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12:20 AM
I have no idea
unless you're asking me to see if I can pull off better code @TheSimpliFire
6 hours later…
6:43 AM
16 hours ago, by EuxhenH
by increasing I meant to say non-decreasing
remember that rant you had :P
1 hour later…
8:12 AM
Hello @Peter
Do you know how to check if an integer is a square number in PARI/GP?
this is easy : issquare(n)
Really, there's a function for that?
yes, and even for a perfect power : ispower(n)
8:14 AM
I used to use type(sqrt(n))=="t_INT"
issquare(n) is 1 , if n is a perfect square and 0 otherwise
ispower(n) is greater than 0 , if n is a perfect power
to get the integer part of the square root, use sqrtint(n)
Thanks, how do you check if an integer is a perfect power of 2?
If the numbers are small, you could factor them.
Best is probably to first check whether they are a perfect power and then factor.
Or you create a vector of the powers of 2 and search the number in this vector.
I have two conjectures.
1. $\sum\limits_{i=1}^n p_i!$ will never be a power of $2$ for all $n>3$
2. $1+\sum\limits_{i=1}^n p_i!$ will never be a square for all $n>3$
This is far more efficient :
? s=0;forprime(p=1,10^4,s=s+p!;if(ispower(s)>0,print(p," ",s)))
3 8
5 128
8:25 AM
ah thanks
I never know how to use the s=0, s=s+1 etc structure
1. is almost surely true. Maybe modulo some number, we can even proof it.
Let's try factoring
2. is true, since from prime p = 11 on, the sum is divisible by 3 , but not by 9.
In 1., the sum is apparently not even a perfect power for n > 3
@Peter But what if it is divisible by something like 7^2?
In a perfect square, every prime must have an even exponent.
8:32 AM
You can ask 1. here on the site. I am pretty sure someone can rigorously proof it.
@Peter Note that $\sum\limits_{i=1}^np_i!$ is always divisible by $2^4$ for all $n>3$
(09:30) gp > s=0;forprime(p=1,10^4,s=s+p!;print(p," ",factor(s)))
2 Mat([2, 1])
3 Mat([2, 3])
5 Mat([2, 7])
7 [2, 4; 17, 1; 19, 1]
11 [2, 4; 2495123, 1]
13 [2, 4; 391683923, 1]
17 [2, 4; 5791, 1; 3838863053, 1]
19 [2, 4; 256939, 1; 29676497657, 1]
23 [2, 4; 7523, 1; 214775843577022801, 1]
29 [2, 4; 227, 1; 1327, 1; 1249519, 1; 2477159, 1; 592686876647, 1]
31 [2, 4; 41, 1; 89, 1; 547, 1; 257755136646760884703698641, 1]
37 [2, 4; 47, 1; 43252507, 1; 423163058829697575683900200085287, 1]
41 [2, 4; 33629, 1; 38425469, 1; 16600230379, 1; 97467959793419596716141137, 1]
Extremely large primes
Spectacular factorizations !
Seems that the sum is never divisible by $2^5$
(2! + 3! + 5!) = 128 divisible by 16
7!, 11! etc contain the factors 2, 4, 6 => divisible by 16
OK, yes, I missed this exception, but this is apparently the last one.
8:40 AM
@Peter $\sum\limits_{i=1}^np_i! - 7!$ is divisible by 2^5
Only 7! misses the final factor of 2
All other p_i! contain enough factors of 2 to make 32
Hence the sum cannot be divisible by $2^5$ except for $p_i=5$, right ?
but then it cannot be a power of 2, but it could still be a square (a cube is impossible). If you can rule that out, we can even strengthen 1.
of course, 8 is again an exception.
Yes, a counterexample would be of the form $(2^2\cdot k)^2$
for $p_i\ge 5$, the sum seems to be congruent to 2 modulo 9, excluding a perfect square (and therefore also a fourth power)
8:46 AM
How did you prove that?
It is enough to go to 2!+3!+5!+7! , every additional summand will be divisible by 9
I see
Therefore, we never have a perfect power for n > 3 , nice !
Yes, you proved it :)
Can you run a routine for me ? Your computer surely is faster than mine.
8:49 AM
first the self-defined function
and now
What am I looking for?
did you use $10^{10}$ or $10^{11}$ ? We search for orders of at least 28 to base 7
I did 10^11
Gresnik's problem
8:54 AM
The power summer?
Q: Some interesting calculation puzzle that I made

Leonard GreenlandSo I'm creating some kind math puzzle that goes like this: 1=-2+3 12=3*4 123=(4*5*6)+7+8+9-10-11 1234=(5*6*7*8)-(9+10+11+12+13+14+15+16+(17*18)+19+20+(-21+22)) ...and so on. So how long (or what is the shortest possible) that kind of equation to get 123456789?

@TheSimpliFire @MrPie
nice puzzle by new user!
@TheSimpliFire Since I had no success with larger bases, I fell back to base 7, which gives small digitsums which should give us good chances to get large orders.
I thought we were only working in base 10
@OmegaKrypton Yes, it seems we are allowed Z(+, -, x)
yes, but Gresnik generalized the problem which made the search much more time-consuming (but also more interesting)
8:58 AM
Right, that's fine, I'll leave the program on for a few hours
@OmegaKrypton What exactly are we allowed to do ? Can we only use consecutive integers with + and * ?
idk :/ think so, and also minus
OK, your examples show it.
The first integer is arbitary ?
No computers :)
only a calculator
@OmegaKrypton I found one but off by 300
In this case, we can add the numbers 13717417 to 13717425 , which is already a quite short expression.
9:06 AM
@Peter You solved it!
But I think there is another pattern that must be fulfilled...
The starting digits go 2, 3, 4, 5,...
So for 123456789, we must start with 10 @Peter
(The OP is rather unclear in the rules)
OK, this makes sense. Then, the solution is far more difficult.
@TheSimpliFire constructed one
Summing 10 to 15713 is 207 too high
Adding 10 to 15715 (except 7908 and 7910 , which we subtract) is a solution, but barely the shortest one !
@OmegaKrypton What is your solution ?
9:18 AM
off by 4k lol
Perhaps, it helps that $$123456789=10821\cdot 11409$$
A: Some interesting calculation puzzle that I made

Omega KryptonMy solution, may not be optimal 12345:

@TheSimpliFire hex. found one with max 81
Done with 60
wow post it then!
a sec
@OmegaKrypton @Peter
9:32 AM
posted answer for 12345
@TheSimpliFire Congratulations! This will be hard to beat
What is current best for 12345 ?
please help with 123456
Do realise that you can use brackets...
9:43 AM
@TheSimpliFire how's everything?
Not bad
10:02 AM
@TheSimpliFire Thanks
@TheSimpliFire First one is easy
@Haran Peter solved both of them
The expression is always divisible by $3$ for $n \geqslant 2$
Oh ok
I need to come up with more interesting ones
What prime with odd power did you guys find as factor for the second one
If that is what you did...
10:06 AM
Oh lol
but not to an odd power
How does that solve the problem?
@Haran Because 2!+3!+5!+7! = 2mod 9 and everything after that is =0 mod 9
So the sum is of the form 3k+2
However, no square number is of that form
No it isnt
10:08 AM
The sum will be of the form 3k+1
$S=1+1!+2!+3Q = 3(Q+1)+1$
We don't have 1!
Then it will be of the form $3k$
@Peter Actually, I think we missed the additive term of 1 at the start for the 2nd conjecture
10:10 AM
Even if you missed it, how did you come with $3k+2$
We have 1+2!+3!+5!+7!+11!+... , right ?
Did you also miss the 1!?
@Haran We solved 2! + 3! + ... but not 1 + 2! + 3!+ ...
@Peter Yes, but before we missed the 1
OH only prime indices very well
so the sum is of the form 9k^2
10:12 AM
I did not miss the 1, I checked 1+2!+3!+5!+7!+11! and it is divisible by 3, but not by 9
Ah that is the argument
1 hour ago, by Peter
for $p_i\ge 5$, the sum seems to be congruent to 2 modulo 9, excluding a perfect square (and therefore also a fourth power)
That is what I suggested at first, prime with odd power, here $3^1$
Now that the $P(x) \mid x!$ sum is done, I can now attempt @Grešnik 's problem
What do you mean with "P(x)|x!" is done ?
10:56 AM
@TheSimpliFire Output for my routine ?
[1000000003, 7, 1]
[1000000009, 7, 3]
[1000000027, 7, 4]
[1000000075, 7, 8]
[1000000111, 7, 9]
[1000000243, 7, 14]
[1000037575, 7, 15]
[1000168075, 7, 17]
[1000232437, 7, 19]
[1001093599, 7, 20]
[1002760799, 7, 21]
[1025529479, 7, 23]
[1171989593, 7, 26]
[1410474721, 7, 27]
OK, so still no order larger than 27
not yet
11:14 AM
seems like i missed quite a lot here
@Mathphile Hi, anything new about Gresnik's problem ?
not atm
i have been focusing more on finishing my game atm
11:38 AM
Okay, what are we gonna try to solve today?
I'm currently working on an integral
Q: Difficult to evaluate $\int_{-\infty}^{\infty}\frac{x(x+a)(x+b)(x+c)}{(x^3+2x^2-x-1)^2+(x^2+x-1)^2}dx$

Power rangerWhere $a,b$ and $c$ are consecutive arithmetic terms. We wish to evaluate this integral, $$\int_{-\infty}^{\infty}\frac{x(x+a)(x+b)(x+c)}{(x^3+2x^2-x-1)^2+(x^2+x-1)^2}\mathrm dx$$ I don't even really know how to make an attempt. If I expanded the denominator it is very messy. I can't factori...

Go Go Power Rangers
Go Go Power Rangers
Go Go Power Rangers
Mighty Morphin' Power Rangers!
11:53 AM
@TheSimpliFire guihnsadjibghfiudsh
I'll be busy most of today computing 8th order partial derivatives of the Beta function
@SimplyBeautifulArt ?!
what you need them for?
For computing $\int_0^1\frac1x\ln^n(x)\ln^{8-n}(1-x)~\mathrm dx$
which is for computing $\int_0^1\frac1x\ln^n(x)\ln^{8-n}(1+x)~\mathrm dx$
which are supposedly unsolved for as of yet
I see
Looks like I'm doing a 'less' complicated integral then
I'm also playing a game called battle cats
which is pretty fun
oh yeah do you want to check my work on a couple simple equations though? If you haven't checked my edit to my newest question...
11:58 AM
Q: Which variables can be solved for?

Simply Beautiful ArtWhile answering another question, I stumbled upon a system of equations, given the form: $$C_{a,b,c}=\sum_{n=0}^{\max\{b,c\}}\left[\binom cn+(-1)^a\binom bn\right](-1)^nS_{a+n,b+c-n}$$ with known $C_{a,b,c}$. The motivation for finding when this is solvable is that it allows marking off some ve...

I take it a,b,c are positive integers
The systems of equations suggest some sort of induction
you want to consider the system for a+b+c = k
That'll give you a system of equations in $S_{p,q}$ with $p+q=k$
namely I just wanted you to check my work on k = 8
How do you define $\binom bn$ when $n>b$?
12:14 PM
@TheSimpliFire This is defined as $0$
@OmegaKrypton Got beat with n=38: puzzling.stackexchange.com/a/86007/43767
Clever use of nested brackets
@TheSimpliFire Still no order greater than 27 ?
the chances are still there
Between 26 and 27 there was a 3x10^8 gap
My current record for 27 is smaller than the current record for 26, same for 29 which is smaller than the record for 28
12:34 PM
@SimplyBeautifulArt I notice that $C_{2p-1,4,1}=-2C_{2p,2,2}$
Unfortunately, trying to get a 2x2 system for the unknowns results in a redundant one
Yeah that tends to happen a lot
:P math would seem to prefer me not to be able to solve these integrals
Also I ran into a fickle with these derivatives
I ran into an expression WA agrees with
but when trying to take the limit it doesn't exist
but when I "fudge" the result, I get the right answer
@Peter What are the records for other bases?
$$\frac1{B(x,y)}\frac{\partial^3}{\partial x\partial y^2}B(x,y)=(\psi^{(0)}(x)-\psi^{(0)}(x+1))((\psi^{(0)}(y+1)-\psi^{(0)}(x+y+1))^2\\+\psi^{(1)}(y+1)-\psi^{(1)}(x+y+1))-2\psi^{(1)}(x+y+1)(\psi^{(0)}(y+1)-\psi^{(0)}(x+y+1))-\psi^{(2)}(x+y+1)$$
@SimplyBeautifulArt Why not just plug them into WA/Mathematica directly?
Because WA says the limit doesn't exist lol
or it just fails to do it
I need to take the limit as $(x,y)\to(0,0)$ of the partial derivatives of $B$
12:42 PM
Maybe ask for a contour plot
it just gives nightmarish stuff that only further justifies it not existing x'D
So to take the limit I divide both sides by $x$. Supposing the limit exists and stuff, you can solve this in a couple applications of L'Hopital's rule
Is your goal just to find S[1,7], S[2,6] and S[3,5]?
(should be y+1 inside of B)
@TheSimpliFire No, I'm trying to find the C's
But there comes the finicky part: $$\frac{\psi^{(1)}(1)-\psi^{(1)}(x+1)}{x^2}$$
If I L'H this twice and ignore the consequences, I get $\frac12\psi^{(3)}(1)=3\zeta(4)$
which gives the correct result $$\int_0^1\frac1x\ln(x)\ln^2(1-x)~\mathrm dx=\zeta^2(2)-3\zeta(4)$$
except this limit doesn't actually exist
and I don't really have anything in there to balance it out as far as I can tell
(also we can directly just let y = 0. it's the x -> 0 part that's troublesome)
12:57 PM
@TheSimpliFire In the room "order records" there are many data about this , I am not sure whether the new records 5 505 517 (base 631 , order 25) and 10 209 377 (base 631 , order 27) are already mentioned there.
1:10 PM
@SimplyBeautifulArt I think it's possible to build a recurrence for that
that's what I'm doing right now actually
well not the C's
but the things that makes up the C's
$$C_{2p,2,2}=2(S_{2p,4}-2S_{2p+1,3}+S_{2p+2,2})\\C_{2p-1,4,1}=-3S_{2p,4}+6S_{2p+1,3}-4S_{2p+2,2}+S_{2p+3,1}=-2C_{2p,2,2}$$ as said above
1:27 PM
@Peter I'm afraid I need to stop the program now
It's been 5 hours and no further results
1:57 PM
Anyone knows something about general theory of fields?
Up to a basic level
I must've made a mistake in my calculations earlier
when I write it out with recurrences it works out much neater
woot woot
what does the recurrence look like?
2:13 PM
Let $\partial^{(m,n)}=\frac1{B(x,y+1)}\frac{\partial^{m+n}}{\partial^m\partial^n}B(x,y+1)$.
We have this recurrence, where the subscript denotes partial derivative w.r.t. that variable
and in the linked post on my newest question, this gives the formulas:
where $J(p,q)=\int_0^1\frac1x\ln^p(x)\ln^q(1-x)~\mathrm dx$
this actually isn't so bad if you work everything down to $\partial^{(1,0)}$ and $\partial^{(0,1)}$ now
You get a bunch of difference quotients and stuff, but when written this way, it's easy to see which parts group together
sometimes multiple parts diverge individually, but by Taylor expanding it's very easy to work out the values
4 hours later…
6:41 PM
@Grešnik Did you have a specific question?
3 hours later…
10:09 PM
Hello, I have a question regarding euclidean geometry.

In https://imgur.com/a/kO2W9cm, the author assumes that the point K is between E and N.

How to prove this assumption?

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