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05:40
@JoseArnaldoBebita-Dris hi, i observed that you are interested in odd perfect numbers
@Ante, hi there. Yes, your observation is correct! =)
what are you trying to find, that there are some or that there is none?
Either way will be satisfying =)
Although admittedly, finding an OPN is much rarer than finding a unicorn!
because, most probably, and evidence suggests that also, there are none
Well, the fact that nobody has still derived a contradiction amidst the myriad of necessary conditions that an OPN must satisfy, if it exists, might say something about its possible existence... =)
05:48
nah, i think surely there is none, just the proof is missing
Yes, only time can tell =)
Permission to post two questions of mine ^_^
1
Q: If $n = km$ is a Descartes number with quasi-Euler prime $m$, then $m < k$.

Jose Arnaldo Bebita-DrisLet $$\sigma(x) = \sum_{d \mid x}{d}$$ denote the sum of divisors of $x \in \mathbb{N}$, where $\mathbb{N}$ is the set of natural numbers or positive integers. Recall that a Descartes number is an odd number $n = km$, with $1 < k$, $1 < m$, satisfying $$\sigma(k)(m+1)=2km.$$ ($m$ is called the ...

and:
2
Q: Why did the Egyptians not represent $2/3$ as a sum of unit fractions in the Rhind papyrus?

Jose Arnaldo Bebita-DrisThe following is taken verbatim from the MathWorld Wolfram page on Egyptian fractions: An Egyptian fraction is a sum of positive (usually) distinct unit fractions. The famous Rhind papyrus, dated to around 1650 BC contains a table of representations of $2/n$ as Egyptian fractions for odd $n$...

can you explain what do you ask exactly here:
1
Q: If $n = km$ is a Descartes number with quasi-Euler prime $m$, then $m < k$.

Jose Arnaldo Bebita-DrisLet $$\sigma(x) = \sum_{d \mid x}{d}$$ denote the sum of divisors of $x \in \mathbb{N}$, where $\mathbb{N}$ is the set of natural numbers or positive integers. Recall that a Descartes number is an odd number $n = km$, with $1 < k$, $1 < m$, satisfying $$\sigma(k)(m+1)=2km.$$ ($m$ is called the ...

500+ is very nice
06:10
I ask if the two hypotheses used in the proof for $m < k$ could be relaxed, if $n = km$ is a Descartes number with quasi-Euler prime $m$. So far, mathlove was able to remove the reliance of the proof on the condition $\gcd(m,k)=1$. Next, I ask if the hypothesis
$$\frac{\sigma(k')}{m'}+\frac{\sigma(m')}{k'}=\frac{670763}{819} \approx 819.002$$
could be similarly relaxed? (That is, could we simply not use it in the proof?)
06:22
i will try to solve that
Note that the inequality $m < k$ is equivalent to the assertion that $k > 1$ is not an odd almost perfect number, which is consistent with the conjecture that there does not exist an odd almost perfect number other than $1$.
@Ante, okay thanks!
06:47
@Mathphile: Actually, we usually define $0^0 = 1$ (as a matter of convention). I think @Peter will be able to give us a convincing explanation.
 
2 hours later…
08:48
@Peter 500+ bounty!
Really high! For what ?
1
Q: If $n = km$ is a Descartes number with quasi-Euler prime $m$, then $m < k$.

Jose Arnaldo Bebita-Dris(Note: This question is tangentially related to this later one.) Let $$\sigma(x) = \sum_{d \mid x}{d}$$ denote the sum of divisors of $x \in \mathbb{N}$, where $\mathbb{N}$ is the set of natural numbers or positive integers. Recall that a Descartes number is an odd number $n = km$, with $1 < k...

Hello, @Peter. =)
@JoseArnaldoBebita-Dris Hi, you are apparently very interested in everything about odd perfect numbers :)
@Peter, you bet! =)
08:53
Are there other Descartes numbers known than the number mentioned in the Wikipedia article ?
Maybe, you can help me in proving that a squarefree composite number n cannot satisfy n | sigma(n)+phi(n) ?
@JoseArnaldoBebita-Dris
There are no other Descartes numbers known other than the number mentioned in the Wikipedia article. I still wonder, to this day, how Descartes found it.
Proving that a squarefree composite number n cannot satisfy n | sigma(n)+phi(n) (or in general, proving that a class of numbers m cannot satisfy m | (f(m) + g(m)), for some multiplicative functions f and g), can be very difficult in practice.
But let me try.
So since n is squarefree and composite, then n takes the form
$$n = pqrstuv \cdots$$
for primes p, q, r, s, t, u, v, $\ldots$.
Then sigma(n)=(p+1)(q+1)(r+1)(s+1)(t+1)(u+1)(v+1) $\cdots$
And phi(n) = (p - 1)(q - 1)(r - 1)(s - 1)(t - 1)(u - 1)(v - 1) $\cdots$
So you are essentially asking when:
$$pqrstuv \cdots \mid \bigg(\bigg((p+1)(q+1)(r+1)(s+1)(t+1)(u+1)(v+1) \cdots\bigg) + \bigg((p - 1)(q - 1)(r - 1)(s - 1)(t - 1)(u - 1)(v - 1) \cdots\bigg)\bigg)$$
Consider the case of n a semiprime.
n = pq
sigma(n) = (p+1)(q+1)
phi(n) = (p-1)(q-1)
2 and 3 factors are easy
Okay, I guess you got that figured out.
09:08
the sum is 2pq + 2 which can obviously not be divisible by pq
Uhm.
Well, gcd(pq, pq+1)=1, is it not?
sum=2pq+2=2(pq+1)
Assuming p and q are odd, you have your desired conclusion.
We would get pq | 2 , but q is surely odd.
Ohh okay, I misread your last assertion.
Yes, the sum 2pq+2 is obviously not divisible by pq.
How about four factors?
For 3 factors, it is not much more complicated
For 3 factors, I get:
09:16
In the case of 4 factors , the sum is 2 (pq+pr+ps+qr+qs+rs+1) + 2pqrs
(p+1)(q+1)(r+1)+(p-1)(q-1)(r-1)=2pqr + 2(p + q + r)
And 2(p+q+r) < 6r <= pqr (As Ante showed)
in TheSimpliFire's Chatroom, 32 mins ago, by Ante
if p<q<r then 2(p+q+r)<2(r+r+r)=6r and pgr>=6r and is equal only if p=2 and q=3
@Ante Even in the case of equality , we are done.
09:18
So for 3 factors, no more problem?
3 factors is solved.
Okay.
For 4 factors, you can reason as follows:
gcd(pqrs,pq+pr+ps+qr+qs+rs+1)=1
I think that can be proved using e.g., the Euclidean Algorithm for computing the GCD.
that is not true if p=2
Maybe, we can solve the problem with induction over the number of prime factors.
I tried to compute the GCD when p=2 using WolframAlpha, it says it is 1:
https://www.wolframalpha.com/input/?i=GCD%282qrs%2C2q%2B2r%2B2s%2Bqr%2Bqs%2Brs%2B1%29
@Peter Yes indeed, maybe!
09:26
it is not 1 always
We have 2(pq+pr+ps+qr+qs+rs+1) < 12 rs because of pq+1 < rs
@JoseArnaldoBebita-Dris But this is a polynomial gcd
We can conclude p = 2
Any way, may I ask what bigger research problem you have in mind that led you to consider squarefree composite integers n that ought to satisfy n | sigma(n) + phi(n) ????
The aim is to classify the solutions , and it is useful if squarefree numbers can be completely ruled out.
@Ante Can we make the estimate sharper to solve the 4-factors-case ?
@Peter what equation do you get for the four numbers case?
modulo pqrs we have 2(pq+pr+ps+qr+qs+rs+1)
09:32
you got that pqrs must divide that?
yes
and my above estimate shows p = 2
@Peter, @Ante: You might want to take a look at oeis.org/A011774 to check if any solutions listed there are squarefree. =)
and q = 3 or 5
@JoseArnaldoBebita-Dris Is there a comment ? And what is the search limit ?
if you got that 2(pq+pr+ps+qr+qs+rs+1)<12rs if p<q<r<s and pqrs is = (pq)rs then p=2 and q=3 or q=5 or p=3 and q=3, upon those substitutions the equations reduce
p=q is impossible, we have a squarefree number
09:37
then substitute that to reduce the equations
you have two cases
There are several references.
The formula for phi and sigma would be different in this case as well.
but what is the full equation you got
?
without mod
the additional term is 2pqrs
@JoseArnaldoBebita-Dris I cannot copy the numbers in Excel because some are too big. How can I copy just the text or the numbers ?
but in the squarefree case phi is $$\prod_{r=1}^{n}(p_r-1)$$
?
09:45
The sum is (p+1)(q+1)(r+1)(s+1) + (p-1)(q-1)(r-1)(s-1)
yes
which is equal to 2pqrs + 2(pq+pr+ps+qr+qs+rs+1)
but it is also equal to 2(p-1)(q-1)(r-1)(s-1) + something if you write p+1=p-1+2 and q+1=q-1+2 and r+1=r-1+2 and s+1=s-1+2
you mean abcd + (a+2)(b+2)(c+2)(d+2) ?
yes
is now easier to estimate something?
you have now (a+1)(b+1)(c+1)(d+1) has to divide that
09:51
What, if we just insert the cases p = 2 , q = 3 and p = 2 , q = 5 and estimate then ?
ah, i wouldn´t, that will hardly solve the general case
but if p=2 necessarily then we are done
or not
Why ?
We can try the induction idea I tried yesterday.
I set P:=(p+1)(q+1) ... and Q=(p-1)(q-1) ... The idea is : If there is a solution for k+1 factors, there is one for k factors. Could this work ?
nope
Then, we need an estimate working for all k
yes
or to see why it cannot be divisible
10:00
I just check all squarefree numbers with no prime factor exceeding 113. Takes still a while, no solutions yet.
Hold on, @Peter.
I already checked all squarefree numbers with no prime factor exceeding 97.
Save this as a text file:
Then export to Excel.
Note that all of the numbers there are even.
It seems that they are all divisible by 4, can you check this ?
One exception, maybe.
Still did not work, Excel interprets it as a number or does not seperate the columns.
i think general case can be solved with AM-GM inequality, perhaps
10:13
@Mathphile Hi, we need your help.
@JoseArnaldoBebita-Dris If I looked right, only one number not divisible by 4, but this is divisible by 9. So, there should be no example within this search limit.
But anyway, I would expect a possible number to be very large.
Okie dokie, sounds good, @Peter.
did you ask a question on the site? @Peter
I could ask one.
0
Q: Can a squarefree composite number $\ n\ $ satisfy $\ n\mid \sigma(n)+\varphi(n)$?

PeterLet $n$ be a squarefree composite number. Can $\ \sigma(n)+\varphi(n)\ $ be disivible by $\ n\ $ ? $\ \varphi(n)\ $ is the totient function and $\ \sigma(n)\ $ the sum of the positive divisors of $\ n\ $. I solved the case with no more than $\ 3\ $ prime factors. Also, I checked all square...

+1
10:28
Thank you
not a single comment here:
7
Q: Is this arithmetic function strictly positive and unbounded?

AnteAs requested by Mathphile, since there have been efforts but no complete solutions to some questions raised when this question was asked on MSE, and since we think that here the question is more probable to be solved completely, I reproduce it here, in a different style. The function $$r(b)=\sum...

11:03
Upvoted both questions.
thanks
(+1)
thanks
@JoseArnaldoBebita-Dris Thank you
11:09
what do you think about this answer?
3
A: Sufficient conditions on $ a_i,b_i$ for $a_1\phi(n)+b_1, \cdots, a_k\phi(n)+b_k$ to be simultaneously prime infinitely often?

AnteMost probably $\gcd(a_i,b_i)=1$ and $a_i \geq 1$ and $b_i$ to be odd for $i=1,...,k$ is enough. Since $\{\varphi(n): n \in \mathbb N\}$ can be partitioned into countably many subsets $C_l=\{\varphi(w_{rl}): r \in \mathbb N\}$ such that $\varphi(w_{ml})<\varphi(w_{nl})$ if $m<n$ the $k$ forms $a...

it is very hard to know anything, since Dirichlet´s theorem is not generalized in direction i mentioned
(+1) The generalized Bunyakovsky conjecture predicits infinit many primes p , such that ap+b is prime as well, whenever gcd(a,b)=1. But I am pretty sure that even this special case is open.
yes, that is special case of Dirichlet´s theorem
its generalization onto subsequences
rather
Heuristically, it should be true. The evidence is comparable with the evidence of the truth of the Goldbach conjecture.
all the primes are relatively prime, that could be some of necessary conditions, perhaps
@Peter i think some of your questions would be better received at MO
not all but some
My problem could be solved, if we could show that for an odd positive integer m > 1 we cannot have 2m | phi(m) + 3sigma(m)
m squarefree
12:21
@Ante I checked all squarefree numbers with no prime factor greater than 113. No examples.
any reactions to question?
+3 , but no reactions yet.
why didn´t you asked it on MO?
MO ? Worse than this site to my experience.
why, did you argue with some of the users there? i have seen you have some highly upvoted questions there
12:47
i have an idea
wanna compute something?
13:00
OK
can you display all the numbers $${\frac {1}{p}} \cdot {p \choose k}$$ for $k=2,...,p-1$
and $p$ prime
so binomial(p,k)/p ?
yes for k=2,...,p-1 and p prime
I have no chatjax
why not?
13:04
since windows crashed.
yes, that´s what i ask (binomial(p,k))/p for k=2,...,p-1 and p prime
? forprime(p=1,20,for(k=2,p-1,print([p,k,binomial(p,k)/p])))
[3, 2, 1]
[5, 2, 2]
[5, 3, 2]
[5, 4, 1]
[7, 2, 3]
[7, 3, 5]
[7, 4, 5]
[7, 5, 3]
[7, 6, 1]
[11, 2, 5]
[11, 3, 15]
[11, 4, 30]
[11, 5, 42]
[11, 6, 42]
[11, 7, 30]
[11, 8, 15]
[11, 9, 5]
[11, 10, 1]
[13, 2, 6]
[13, 3, 22]
[13, 4, 55]
[13, 5, 99]
[13, 6, 132]
[13, 7, 132]
[13, 8, 99]
[13, 9, 55]
[13, 10, 22]
[13, 11, 6]
[13, 12, 1]
[17, 2, 8]
[17, 3, 40]
[17, 4, 140]
[17, 5, 364]
[17, 6, 728]
[17, 7, 1144]
[17, 8, 1430]
[17, 9, 1430]
[17, 10, 1144]
i see no pattern for MP
13:24
another idea
there has to be some pattern!
@Peter are you there?
pattern for what ?
for Mersenne primes, i have another idea
Mersenne primes are very well known, why do you search a pattern for them ?
to find necessary conditions on p
A criterion when 2^p-1 , with p prime , is prime ?
13:38
yes, but only necessary
it is only known that p has to be prime, nothing else
in fact , apart from a weird conjecture that is true for all known cases.
which one?
In mathematics, the Mersenne conjectures concern the characterization of prime numbers of a form called Mersenne primes, meaning prime numbers that are a power of two minus one. == Original Mersenne conjecture == The original, called Mersenne's conjecture, was a statement by Marin Mersenne in his Cogitata Physico-Mathematica (1644; see e.g. Dickson 1919) that the numbers 2 n − 1 {\displaystyle 2^{n}-1} were prime for n = 2, 3, 5, 7, 13, 17, 19, 31, 67, 127 and 257, and were composite...
comes later, not displayed here.
compute (1/p)* sum of binomial(p,k) for k=1,...p-1 and p prime and factor those numbers
in other words, factor (2^p-2)/p
Hi all! Whats the difference between the (general) Mathematics chatroom and this one?
13:51
some guys there know about some topics more than we who are usually here
@Rudi_Birnbaum
and we are not often there
If p has the form 4k+3 and 2p+1 is prime as well, then 2p+1 is a factor of 2^p-1
what is else known about factors of (2^p-2)/p
? forprime(p=1,100,print(p," ",factor((2^p-2)/p)))
2 matrix(0,2)
3 Mat([2, 1])
5 [2, 1; 3, 1]
7 [2, 1; 3, 2]
11 [2, 1; 3, 1; 31, 1]
13 [2, 1; 3, 2; 5, 1; 7, 1]
17 [2, 1; 3, 1; 5, 1; 257, 1]
19 [2, 1; 3, 3; 7, 1; 73, 1]
23 [2, 1; 3, 1; 89, 1; 683, 1]
29 [2, 1; 3, 1; 5, 1; 43, 1; 113, 1; 127, 1]
31 [2, 1; 3, 2; 7, 1; 11, 1; 151, 1; 331, 1]
37 [2, 1; 3, 3; 5, 1; 7, 1; 13, 1; 19, 1; 73, 1; 109, 1]
41 [2, 1; 3, 1; 5, 2; 11, 1; 17, 1; 31, 1; 61681, 1]
43 [2, 1; 3, 2; 7, 2; 127, 1; 337, 1; 5419, 1]
It is even unknown whether there are infinite many composite Mersenne numbers with prime exponent !
Although this "must be the case"
:53995605, OK!
The only possible method is to rule out primes with trial division (using that every factor must have the form kp+1) and then testing it.
It is incredible how large the verified range without gaps is !
Moreover, the complete factorizations are known upto a high exponent (not necessarily prime)
14:08
Mersenne numbers with prime exponent are finite, wouldn’t this imply a different prime density than the one implied in the prime number theorem?The one I don’t know if my heuristic here is correct. But if composite
If* mersenne numbers....
It is conjectured that there are infinite many Mersenne primes, but much stronger conjectured that infinite many Mersenne numbers with prime exponent are composite.
The prime number theorem can neither prove nor disprove this.
Of course, there are infinite many Mersenne numbers with prime exponent, you seem to have misunderstood my statement.
That there are infinite many primes, is sure since Euklid.
can you factor (2^n-2)/n in some reasonable range for all n, both composite and prime
Sorry about that my ipad had a stroke while writing that sentence. I'm on my laptop now. I meant to write "I don't know if my heuristic is correct. But if composite Mersenne numbers with prime exponent are finite, wouldn't this imply (or at least) suggest a prime number density, probably different from the prime number theorem?"
i think no, if PNT suggest that there is an infinite number of them, it doesn´t prove there is an infinite number of them
yeah, I'm not talking about proof. What I mean by this is as follows:
If composite Mersenne numbers of prime exponent are finite then there exists a (large) N such that for all Mersenne numbers 2^p - 1 larger than N, 2^p-1 is prime
14:19
yes, that´s sure observation
from that (and here is the part I might be wrong) this impacts the asymptotic behaviour of primes
yes, it impacts behavior of primes only of the form 2^p-1, not all primes
@palindromicprime This would be strange (to say the least), but we cannot rule it out. But the mersenne numbers form a very small subset of the positive integers. So , the prime number theorem does not apply directly. But indirectly, it does apply, since we can estimate the "probability" that a number is prime.
what I was thinking maybe is to feedback those large mersenne primes back into the mersenne expression. So taking a mersenne prime 2^p - 1 larger than mentioned N, 2^(2^p - 1) is prime as well. But yeah, I think im dead wrong here, since I think the observations above still apply
In fact, M(M(61)) or something like that has been partially verified. This is a special variation of Mersenne-chains. As far as I know, for some exponents p , M(M(p)) could be prime, but the numbers are too large to check it.
The growth rate of such numbers is double-exponential ! So , only very small cases are feasible.
14:30
@palindromicprime no, that´s a good idea, i had exactly the same idea
@Peter Do you know, how is it partially verified? Wouldn't a single digit, change the whole result completely?
For numbers so huge, we can only find a factor , or we won't ever know whether they are prime.
@Ante haha yeah, I guess it must be a common pitfall one falls into after one learns that 2^p - 1 prime implies p prime
in fact, the converse is false.
@palindromicprime well, it seems to be a pitfall, until you actually succeed to prove something new with that idea, as i am going to try in a few moments
14:33
@Peter then my question becomes, how can we tell a number is a factor of a really huge number?
@Ante oh nice! would be happy to check it out if/when you are ready to share
modulo calculation allows to find a factor, if it is small enough. We can check 2^m mod n , even if 2^m is huge.
only m and n have to have a reasonable magnitude.
@Peter oh i see so you exploit the form of the number in the modulo, makes sense
Without this, we would not know that the Fermat numbers F_n are composite for 5<=n<=32
yes of course, it flew straight over my head, haven't slept at all
by the way, I have a problem I'm trying to solve. is it okay if i share it here?
Why not ? Interesting problems are always welcome here !
14:38
3
Q: Contour Integral involving Zeta function

palindromicprimeI'm trying to compute the contour integral $$\frac{1}{2 \pi i} \int_{c - i \infty}^{c + i \infty} \zeta^2(\omega) \frac{8^\omega}{\omega} \ d \omega$$ where $c > 1$, $\zeta(s)$ is the Riemann zeta function. Using Perron's Formula and defining $D(x) = \sum_{k \leq x} \sigma_0(n)$, where $\sigma_0...

Seems complicated (+1) @Ante Do you have an idea ?
nope, i haven´t done much contour integration
Me neither, and in particular with the zeta-function, I have almost no experience.
yeah me neither, but it has been an interesting path to take. the zeta function has a really beautiful laurent series expansion, so I worked with that and got most of it
just need to take the line integral over over the critical strip and that just break my head every time I try
took me years to find that integral expression for D(x), only to find it has already been found and its even generalized for any arithmetic function
you can imagine the frustration of having missed it during preliminary research on the topic
In fact, frustrating !
14:44
oh well, thanks for the upvote anyway it helps
Suppose, someone could prove the Riemann-hypothesis. Would it then be easy to prove the generalized version ?
Some important results are not implied by RH alone.
Well depends how the proof looks like. But mostly likely I think not.
In particular, if the proof is done with an abstract point of view of L-functions, then wouldn't skip my imagination that one could generalize this to other L-functions
I read an article about someone skeptical about RH (I think it is Littlewood). He mentioned some strong indications against RH , but apparently those arguments are not taken serious. I wonder why.
do you think i got too much upvotes with this question:
19
Q: Is every function $f: \mathbb R \to \mathbb R$ differentiable at at least one point when restricted to some everywhere dense subset of $\mathbb R$?

AnteI was doing some fairly simple research a few hours ago and I almost asked a similar question with the word continuous instead of differentiable in the title, but then I found this question asked by Gro-Tsen where there is an affirmative answer to that question. Apparently, that is the result of...

15:01
I would say that there are much worse and much more upvoted questions. To be honest, it would not belong to my favourite questions, but this does not mean it is bad. It just does not hit my interests. Anyway, be glad about the good score.
yes, good point
Shall I post my question on MO ?
ah, i think you should be patient, and post it in a day or two, if you do not receive answer on MSE
OK
generally, you could observe when your questions are for MSE and when for MO, that one was immediately for MO, i think
unless we do not see something trivial
as some bounds
some guy ran into problems, i gave him the best advices i could:
2
A: Asking for advice (references/papers) in masters project in analytic number theory as i have no other source of help and i am stranded midway

AnteAt least theory of numbers is full of open problems, if you really much want to solve some of them, or at least one. But if you lost guidance of some professor that does not mean that you lost your own "guidance of yourself", so, when you see that you can "smoothly" read some topic that really i...

the question was closed, apparently not suitable for MO
15:48
@Peter what problem (open) would you like to see solved first?
twin prime conjecture
Can you explain this strange ouput ?
? for(n=10^7,10^8,if(issquarefree(n)==1,if(poldegree(f(n))==1,w=factor(f(n));w=component(w,1);w=w[1];if(polcoeff(w,1)==1,if(isprime(-polcoeff(w,0))==1,if(gcd(polcoeff(w,0),n)==1,print(n," ",w)))))))
^C *** at top-level: ...degree(f(n))==1,w=factor(f(n));w=component(w,
*** ^--------------------
*** in function f: eulerphi(m)*(p-1)+si
*** ^--------------------
*** eulerphi: user interrupt after 12min, 2,429 ms
*** Break loop: <Return> to continue; 'break' to go back to GP prompt
? f
%5 = (m)->eulerphi(m)*(p-1)+sigma(m)*(p+1)-4*m*p
?
nope
How could n overshoot ?
i do not know
do you remember phi(pm+1)=phi(pm+p+1)?
yes
15:58
well, it works for sigma instead of phi also !!!
no problems at all
what a "duality"
!
what do you mean with "it works" ?
well, choose m, for example, m=1,...,100 and try to find p such that sigma(pm+1)=sigma(pm+p+1)
and you will find them!
ah,OK
i thought of asking a question about that also
OK, you can do it.
16:04
only if i find some other arithmetic functions for which it works
sum of the squares of the divisors ?
ah, that is not linear, i am not sure that will work
but Moebius function could work
moebius is too trivial
well, you can try for sum of squares of divisors
I am busy with finding a solution for squarefree n, and I am confused about the pari-issue. Hope, it is just an issue with calling a variable after a break.
16:10
how would code go for sum of squares
v=divisors(n);s=sum(j=1,length(v),v[j]^2)
@Knight Do you have a fast computer to run pari/gp ?
i will run it, send me the code
f(m)=eulerphi(m)*(p-1)+sigma(m)*(p+1)-4*m*p
or "-3*m*p" (you can use a second window)
for(n=1,10^10,if(issquarefree(n)==1,if(poldegree(f(n))==1,w=factor(f(n));w=component(w,1);w=w[1];if(polcoeff(w,1)==1,if(isprime(-polcoeff(w,0))==1,if(gcd(polcoeff(w,0),n)==1,print(n," ",w)))))))
what are you researching?
squarefree n with n | phi(n) + sigma(n)
16:20
why don´t you just compute (phi(n)+sigma(n))/n for every n?
send me a code for that
i will search
An example will be very large, therefore my other version. OEIS has an entry with a huge search limit, no examples.
that way we will see near misses, no worries, i will search it upto a large limit
Here I search some suitable m, such that there is some prime p not dividing m, such that m*p is a solution.
i understand, but i would like to see fractions for all n, upto a very large n
to see what is happening
OK, just define f(n)=denominator((eulerphi(n)+sigma(n))/n)
? forcomposite(m=1,10^6,if(f(m)==1,print1(m," ")))
312 560 588 1400 23760 59400 85632 147492 153720 556160 569328
?
shall we also count the solutions ?
16:26
nah, that´s not interesting, i would like to see all the fractions (phi(n)+sigma(n))/n, not just only with denominator 1
? forcomposite(m=1,10^6,if(f(m)==1,print(m," ",(eulerphi(m)+sigma(m))/m)))
312 3
560 3
588 3
1400 3
23760 4
59400 4
85632 3
147492 3
153720 4
556160 3
569328 3
?
the OEIS entries give no 5 or larger
With great effort, I found an example with fraction 5.
update the OEIS
16:50
@Peter What kind of computer do we need? Would Mac Book Pro do the job?
3
Q: Is this a known method regarding expressing a prime as the sum of two squares?

S. DolanSuppose that the quadratic $x^2+bx+c$ has integer coefficients and non-zero discriminant and constant term. Then it is a simple exercise for students to prove that the number of positive integer roots of $x^2+bx+c$ added to the number of positive integer roots of $x^2-bx+c$ is either 0 or 2. Thi...

Is this a valid proof ?
@Ante
@Knight It must only be fast and being able to run, for example pari
Do you have a room available for me?
@Peter seems valid
@MatsGranvik this is that room, welcome
17:05
@Ante Ok, I thought you were fully booked.
@MatsGranvik not really, but we mostly here do some non-advanced number theory, if that interests you
non-advanced number theory is my thing
for example, we found that if we choose m in some relatively large range, say, m=1,2,...300, then if for every such m we try to find p such that phi(pm+1)=phi(pm+p+1), then almost always there are solutions, with rare exceptional cases, as though totient has some "hidden laws", p is not necessarily prime, but a natural number (although p often denotes prime in books)
more generally, we researched the expression phi(pm+k)=phi(pm+p+k)
observe that (pm+k)/(pm+p+k) is very close to 1
@Ante Sorry that I interrupt, but how does the discriminant show that the prime has such a representation ? It only guarantees real solutions, not integer solutions.
Hey guys
17:15
@Mathphile Look at the above proof that a prime of the form 4k+1 is the sum of two squares
@Peter it is only important that it is non-zero
@Mathphile hey
and an integer @Peter
the discriminant must be a square, not just an integer.
@Mathphile Does the proof convince you ?
@Peter where does the proof start?
3
Q: Is this a known method regarding expressing a prime as the sum of two squares?

S. DolanSuppose that the quadratic $x^2+bx+c$ has integer coefficients and non-zero discriminant and constant term. Then it is a simple exercise for students to prove that the number of positive integer roots of $x^2+bx+c$ added to the number of positive integer roots of $x^2-bx+c$ is either 0 or 2. Thi...

After a few lines
ah
I don't understand it
1
A: What is the antiderivative of $x^{x^{x^{x...}}}$?

MathphileAlthough Szeto's answer: $$\int x^{x^{...}} dx= \sum^{\infty}_{n=0}\frac{(-n)^{n-1}}{n!}\gamma(n,-\ln(x))$$ is correct. It does not converge for the complete domain of $x^{x^{x...}}$ which is $e^{-e}\le x \le e^{1/e}$ and instead only converges for $e^{-1/e}\le x\le e^{1/e}$. This is because the ...

17:34
Masterphile.
17:46
@Ante I have doubts about the proof above.
18:00
then comment the question?
 
4 hours later…
21:31
@Peter I want to prove that the area under a function ( say from $0$ to $a$) can be equal to the value of the function at $a$. The most important and foremost question is Can we prove it? and if yes do we need any rigorous analysis or some simple hypothesis would do the job?
@Ante ^
what do you suppose of properties function needs to have?
I didn’t get your question
i understand, you want to construct some function that has that property
or just prove existence?
Just the existence
I want to prove that it is possible
I want to prove that it is possible $$ \int_{a}^{b} f(x) dx = f (b)$$
define f(x)=1/2 in [0,1/2] and f(x)=1/4 in [1/2,1/2+1/4], and f(x)=1/8 in [1/2+1/4,1/2+1/4+1/8] and proceed like that, the integral in [0,1] of g=1-f will be 1, and you have g(1)=1, as desired
ah, nope, the integral of g=1-f will be 0
but for this f you have f(a)=int_a^{b}f(x)dx
almost!
yes, it is possible
i will try now to find a more suitable construction
21:47
@Ante I got some more information about $f(x)$
for example?
$f(x)$ assumes the value of zero at least once in its interval
But I don’t think it’s helpful in anyway, ha?
it is easy to construct it for some special choices of a and b but i am thinking about general case
05:00 - 22:0022:00 - 00:00

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