06:00 - 14:0014:00 - 20:00

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

@Grešnik haha, not any time soon though ;)
@MrPie time to checkmark the answer :P

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

"only" :)

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

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

2:15 PM
Okay, but first I need to have some new results, I basically only have ideas and conjectures.

just assume your conjectures are true
and then write in a notation that nobody understands (yet)

hahah that´s a strategy!

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

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

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

2:26 PM
What is NP-hard to determine exactly?
In that problem.

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...

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*

There are many undecidable problems under ZFC.

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.

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.

@TheSimpliFire yup, checkmarking now ;)
I was watching 'Death Note'

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

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

@Grešnik haha
That goes longer than any movie ever made

2:51 PM
I used to put it on background

Depends on the match, some really are long.

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

Why you don´t watch it on TV?

because other family members wanted to watch something else :P

@TheSimpliFire imagine if you had 14159 rep ;)

2:54 PM
hehe

Did you guys watch THE LION KING (1994)?

yeah

The new one comes out in 6 days

And did you watch all the TRANSFORMERS?

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.

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

Maybe, I am lucky with ECM

That one looks like it SHOULD be divisible by 7.

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

Then it just ends in 7?

3:16 PM
@Grešnik The exponents are distinct. There is no small factor.

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

it will end in 9
@Grešnik It has 124 digits

But why it will end in 9?

This is the decimal expansion of the 125-digit number above :
14977759353271484316515730576360258371517529622286995752615458818535120089616673
280629182422244615480405974473791137416738897

Nvm, 67^68 ends in 1
so the sum will end in 7

3:21 PM
Enzo Creti could run yafu, but he has been banned :(

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

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

I cannot find some table there.

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

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

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.

@Haran You may want to try this as well

@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 ?

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

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.

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

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.

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

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"

More than polynomial with what degree?

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

4:09 PM
I do not understand, is the degree a function of the number of digits of n? Or it shouldn´t be?

4:31 PM
@Grešnik what sin did you commit? :P

Haha, some of them surely did.

@Mathphile Hi, do you have yafu ?

@Peter no

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

Yes, what?

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.

Time or steps involved?

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

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

it can, but it is much slower
And this number seems to be a hard case.

is it a semiprime with big prime factors?

4:44 PM
But if the number is n you need to do only n-2 divisions maximally to find a factor?

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.

no factor till now on pari
any reason you try to factor this number?

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.

any semiprime of this form?

Proven that cannot be prime for odd n?

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

I do not see some crucial difference between odd and even n.
@Mathphile Do you know anything in Python?

@Grešnik a bit
i mainly program in C++

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)

5:15 PM
did you write this?

Of course not.

i am getting an error
do you have the file "factors.txt"?

Nope. :D

i think that might be the reason for the error
although i am not sure as i haven't done file handling in python

5:33 PM
Not a problem.
Can you give some screenshot of your game?

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?

@Mathphile I am back, sorry
concat concatenates vectors, right ?

6:02 PM
i'm not sure
let me read the function information

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.

@Peter okay thank you
@Grešnik here are 2 more pictures of the game

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

@Mathphile Did you program the grass/flowers?
@MrPie See above shots ^^

@Mathphile i think that you´re expert.

6:13 PM
@TheSimpliFire well it was more of 3d modelling instead of programming
but a bit of easy coding is required

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

@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?

How much have you tested this?
And what role does k play in the equation?

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

This is very simple

6:25 PM
@TheSimpliFire seems to work upto 10k

concat(a,b)-b=concat(a, bunch of zeros)
and ab + a = a(b + 1) = a * 10^k
qed

i feel dumb now lol

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

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

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

6:35 PM
If that´s not too personal question?
What are the major differences between them, if any?

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

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".

No, you need to get past the interviews as well

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?

7:05 PM
They probably interview something like 500/10000 applicants

@Mathphile beautiful!
The graphics look wonderful! :D

@Mathphile seems to be our future of game development.

06:00 - 14:0014:00 - 20:00