TheSimpliFire's Chatroom - 2019-07-10 (page 2 of 2)

2:02 PM

@YODA In non-precise terms, integral of the derivative of f is just f itself (+ some constant)

@TheSimpliFire You could become one of the next leaders in UK mathematics, it´s been much time since Hardy was active. :D

TheSimpliFire

@Grešnik haha, not any time soon though ;)

@MrPie time to checkmark the answer :P

Grešnik

Haha, well, you only need to beat Heath-Brown and Wiles, if Number Theory is your major interest.

TheSimpliFire

"only" :)

Grešnik

Yes, and if you do that I could come in a visit with my notebooks and die there, like Ramanujan hahahah! :(

TheSimpliFire

You could come in a visit with your notebooks and not die :P

Grešnik

2:15 PM

Okay, but first I need to have some new results, I basically only have ideas and conjectures.

TheSimpliFire

just assume your conjectures are true

and then write in a notation that nobody understands (yet)

Grešnik

hahah that´s a strategy!

TheSimpliFire

Here's one: $$\phi_{\epsilon-\kappa_{|\mathfrak p|^x}}^{\Sigma^{-a-tx\sum\limits_{i=k}^{k_a}}\odot t_{\eta}}=\,\,\,\style{display: inline-block; transform: rotate(90deg)}{\beta}$$

a shame the margin's too small for me to write a proof

Grešnik

Yes, the problem is never in us, it´s the margines we have to blame for.

You know about Mochizuki, right?

TheSimpliFire

yes

the abc conjecture

@Peter Do you think the problem is NP-hard?

Grešnik

2:26 PM

What is NP-hard to determine exactly?

In that problem.

TheSimpliFire

either one of the two questions

though it is much easier to check for a cyclic prime than to find one... but then again if there are infinitely many...

Grešnik

No matter the hardness, I think that all the mathematical problems, stated clearly and unambiguosly, so that they are either true and false, can be proved to be true or false, under some set of axioms that is weaker than the problems itselves.

true or false*

TheSimpliFire

There are many undecidable problems under ZFC.

Grešnik

Of course there are, ZFC axiomatizes only some "part" of mathematics, there are other set theories.

ZFC seems to be even not enough to settle CH, by some works of Godel and Cohen.

Peter

2:46 PM

@TheSimpliFire You mean : "Is deciding whether the abc-conjecture is true NP-hard" ? The abc-conjecture could be independent of ZFC.

I do not know whether "NP-hard" makes sense for the difficulty of deciding some statement.

Mr Pie

@TheSimpliFire yup, checkmarking now ;)

I was watching 'Death Note'

Grešnik

@Peter No, he asked that about his problem with lists of primes.

TheSimpliFire

A shame I don't have enough time to watch movies nowadays...

Grešnik

Watch Wimbledon instead?

Mr Pie

@Grešnik haha

That goes longer than any movie ever made

TheSimpliFire

2:51 PM

I used to put it on background

Grešnik

Depends on the match, some really are long.

TheSimpliFire

but then I kept checking every 10 secs so I ended up doing nothing except watching matches xD

Grešnik

Why you don´t watch it on TV?

TheSimpliFire

because other family members wanted to watch something else :P

Mr Pie

@TheSimpliFire imagine if you had 14159 rep ;)

TheSimpliFire

2:54 PM

hehe

Grešnik

Did you guys watch THE LION KING (1994)?

Mr Pie

yeah

TheSimpliFire

yes

Mr Pie

The new one comes out in 6 days

Grešnik

And did you watch all the TRANSFORMERS?

Peter

3:03 PM

@TheSimpliFire O sorry, misunderstanding. Determining the sum as well as the factorization will be hard at some point. So, the problem can well be NP-hard.

Can someone download yafu and factor my above number with "siqs(66^67+67^68)" ? With my computer, I guess, it would take me two weeks or more. With a fast computer, maybe only two days.

TheSimpliFire

I'm currently running quite a big project in RStudio so I'm not sure if my laptop is fast enough to run that as well

Peter

Maybe, I am lucky with ECM

Grešnik

That one looks like it SHOULD be divisible by 7.

TheSimpliFire

66 = 3 mod 7 and 67 = -3 mod 7 and 4 * 3^67 is clearly not divisible by 7

Grešnik

Then it just ends in 7?

Peter

3:16 PM

@Grešnik The exponents are distinct. There is no small factor.

Grešnik

But can that be done with paper and pencil "quickly"?

TheSimpliFire

it will end in 9

@Grešnik It has 124 digits

Grešnik

But why it will end in 9?

Peter

This is the decimal expansion of the 125-digit number above :

14977759353271484316515730576360258371517529622286995752615458818535120089616673
280629182422244615480405974473791137416738897

TheSimpliFire

Nvm, 67^68 ends in 1

so the sum will end in 7

Peter

3:21 PM

Enzo Creti could run yafu, but he has been banned :(

Grešnik

Is there any page on the internet where all the numbers are factored, at least up to 10000?

Peter

@Grešnik factordb is a database of many numbers with factorizations. Upto $10^4$, all factorizations should be there.

Grešnik

I cannot find some table there.

Peter

You can either enter a single number or an expression depending on n

Grešnik

Okay, but I was thinking that there is some table somewhere.

Peter

3:29 PM

@Grešnik If you enter "n" in the box , you see the first few factorizations, you can increase the number of numbers displayed upto 200, then click "next page" to continue.

TheSimpliFire

@Haran You may want to try this as well

Peter

@Grešnik I would not be surprised, if someone would have displayed somewhere the factorizations upto $10^6$ or so. But I am not aware of such a table. Are you just curious or do you need such a table ?

Grešnik

@Peter I was just thinking of investigating why the "factorization problem" is considered "hard"?

Peter

it is only hard for large numbers, very small numbers can even be factored by trial division.

and with some luck, even large numbers can be factored.

Grešnik

But that´s why I need a table, to see more clearly what and where are the issues?

Peter

3:38 PM

Clearly, trial division is useless at some point. So we need other algorithms.

Small factors (upto 30 digits) can be efficiently found with ECM, numbers upto about 100 digits can be relatively fast factored with SIQS, but no efficient method is known for arbitary numbers. The best known are subexponential, but worse then polynomial.

The problem is not to find the factors in a finite time (which is obviously possible) , but to find them in a reasonable time.

Grešnik

So, it isn´t known can n be factored in P(n) steps? P is a polynomial

With fixed degree.

Peter

3:54 PM

not polynomial with n , but with the number of digits of n.

The time needed should not grow more than polynomial with the input length

this would be "efficient"

Grešnik

More than polynomial with what degree?

Peter

With any fixed degree. We know no method which does not grow more than a polynomial with some fixed degree of the input length.

Grešnik

4:09 PM

I do not understand, is the degree a function of the number of digits of n? Or it shouldn´t be?

Mathphile

4:31 PM

@Grešnik what sin did you commit? :P

Grešnik

Haha, some of them surely did.

Peter

@Mathphile Hi, do you have yafu ?

Mathphile

@Peter no

Peter

@Grešnik Let us start from beginning. What is meant with an "efficient" (polynomial) algorithm ?

Grešnik

Yes, what?

Peter

4:39 PM

It is meant that the worst case running time grows not more than a polynomial (which can have any degree) with the input length as tha argument.

Grešnik

Time or steps involved?

Peter

So, if we would need at most l^3 steps when l is the input length, this would be efficient (polynomial)

We can usually also take the steps.

@Mathphile Because I try to factor 66^67+67^68

Mathphile

@Peter i presume pari will not be able to factor this?

Peter

it can, but it is much slower

And this number seems to be a hard case.

Mathphile

is it a semiprime with big prime factors?

Grešnik

4:44 PM

But if the number is n you need to do only n-2 divisions maximally to find a factor?

Peter

I do not know whether it is a semiprime, but it very likely has no prime factor with less than 30 digits

@Grešnik Yes, but the definition is with the input length.

Let us say, a 200 digit number would need twice the time to factor than a 100 digit number (and also with larger numbers). This would be a linear growth (linear with the input length)

If it would take four times longer, it would be quadratic etc.

@Mathphile Chances that the number is semiprime are good, but we only know it when we know the factorization.

Mathphile

no factor till now on pari

any reason you try to factor this number?

Peter

It is the smallest number of the form $$n^{n+1}+(n+1)^{n+2}$$ for which no factor is known in the factoring database "factordb"

By the way, I wonder why the expression cannot be prime for odd n , which apparently is the case.

Mathphile

any semiprime of this form?

Grešnik

Proven that cannot be prime for odd n?

Peter

4:59 PM

@Grešnik No, so far a conjecture.

In factordb I found no odd case without a known factor

- FF 1 1 (3)^2 = (3)^2
- P 2 2 89 = 89
- FF 3 4 1105 = 5 · 13 · 17
- P 4 5 16649 = 16649
- FF 5 6 295561 = 7 · 42223
- P 6 7 6044737 = 6044737
- FF 7 9 139982529 = 3 · 46660843
- FF 8 10 3621002129<10> = 1249 · 2899121
- FF 9 12 103486784401<12> = 7 · 11^2 · 13 · 157 · 59863
- P 10 13 3238428376721<13> = 3238428376721<13>
- FF 11 15 110131633755793<15> = 13 · 8471664135061<13>

The first factorizations. We have also some semiprimes.

- P 2 2 89 = 89
- P 4 5 16649 = 16649
- P 6 7 6044737 = 6044737
- P 10 13 3238428376721<13> = 3238428376721<13>
- P 392 1023 393^394+392^393<1023> = 1552639658...61<1023>
495 numbers not shown. Reasons: 89 times fully factored 381 times incompletely factored 25 times composite

five primes in the range [1,500]

And the cases without a known factor :

5 C 66 125 67^68+66^67<125> = 1497775935...97<125>
5 C 128 275 129^130+128^129<275> = 2387230121...09<275>
5 C 164 369 165^166+164^165<369> = 1268524961...49<369>
5 C 178 406 179^180+178^179<406> = 3269149076...93<406>
5 C 190 438 191^192+190^191<438> = 9104162450...81<438>
5 C 192 444 193^194+192^193<444> = 2505779361...61<444>
5 C 198 460 199^200+198^199<460> = 5907658431...13<460>
5 C 232 554 233^234+232^233<554> = 9161555295...61<554>
5 C 233 557 234^235+233^234<557> = 5839924941...53<557>

@Grešnik Oh, I just see there are also odd cases without a known factor, so my conjecture might be false.

Maybe, the expression can also be prime for odd n

Grešnik

I do not see some crucial difference between odd and even n.

@Mathphile Do you know anything in Python?

Mathphile

@Grešnik a bit

i mainly program in C++

Grešnik

Can you try to execute this:

def get_factors(n):
factors = {2:[2]}
primes = [2]
for i in range(3,n):
factors[i] = []
for p in primes:
if i%p == 0:
factors[i].append(p)
factors[i].extend(factors[i/p])
break
if not factors[i]:
factors[i].append(i)
primes.append(i)
print("factored")
file = open("factors.txt","w")
for k in factors:
file.write(str((k,factors[k])))
file.write("\n")
file.close()
print("done")
get_factors(1000000)

Mathphile

5:15 PM

did you write this?

Grešnik

Of course not.

Mathphile

i am getting an error

do you have the file "factors.txt"?

Grešnik

Nope. :D

Mathphile

i think that might be the reason for the error

although i am not sure as i haven't done file handling in python

Grešnik

5:33 PM

Not a problem.

Can you give some screenshot of your game?

Mathphile

5:46 PM

for(a=1, 1000, for(b=1, 1000, if((a*b+a+b)==concat(a,b), print([a, b]))))

@Peter i think my code isn't working

i am trying to find $a, b$ such that $ab+(a+b)=$concat$(a, b)$

$9 \times 9 + (9 + 9) = 99$ is an example but i am not getting it as an output

@Grešnik okay

@Peter you there?

Peter

@Mathphile I am back, sorry

concat concatenates vectors, right ?

Mathphile

6:02 PM

i'm not sure

let me read the function information

Peter

For Enzo Creti, I wrote a function that concatenates the digits of two numbers.

c(zahl1,zahl2)={zahl1*10^length(digits(zahl2))+zahl2}

With "a*b+a+b==c(a,b)" , you should get what you want.

Mathphile

@Peter okay thank you

@Grešnik here are 2 more pictures of the game

Peter

@Mathphile By the way, the solutions of the equation a * b + a + b = c(a,b) can easily be found out without a computer.

TheSimpliFire

@Mathphile Did you program the grass/flowers?

@MrPie See above shots ^^

Grešnik

@Mathphile i think that you´re expert.

Mathphile

6:13 PM

@TheSimpliFire well it was more of 3d modelling instead of programming

but a bit of easy coding is required

TheSimpliFire

So at least you didn't need to write code for making each blade of grass/each leaf of a tree

Mathphile

@Grešnik nah, i far from an expert

@TheSimpliFire no lol

@TheSimpliFire that's where the coding comes in

hmm

if $b$ is a number of the form $10^k-1$, and $a, k \in \Bbb{N}$, then $ab+(a+b)=$concat$(a,b)$

can we prove this?

TheSimpliFire

How much have you tested this?

And what role does k play in the equation?

Mathphile

Ex. $2 \times 9 +(2+9) =29$

TheSimpliFire

This is very simple

Mathphile

6:25 PM

@TheSimpliFire seems to work upto 10k

TheSimpliFire

concat(a,b)-b=concat(a, bunch of zeros)

and ab + a = a(b + 1) = a * 10^k

qed

Mathphile

i feel dumb now lol

TheSimpliFire

Nah, that's how you make good conjectures, you start from basic things and build up

Grešnik

So, would you more like to be at Cambridge or at Oxford?

TheSimpliFire

@Grešnik I wouldn't mind; I'd be lucky if I got an offer by either of them!

Grešnik

6:35 PM

If that´s not too personal question?

What are the major differences between them, if any?

TheSimpliFire

ox.ac.uk/admissions/undergraduate/applying-to-oxford/teachers/…

From Oxford, so most likely to be slightly biased towards them

Oxford–Cambridge rivalry

Rivalry between the Universities of Oxford and Cambridge is a phenomenon going back many centuries. During most of that time, they were the only two universities in England and Wales, making the rivalry more intense than it is now. The University of Oxford and the University of Cambridge, sometimes collectively known as Oxbridge, are the two oldest universities in the United Kingdom. Both were founded more than 800 years ago, and between them they have produced a large number of Britain's most prominent scientists, writers and politicians, as well as noted figures in many other fields. Yet for...

@Grešnik Are you a graduate student by the way?

Grešnik

What do you mean by "an offer"? Suppose that you have good grades at the high school, and that you go at one of those two colleges (or both) and write some preliminary test and have a good score, isn´t that enough? No, math is my "hobby".

TheSimpliFire

No, you need to get past the interviews as well

Grešnik

Well, that seems good, if they also want to talk with you before they make the decision, I think.

How they can talk with every single applicant, what if 10 000 people come to write a preliminary test for mathematics?

TheSimpliFire

7:05 PM

They probably interview something like 500/10000 applicants

Mr Pie

@Mathphile beautiful!

The graphics look wonderful! :D

Grešnik

@Mathphile seems to be our future of game development.

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$Mathematics$ TheSimpliFire's Chatroom