456 = 1 * 100000 + 2 * 10000 + 3 * 1000 + 4 * 100 + 5 * 10 + 6 = 1 * (99999+1) + 2 * (9999+1) + 3 * (999+1) + 4 * (99+1) + 5 * (9+1) + 6
When checking whether it is a multiple of some k, you can add/subtract multiples of k without changing the result, and those 99...9 are multiples of both 3 and 9.
So 123456 is a multiple of 3 (or 9) iff
1 * 1 + 2 * 1 + 3 * 1 + 4 * 1 + 5 * 1 + 6 * 1 = 1 + 2 + 3 + 4 + 5 + 6
is. Apply the same rule as often as you want -- that is, until you only have one digit left, because then it won't get simpler anymore.
That is it. That is all. Pish posh.
If a number is not prime, then it is the product of at least two numbers smaller than itself.
If any of them are larger than its square root, all others must be smaller, or their product would be larger than the candidate prime.
Ergo, just check that the candidate is not evenly divisible by any number equal or lower than its square root.
This reasoning holds, independent of scale.
QED. Check mate. Shazam.
If you have the time.
Or if we can expand quantum superposition algorithms from 2^N states, for quantum circuits with N control qubits, to 2^(T*N) superpositions over T time steps, via some kind of superposition tree recursion. The number of superpositions increasing exponentially for T steps (and then reducing for another T steps) on a single recursive physical circuit.
That is not supported by the physical laws we have, but it is an interesting idea.
https://archive.is/8R0Fq
The community has converged on this being the least-worst approach after wrestling with the issue for well over a decade, and it's sufficient to helpfully post the archive link without a snarky editorial comment :)
In other words, an icon showing whatever-wall status, submitter can add an alternate link, etc.
if you are using firefox, just enter reading mode and you can read the entire article without popups in your preferred background, text color, font, etc