12:20 AM
Let $Q_1$ be the set of numbers that have an even number of odd quotients. Let $Q_2$ be the set of numbers that have an odd number of odd quotients. I conjecture that number of elements in set $Q_1$ is greater than the number of elements in set $Q_2$

2 hours later…
1:55 AM
Unless you are conjecturing both sets are finite, my intuition tells me you can draw a 1-1 correspondence
@ℋolo what's up with you these days anyway

@Quintec The stronger version of this conjecture seems true too
Let $Q_1$ be the set of numbers less than $n\in \Bbb{N}$, that have an even number of odd quotients. Let $Q_2$ be the set of numbers less than $n\in \Bbb{N}$, that have an odd number of odd quotients. I conjecture that number of elements in set $Q_1$ is greater than the number of elements in set $Q_2$ for any $n$.

2:14 AM
@Quintec hi, sadly I am working(programmer), so I have less time for maths(although my plan is still to be a researcher eventually)
@Mathphile you can rather simply show that @Quintec is correct and there is 1 to 1 correspondence between Q1 and Q2
Although idk about the bounded version, which is probably more interesting

@ℋolo by 1 to 1 correspondence do you mean that the number of elements are the same in both the sets?

2:37 AM
Yes. It is both countable infinity

1 hour later…
3:48 AM
hello

Hollo

@ℋolo you're back :P
@Mathphile 1 to 1 correspondence is just like ratio between cardinality of set A and cardinality of set B, being 1:1 so the same.
@ℋolo how is your programming coming along?

If is pretty boring work tbh. Although it does related to maths(my job relates to information retrieval)
Ratio between cardinals is not really... A thing

oh
well, that's how I've interpreted it. but if we want to have a specific definition of 1 to 1 correspondence, we need to include function mapping

Well... It just means... Equal sizes...
I find this pretty nice way to interpret it, don't you think?

3:58 AM
I mean what m to n correspondence is for all non-negative m and n
and then after defining that, we let m=n=1
@ℋolo but yes, you are right :P

M to n map means that we can have a function F such that for every A subset of the range of F, if |A|<=n then |F^{-1}[A]|<=m
If you add "correspondence" we get:
M to n correspondence map means that we can have a function F such that for every A subset of the range of F, if 0<|A|<=n then 0<|F^{-1}[A]|<=m
Which in the special case of n=m=1 we get a nice result

ye

Or rather, F is the map from the start of the line

I'm working on sums of fourth powers lol

Hmm

4:05 AM
I have found heaps of general diophantine formulas
but goddamn fourth powers are really hard
even with one parameter xD

All finite cases are boring, so who cares. For infinite cases, it is just equal to the sum of the cardinals themselves, or the cardinality of the Union of the cardinals
Done :>

yes hahaha
that was neatly said

Thank you, thank you

I am looking at Ramanujan's equation $$(a+b+c)^2+(b+c+d)^2+(a-d)^2=(c+d+a)^2+(d+a+b)^2+(b-c)^2$$ such that $\frac ab = \frac cd$
Of course, Ramanujan's equations always have another fun fact about them: in this equation, if you replace $2$ by $4$, it is still true.

Oops I forgot to add an important condition in my conjecture

4:12 AM
hello @Mathphile

Let $Q_1$ be the set of numbers less than $n\in \Bbb{N}$, that have an even number of odd quotients $\ne 1$. Let $Q_2$ be the set of numbers less than $n\in \Bbb{N}$, that have an odd number of odd quotients $\ne 1$. I conjecture that number of elements in set $Q_1$ is greater than the number of elements in set $Q_2$ for any $n$
@MrPie hi

@MrPie Hmm... You really did study maths while I wasn't here

@Holo any idea how to solve this one?

@Mathphile wdym by quotients?
Like, if a number has a certain amount of quotients. Is that like saying 2 = 4/2 = 6/3 etc or something?

@Mathphile this is too much number theoretical for me

4:18 AM
@MrPie For some integer $x$, all the natural numbers $q$ which satisfy $x=qd+r$ are the quotients of $x$

oh

@ℋolo :\

I assume $0\ge d, r \ge x$?

@MrPie $d\gt 0$ and $r\ge 0$
Oh and $d, r$ must be whole numbers too

okie
welp I got no idea xD

4:23 AM
I should phrase my conjecture better to avoid confusion
Though I don't know how should I do it without making it too long

When is $7x^2-14x+100$ a square number?
hm, I think I can write it as difference of two squares

@MrPie Here's a challenge for you
Is $n!+n+1$ ever a perfect power?

hmmm
I don't think so
$7x^2-14x+100=(4x-7)^2-(3x-7)^2+10^2$

5:28 AM
@Mathphile I have a conjecture
Consider the equation $4a^4+b^4$
Conjecture: the only prime of this form is $5$.

5:42 AM
Did you run a program to verify this?

no

why do you suspect this is true?
It is actually pretty easy to prove this
Did you figure it out? @MrPie

6:07 AM
$4a^4+b^4=(2a^2+b^2)^2-4a^2b^2$
$4a^4+b^4=(2a^2+b^2+2ab)(2a^2+b^2-2ab)$
$2a^2+b^2-2ab\ge 1$ when $a,b \in \Bbb{N}$
And we are done

1 hour later…
7:11 AM
Hello
@Mathphile oh my XD
Well, that was simple
I was trying to find a cubic Diophantine equation and I finally did
for all $a$ and $b$, it follows
$$(18a^3+2b^3)^3+(19a^3+b^3)^3+(21a^3-b^3)^3+3(10a^2b)^3=(28a^3+2b^3)^3$$

7:45 AM
The 15 theorem or Conway–Schneeberger Fifteen Theorem, proved by John H. Conway and W. A. Schneeberger in 1993, states that if a positive definite quadratic form with integer matrix represents all positive integers up to 15, then it represents all positive integers. The proof was complicated, and was never published. Manjul Bhargava found a much simpler proof which was published in 2000.Bhargava used the occasion of his receiving the 2005 SASTRA Ramanujan Prize to announce that he and Jonathan P. Hanke had cracked Conway's conjecture that a similar theorem holds for integral quadratic forms, with...
there are theorems based on numbers: like "The 15 Theorem" and "The 290 Theorem" XD
I just learnt that now lol

8:10 AM
@MrPie nice
@Haran new conjecture to prove
4 hours ago, by Mathphile
Let $Q_1$ be the set of numbers less than $n\in \Bbb{N}$, that have an even number of odd quotients $\ne 1$. Let $Q_2$ be the set of numbers less than $n\in \Bbb{N}$, that have an odd number of odd quotients $\ne 1$. I conjecture that number of elements in set $Q_1$ is greater than the number of elements in set $Q_2$ for any $n$

I starred it

^^starred orig

8:39 AM
oh
that's the third time I have said "oh" today :/
...
...oh

2 hours later…
10:29 AM
Found something cool :)
$$\sqrt{\cfrac{7-\sqrt 5 + \sqrt{6(5+\sqrt 5)}}{9-\sqrt 5 + \sqrt{6(5+\sqrt 5)}}}=\cfrac 14\left\{\sqrt 3 (\sqrt 5 + 1) - (\sqrt 5 - 1)\sqrt{\cfrac{5-\sqrt 5}{2}}\right\}$$

2 hours later…
12:55 PM
@MrPie Re your question: it doesn't equal G as your continued fraction per iteration is monotonically decreasing and ends up being around 0.894 (Python 3), which is past G = 0.91.

Dammit!
Well, thanks for the python code :)

2 days ago, by TheSimpliFire
Python code stolen from Quintec :P

haha, lucky :P

3

I don't know if this conjecture is well know but let me try it : Let $f(x)$ be a continuous , differentiable function on $[0;+\infty[$ with $f(x)\geq 0 \quad\forall x\geq 0$ and such that :$$\int_{0}^{\infty}\frac{\sin(f(x))}{f(x)}dx=\frac{\pi}{2}$$ then the equation $f(x)=x$ have at ...

I saw that
In fact, it's just that and some other question that are there only good ones on the homepage atm :/

1:01 PM
I'm tempted to use Laplace transforms but I know it won't work as f is in the denom too

wait hang on
ur continued fraction in the python code is finite

it's meant to be infinite

But in each iteration the fraction gets smaller

no it doesn't
it oscillates

1:03 PM
Let me check again

it gets smaller then bigger then smaller then bigger, I am pretty sure

Well at 21^4, 23^4, 25^4 it stays the same
0.8940270239874368
It's probably due to the large magnitude of increasing fourth powers, increasing the denominator so there is little variance

I got different results on Wolfram alpha

What are they

1:07 PM
Hmm

That looks much closer to Catalan's constant

Yes
I'm trying to find why they differ

well I just deleted my question on it lol
I'll just leave it as is
more suspenseful that way ;)

I get the same result on my calc

same result as WA or python?

1:11 PM
WA
@MrPie I know why
Python treats a^b as 1. To evaluate a power it is a**b :/

@TheSimpliFire that's why the python is one of the most dangerous snakes

hehe

ok

Now it agrees

okie dokie
time to manually stretch this out

1:20 PM
@MrPie I think I can code this in PARI
e.g.
k=contfracpnqn([1,1,3^4,5^4,7^4,9^4;1,8,16,24,32,40])

is there a concise way to code it to, say, 1001^4 and 8*501?

k[2,1]/k[1,1]
Let's see

ok
up to $101^4$ and $408$ we get: 0.9159193791787675
that is the first four decimal digits of Catalan's constant
I've got a gut feeling it's true

1:36 PM
Got part of it
contfuntop(N)={for(n=1,N,print1((2*n+1)^4","));print((2*N+3)^4);}

it's definitely in between 0.9159 and 0.916
At $119^4; 480$ we get: 0.9159991805301483
and then it oscillates back down. At $121^4; 488$ we get: 0.9159330823081906
that's an average of about 0.915966
first 5 dp of Catalan's constant now

contfun(N)={print("[1,1,");for(n=1,N,print1((2*n+1)^4","));print((2*N+3)^4";");for(n=1,N,print1(8*n","));print(8*N+8"]");}
@MrPie Nah it's ok
Almost there

are you sure it's ok?

maybe do 8*(N+1) instead of 8*N+8?
same thing but less confusing

1:53 PM
@TheSimpliFire i think i´m not gonna do complex integration today

@MrPie I can't figure it out with pari
ok @Masterphile

2:09 PM
@Masterphile I have worked out some interesting conditions , but I could still not solve the k = 1 - case completely.
If n + phi(n) = sigma(n) , then we must have phi(n) > n/2

s 2 must be a factor

If I interrupt pari and ask for values, this can cause errors. Neverhteless, I have a (hopefully) complete list of all solutions n + phi(n) | sigma(n) upto about 1.5 * 10^9. Maybe, you can doublecheck it ? You will have to be patient however.

paste them in another room, i won´t do it now but will try as soon as possible

No, n can be odd, I still could not prove that there is no odd solution.
How can I create a phi-sigma room specially for this purpose ?

i think for that phi(n)>n/2+1 is needed
paste them here, only i am in this room:

https://chat.stackexchange.com/rooms/96072/hilberts-hotel

2:25 PM
The even case is done anyway for phi(n) + n = sigma(n) (only solution is 2), but the odd case is not completely solved.
I think, I ask this. Is there an informative OEIS-entry because users always look at this source ?
@Masterphile

rather ask is 2 only solution

Shall I show my argument for the even case ?

you can

2:55 PM
0

I want to show that the only solution of $$\varphi(n)+n=\sigma(n)$$ for a positive integer $n$ is $n=2$. What I worked out is that we must have $$\varphi(n)>\frac{n}{2}$$ To show this assume $n$ is composite and its prime factors are $$p_1,\cdots,p_k$$ Then, the numbers 1,\frac{n}{p_1},\cdots...

A quite long question !
@TheSimpliFire
@Mathphile
@MrPie
@Haran
@Masterphile

I'll take a shot @Peter

+1
@Peter I don't understand how you have $\varphi(n) \leqslant \frac{n}{2}$
It is true that $\varphi(n)+\sigma(n) \geqslant 2n$ and $\sigma(n)-\varphi(n)=n$
You only have, $2 \varphi(n) \geqslant n \implies \varphi(n) \geqslant \frac{n}{2}$

For even n , we have phi(n) <= n (1-1/2) = n/2

Oh sorry, you have put that $\varphi(n) \geqslant \frac{n}{2}$ as well. I misread it. My bad :P
Ok, so we first need to eliminate all powers of $2$
$\sigma(2^k)-\varphi(2^k)=2^k$
As $\sigma(2^k)$ is odd and $2^k$ is even, we have $\varphi(2^k)=2^{k-1}$ to be odd. This concludes that the only even solution is $2$

3:11 PM
you can assume that n is odd since I have proven the even case. In this case, n = 1 is not a solution and for n > 1 , sigma(n) is odd, hence n a perfect square.

3:22 PM
Yeah, got it
Not the solution lol

4:16 PM
@Masterphile Do you know a useful criterion when a Hall-subgroup is guaranteed to be normal ? (I only know the case of a subgroup with index equal to the smallest prime factor of the group-order).
@Haran

@Peter Ya?

4:54 PM
f(m)=denominator(sigma(m)/(eulerphi(m)+m))
? n=1;while(f(n^2)<>1,n=n+2;if(Mod(n,10^6)==1,print1((n-1)/10^6," ")));print(n)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110
passed 10^8 , hence a square satisfying the equation must exceed 10^16

5:19 PM
Hello @Peter
+1

@Mathphile Hi, any idea ?

nope
@Peter did you see my new conjecture?

no

5:34 PM
13 hours ago, by Mathphile
Let $Q_1$ be the set of numbers less than $n\in \Bbb{N}$, that have an even number of odd quotients $\ne 1$. Let $Q_2$ be the set of numbers less than $n\in \Bbb{N}$, that have an odd number of odd quotients $\ne 1$. I conjecture that number of elements in set $Q_1$ is greater than the number of elements in set $Q_2$ for any $n$

How far did you check this ?
And in which range runs a in n\a ?

I think that if $Z(n)\ge 1$ holds then it implies my new conjecture

This is actually the case.

@Peter so we just have to prove $Z(n) \ge 1$ to prove my conjecture?

A moment, what is an "odd quotient" ? n\a odd ?

5:41 PM
@Peter yes

And Z(n) goes upwards if the number of odd quotients is even and downwards otherwise ?
In this case, Z(n) >=1 for all n>=1 is sufficient

@Peter yes it goes upwards I think
This is because even quotients do not affect the parity of $q(b)$, therefore the parity depends only upon the number of odd quotients

6:02 PM
makes sense
In which range did you calculate Z(n) ?
a runs from 2 to n-1 , right ?

$a$ runs from $1$ to $\lfloor \frac{n-1}{2} \rfloor$ in $q(n)$
and $k$ runs from $1$ to $x$ in $Z(x)$
@Peter

6:48 PM
@Peter A similar problem would be $\tau(n)+n=\sigma(n)$ or $n=\sum\limits_{d\mid n}(d-1)$
@Peter @Haran On the Equation $\sigma(n)=n+\phi(n)$ (2017) by Iannucci