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October 17

Names of n-1 dimensional parts of Pascal's simplexes

The points in Pascal's line are simply points. (Pascal's line, the 1-dimensional version, is simply 1-1-1-1-1-1-1-1-1-1-1-1-1-1 infinitely long. The 0-dimensional version, Pascal's point, is simply the number 1.)

The lines in Pascal's triangle are rows. The triangles in Pascal's tetrahedron are layers.

What are the tetrahedrons in Pascal's pentachoron?? What are the pentachorons in Pascal's hexateron?? And so on. Georgia guy (talk) 14:59, 17 October 2024 (UTC)[reply]

While we do have an article on Pascal's simplex, it's completely unsourced and almost all the work of a single editor. So there's really no evidence that there's any kind of standard terminology for these ideas. The name "Pascal's triangle" is well-established, but higher and lower analogues not so much. The good news is that you can define your own terminology with little chance of confusing anyone by going against existing usage. I personally don't see anything wrong with using "layers" for the tetrahedron and above. In some contexts you can talk about "slices" of multidimensional objects, and that may be applicable here as well. Either way, you'd need to define your terms to be clear about what you're talking about. --RDBury (talk) 16:23, 19 October 2024 (UTC)[reply]

October 21

52nd perfect number

How many digits (I want an exact figure) does the 52nd perfect number have?? Georgia guy (talk) 13:11, 21 October 2024 (UTC)[reply]

If you read the perfect number article you will see that only 51 perfect numbers are known. So nobody knows. 196.50.199.218 (talk) 13:38, 21 October 2024 (UTC)[reply]

Please, I learned this morning that a new perfect number has been discovered. Georgia guy (talk) 13:41, 21 October 2024 (UTC)[reply]

Although a possible 52nd Mersenne prime has been discovered, its primality has not been ascertained and its identity has not been released, so we cannot construct a perfect number from it yet. Also, after the 48th Mersenne prime, we get into unverified territory, meaning that there may be additional Mersenne numbers between the Mersenne primes we know about that are also prime, but that we missed. GalacticShoe (talk) 13:42, 21 October 2024 (UTC)[reply]

It was revealed this morning to be prime. Georgia guy (talk) 13:44, 21 October 2024 (UTC)[reply]

Well, do you have the value of

n

that they found produces the new prime

2^{n}-1

? If so then the number of digits is going to be

\lfloor \log _{10}(2^{n-1}(2^{n}-1))\rfloor +1=\lfloor (n-1)\log _{10}(2)+\log _{10}(2^{n}-1)\rfloor +1\approx \lceil (2n-1)\log _{10}(2)\rceil

. GalacticShoe (talk) 13:53, 21 October 2024 (UTC)[reply]

GalacticShoe, I don't want a formula; I want an answer; I believe it's more than 80 million but I want an exact figure. Georgia guy (talk) 13:55, 21 October 2024 (UTC)[reply]

I see someone has updated the Mersenne prime page with the value

n=136279841

. If you plug that into the formula I provided, you get

82048640

digits. GalacticShoe (talk) 14:00, 21 October 2024 (UTC)[reply]

@GalacticShoe: I added your figure to List of Mersenne primes and perfect numbers. Still need the digits of the perfect number, though. :) Double sharp (talk) 14:29, 21 October 2024 (UTC)[reply]

Thanks, Double sharp. Unfortunately, I don't think my computer could handle that kind of number so I'll have to deign to someone else for this one :) GalacticShoe (talk) 14:41, 21 October 2024 (UTC)[reply]

Well, we only need the first six and last six digits for consistency in the table. Wolfram Alpha is giving me 388692 for the first six digits, and it must end in ...008576 by computing modulo 10⁶.

And now I realise that the GIMPS press release links to a zip file containing the perfect number as well. Oops. Well, nice to know for sure that the above is correct. Double sharp (talk) 14:52, 21 October 2024 (UTC)[reply]

Now that I think further, it's actually pretty simple to find the first 6 digits, since all you have to do is take

136279841

, plug it into

a=(2n-1)\log _{10}(2)

to get the approximate base-10 exponent of the perfect number, then find the first six digits of

10^{a-b}

where

b

is an integer offset that allows us to scale the perfect number down by an arbitrary power of 10. Doing so with

b=82048634

yields the aforementioned

388692

. GalacticShoe (talk) 15:08, 21 October 2024 (UTC)[reply]

Using home-brewed routines, I get 3886924435...7330008576. I can produce some more digits if desired, up to several hundreds, but not all 82048640 of them. --Lambiam 17:01, 21 October 2024 (UTC)[reply]

October 25

Why does splitting extension field’s elements into several subfields doesn’t help solving discrete logs despite it helps computing exponentiations and multiplications ?

Let’s say I have 2 finite fields elements $A$ and $B$ in $GF_{p}^{6}$ having their discrete logarithm belonging to a large semiprime' suborder/subgroup $s$ such as $p>s$ .

$A$ and $B$ can be represented as the cubic extension of $GF_{p}^{2}$ by splitting their finite field elements. This give $A.x\in GF_{p}^{2}$  ; $A.\in GF_{p}^{2}$  ; $A.z\in GF_{p}^{2}$  ; and $B.x\in GF_{p}^{2}$  ; $B.y\in GF_{p}^{2}$  ; $B.z\in GF_{p}^{2}$ . This is useful for simplifying computations on $A$ or $B$ like multiplying or squaring by peforming such computations component wise. An example of which can be found here : https://github.com/ethereum/go-ethereum/blob/24c5493becc5f39df501d2b02989801471abdafa/crypto/bn256/cloudflare/gfp6.go#L94

However when the suborder/subgroup $s$ from $GF_{p}^{6}$ doesn’t exists in $GF_{p}^{2}$ , why does solving the 3 discrete logarithm between each subfield element that are :

dlog of $A.x$ and $B.x$
dlog of $A.y$ and $B.y$
dlog of $A.z$ and $B.z$

doesn’t help establishing the discrete log of the whole $A$ and $B$  ? 82.66.26.199 (talk) 13:30, 25 October 2024 (UTC)[reply]

Supposing that you can solve the discrete log in GF(q), the question is to what extent this helps to compute the discrete log in GF(q^k). Let g be a multiplicative generator of

GF(q^{k})^{\times }

. Then Ng is a multiplicative generator of

GF(q)^{\times }

, when N is the norm map down to GF(q). Given A in

GF(q^{k})^{\times }

, suppose that we have x such that

NA=Ng^{x}

. Then

Ag^{-x}

belongs to the kernel of the norm map, which is the cyclic group of order (q^k-1)/(q-1) generated by g^{q-1}. Therefore it is required to solve an additional discrete log problem in this new group, the kernel of the norm map. When the degree k is composite, we can break the process down iteratively by using a tower of norm maps. If (a big if) each of the norm one groups in the tower has order a product of small prime factors, then Pohlig-Hellman can be used in each of them. Tito Omburo (talk) 14:53, 25 October 2024 (UTC)[reply]

And when the order contains a 200‒bits long prime too large for Pohlig‑Hellman ? 82.66.26.199 (talk) 15:39, 25 October 2024 (UTC)[reply]

Well, the basic idea is that if k is composite, then the towers are "relatively small", so they would be smoother than the original problem, and might be a better candidate for PH than the original problem. It seems unlikely that a more powerful method like the function field sieve would be accelerated by having a discrete log oracle in the prime field. The prime field in that case is usually very small already. For methods with p^n where p is large, an oracle for the discrete log in the prime field also doesn't help much (unless you can do Pohlig-Hellman). Tito Omburo (talk) 16:06, 25 October 2024 (UTC)[reply]

October 29

If the white amazon (QN) in Maharajah and the Sepoys is replaced by the fairy chess pieces, does black still have a winning strategy? Or white have a winning strategy? Or draw?

QNN (amazon rider in pocket mutation chess, elephant in wolf chess)
QNC (combine of queen and wildebeest in wildebeest chess)
QNNCC (combine of queen and “wildebeest rider”)
QNAD (combine of queen and squirrel)
QNNAD (combine of amazon rider and squirrel)
QNNAADD (combine of queen and “squirrel rider”)

218.187.64.154 (talk) 17:38, 29 October 2024 (UTC)[reply]

October 30