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01/28/2023, 01:06 PM
(This post was last modified: 01/28/2023, 08:19 PM by Gottfried.)
In MO the user Caleb Briggs brings to discussion an old attempt of mine to the series \( T(x,2) = 1^{1^x}  2^{2^x} + 3^{3^x}  \cdots + \cdots \) .
I've tried this on 2007, and only later learned some techniques with which I might have assessed this with more success. But my basic observations and also the computations that I'd been able to do (over a small range of the exponent \(x\) ) come out to be correct.
For the friends of visual data  there are some nice pictures to see there.
Here is the link :
https://mathoverflow.net/questions/43666...etaseries
Here the link to my fiddlings:
http://go.helmsnet.de/math/tetdocs/Tetra_Etaseries.pdf
Have fun...
Gottfried
Gottfried Helms, Kassel
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Have you tried my summability method and/or the ETA series in the link ?
Regards
Tommy1729
Posts: 898
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(01/28/2023, 01:42 PM)tommy1729 Wrote: Have you tried my summability method and/or the ETA series in the link ?
Regards
Tommy1729
??? . Well, no idea... Unfortunately I don't think I'll step into this... growing a bit tired over the years... Just had my 70'th in December 2022 ...
I think I'll enter only punctually at some interesting and easyenough stuff. But I'd like if you or someone else would give things a try...
Regards
Gottfried
Gottfried Helms, Kassel
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01/29/2023, 09:50 PM
(This post was last modified: 01/30/2023, 02:13 AM by JmsNxn.)
This is actually reducible to a zeta function, in case you guys didn't know. I'm going to switch to \(x\) rather than \(x\).
$$
n^{n^{x}} = \sum_{k=0}^\infty \log(n)^k \frac{n^{kx}}{k!}\\
$$
Then interchanging the sum:
$$
T(x,2) = 1 + \sum_{n=2}^\infty \sum_{k=0}^\infty (1)^{n+1}\log(n)^k \frac{n^{kx}}{k!}\\
$$
This becomes a counting function with \(g(m)\) such that:
$$
T(x,2) = \sum_{m=1}^\infty g(m) m^{x}\\
$$
The counting function \(g(m)\) is definitely "power related", where it has a weighted count of some log series at each point. So let's write:
$$
g(m) = \sum_{k,n = 1\,\,m = n^k} (1)^{n+1}\frac{\log(n)^k}{k!}
$$
Then there's your zeta function. I forget the name of this type of zeta function, but similar ones exist, a famous one, is actually really close to this:
$$
h(m) = \sum_{k,n = 1\,\,m = n^k} (1)^{n+1} = (1)^{m+1} \sum_{k,n = 1\,\,m = n^k} 1
$$
This is actually counting how many even/odd powers exist of a numberand finding out if there are more even/odd power functions. And I believe it's open whether this function is analytically continuable to \(1/2 < \Re(x)\)... Gottfried's function is basically just a weighted version of this function.
Pick up your Analytic number theory book and get cracking Gottfried It's probably equivalent to the Riemann Hypothesis to prove some specific thing about this function, lmao.
Is there anything specific you would like to know? I believe these types of zeta functions appear often in "smooth number theory". https://en.wikipedia.org/wiki/Smooth_number
I'm a little bored, so I thought I'd apply some basic analysis; when we write:
$$
q(m) = \sum_{k,n=1\,\,m=n^k} 1\\
$$
A cleaner way to write this, and something more common in analytic number theory is:
$$
q(m) = \sum_{n \le m\,\,m=n^k} 1\\
$$
I was mistaken with smooth number theory; smooth number theory would restrict the primes available. So I screwed that up, lmao. This is much more standard than smooth numbers. This is just guessing the powers. I believe \(q(m)\) has a name, I just can't remember it off the top of my head. Your weighted version is easily discovered from this form though (though it requires a bit more finesse). Essentially it becomes:
$$
q(m) = \sum_{n\le m} \chi_m(n)\\
$$
Where \(\chi_m(n) = 1\) if there exists a \(k\) such that \(n^k = m\) and \(0\) other wise. From this we are weighting \(\chi_m(n)\) with \(\log\) weight. Getting a bound is easy from this...
The function \(q(m)\) asks how many natural number roots \(m\) has. Gottfried's functions are weighing the solutions, which complicates the matter, but not too much.
This is a very common function, and adding weights is nothing new. I don't even think Gottfried's weights are anything too complicated...
We have to guess how fast \(q(m)\) grows. I think a modest estimate is \(q(m) = m^{1/2+\delta}\), whereby, the estimate of:
$$
g(m) = (1)^{m+1} \sum_{k,n=1\,\,m=n^k} \frac{\log(n)^k}{k!}\\
$$
is about \(O(m^{3/2}\log(m))\)... Which is accomplished through Abel's summation technique. It's probably a good amount smaller, I think a decent estimate would be \(g(m) = O(m\log(m))\). If you could prove \(g(m) = O(\sqrt{m}\log(m))\) you'd probably have proven the Riemann Hypothesis.
I double checked my analytic number theory library of papers, and I believe I can enlighten you on some good "guesstimates" on the behaviour of your function Gottfried. So again, anything specific you are interested in? The function \(q(m)\) is fairly well studied, and I believe there is an expression involving it and the zeta function. In which your function would only be a weighted versionwhich means we can relate the two through convolution and Abel's summation technique.
Oh my god. I forgot how easy it is:
$$
g(m) = \sum_{n\le m} (1)^{n+1} \frac{\log(n)^{\frac{\log m}{\log n}}}{\frac{\log m}{\log n}!}\chi_m(n)
$$
Where \(\chi_m\) was defined as above; if \(n^k = m\) for some \(k\), then \(\chi_m(n)= 1\)otherwise zero. .... The value \(\log(m)/\log(n) = k\), and we are just reorganizing a sum...
We can fucking pull a lazy cauchy bound which is
$$
g(m) \le \log(m) \sum_{n\le m} \chi_m(n) = \log(m) q(m)\\
$$
This means:
$$
T(x,2) = \sum_{m=1}^\infty g(m) m^{x} = \sum_{m=1}^\infty O\left(\log(m) q(m)\right)m^{x}\\
$$
The function \(\chi_m(n)\) is called something. I can't remember right now. It's on the tip of my tongue...
Regards, James
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01/31/2023, 10:09 PM
(This post was last modified: 01/31/2023, 10:24 PM by JmsNxn.)
Well, I'm bored again, so let's write some more about this, as this is an interesting problem. Let's see what happens for coprime pairs in a certain manner.
So let \(\chi_m(n) = 1\) if there exists a \(k\) where \(m = n^k\), and zero other wise. We want to decompose this function into a simpler form than this. A good way to do this is to work with prime powers initially; in which we set \(m=p^r\) for some \(p\), then:
$$
\chi_{m}(n) = 1\,\,\text{if and only if}\,\, \log(n)/p \mid r\\
$$
Where \(a \mid b\) means \(a\) divides \(b\).
So for prime powers we are given:
$$
\sum_{n\le p^r} \chi_{m}(n) = \sum_{d \mid r} 1 = \sigma( r)\\
$$
Where \(\sigma\) is the divisor function. Then Gottfried's counting function for prime powers, is precisely:
$$
\sum_{n\le p^r} \frac{\log(n)^{r \log(p)/\log n}}{(r \log(p)/\log n !)}\chi_{m}(n) = \sum_{d \mid r} \frac{\log(d)^{\frac{r}{d}}}{\frac{r}{d}!}\\
$$
Now let's see what happens when we have two primes, rather than one. So let's write \(m = p_1^{r_1} p_2^{r_2}\). And let's see if we can relate this to the single prime case.
$$
\chi_{m}(n) = 1\,\,\text{if and only if}\,\, n = p_1^{t_1} p_2^{t_2}\,\,\text{where}\,\,kt_1 = r_1\,\,kt_2 = r_2\\
$$
This is a little trickier, because now we are testing two ratios, and seeing if they equal. To simplify this discussion, I am going to introduce the \(\nu\) character. This can be defined as:
$$
\begin{align}
m &= p_1^{a_1}p_2^{a_2} \cdots p_c^{a_c}\\
\nu_{p_j}(m) &= a_j
\end{align}
$$
This states that, in the Fundamental theorem of arithmeticin the decomposition of \(m\), \(\nu_p\) counts how many times the prime \(p\) occurs. Nothing more nothing less. And from this we can write:
$$
\chi_{p_1^{r_1}p_2^{r_2}}(n) = 1\,\,\text{if and only if}\,\, \frac{r_1}{\nu_{p_1}(n)} = \frac{r_2}{\nu_{p_2}(n)}\,\,\text{and}\,\, \nu_{p_1}(n) \mid r_1,\,\,\nu_{p_2}(n)\mid r_2\\
$$
Now the average mathematician might gawk that this is useless and pointless and looks ugly as fuck. I don't like that line of reasoning. It just means we are looking at it wrong. Instead. we are going to look at the expression:
$$
A(n) = \chi_{p_1^{r_1}}(p_1^{\nu_{p_1}(n)}) \chi_{p_2^{r_2}}(p_2^{\nu_{p_2}(n)})[\frac{r_1}{\nu_{p_1}(n)} = \frac{r_2}{\nu_{p_2}(n)}] \\
$$
Where \([a=b]\) is an Iverson bracket which produces \(1\) if true, and \(0\) if false...
We want to prove that \(A(n) = \chi_{p_1^{r_1}p_2^{r_2}}(n)\). We can do this by doing the first implication, by proving that \(A(n) = 1\) if \(n^k = p_1^{r_1}p_2^{r_2}\) for some \(k\). So take such an \(n\). Then since we've projected into the first prime number, and we know that \(\nu_{p_1}(n) k = r_1\) we have a \(1\). Multiply this by the \(1\) you get for the projection into \(p_2\). And we get \(1\). This proves the implication one way. To prove the opposite direction we have to pay close attention to \(k\). Suppose that \(A(n) = 1\), but \(\chi_{m}(n) = 0\). (Literally checking for false posiives). For this we'll note that the first part of our expression equals \(1\) if \(k_1 \nu_{p_1}(n) = r_1\) and \(k_2 \nu_{p_2}(n) = r_2\) where \(k_1 \neq k_2\). Which is the entrance of the Iverson bracket; which tests if the exponents are the same.
So........................................................
To deconstruct the power function:
$$
\begin{align}
m &= p_1^{r_1} p_2^{r_2}\cdots p_c^{r_c}\\
\chi_m(n) &= \prod_{j=1}^c \chi_{p_j^{r_j}}(p_j^{\nu_{p_j}(n)}) \cdot [\frac{r_1}{\nu_{p_1}(n)} = \frac{r_2}{\nu_{p_2}(n)} = ... = \frac{r_c}{\nu_{p_c}(n)}]
\end{align}
$$
Now this may seem stupid to talk about on a tetration forum, but now, we can take the sum:
$$
\sum_{n\le m} \chi_m(n)\\
$$
Which much much much more finesse. Especially because we know the prime power sum is just the divisor function.
I'm going to read more Abel summation shit, because I haven't even really introduced this yet. I'm just fuzzy on some things I remember. Also, in this instance the Iverson bracket is an arithmetical function. You can encode it using basic divisor sums! So the Iverson bracket looks ugly, but it's actually entirely plain mathematical operations to find it. A benefit of the bracket on natural numbers. We can test the equality of two natural numbers by summing numbers! (Fourier/orthogonality relationships abound).
So the root of Gottfried's problem \(\chi_m(n)\), is almost a multiplicative function, upto this stupid Iverson Bracket, lol.
I'll post more as I work through it. I have a sneaking suspicion this zeta function should have an Euler expansionor something like it. I have to refresh myself on decomposing Euler sums.
******
The Iverson bracket in our case can be written as:
$$
[a = b] = \frac{1}{m} \sum_{j=1}^m e^{2 \pi i j \frac{ab}{m}}\\
$$
Since all the terms in our bracket are less than \(m\). And if \(ab\) = 0, then we are summing \(\sum_{j=1}^m 1\). Otherwise, we are summing all the m roots of unity, which is famously the value zero. The function:
$$
[a_1 = a_2 = a_3 =...=a_c]\\
$$
Is a permutation problem; and we get:
$$
\frac{1}{m^c}\prod_{i < l \le c} \left(\sum_{j=1}^m e^{2 \pi i j \frac{a_{i}a_{l}}{m}}\right)\\
$$
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01/31/2023, 11:14 PM
(This post was last modified: 01/31/2023, 11:41 PM by JmsNxn.)
Let's go full Rambo on this. We're going to prove by example with prime power pairs. So let's write \(m = p_1^{r_1}p_2^{r_2}\):
$$
q(m) = \sum_{n\le m} \chi_m(n) = \sum_{n \le p_1^{r_1}p_2^{r_2}} \chi_{p_1^{r_1}}(p^{\nu_{p_1}(n)})\chi_{p_2^{r_2}}(p^{\nu_{p_2}(n)}) [\frac{r_1}{\nu_{p_1}(n)} = \frac{r_2}{\nu_{p_2}(n)}]\\
$$
But we are only summing over \(0 \le t_1 < r_1\) and \(0 \le t_2 < r_2\) here. So we can break this into:
$$
\sum_{t_1 \le r_1} \sum_{t_2 \le r_2} \chi_{p_1^{r_1}}(p^{t_1})\chi_{p_2^{r_2}}(p^{t_2})[\frac{r_1}{t_1} = \frac{r_2}{t_2}]\\
$$
This is ALMOST actually a product sum:
$$
\left( \sum_{t_1 \le r_1} \chi_{p_1^{r_1}}(p^{t_1})\right) \left(\sum_{t_2 \le r_2}\chi_{p_2^{r_2}}(p^{t_2})\right)\\
$$
Where our solution is less than this sum, as the indicator will eliminate some of the values...
But each of these are just the divisor sum:
$$
\sigma(r_1) \sigma(r_2)\\
$$
Where, again, \(\sigma(r_1)\) is how many divisors \(r_1\) has...
$$
\sigma(r ) = \sum_{d \mid r} 1\\
$$
AND THERE YOU FUCKING HAVE IT!!! We can find a non trivial bound of the power indicator function!
$$
\begin{align}
m &= p_1^{r_1} p_2^{r_2}\cdots p_c^{r_c}\\
\sum_{n\le m} \chi_m(n) &\le \sigma(r_1)\sigma(r_2)\cdots\sigma(r_c) = \Pi(m)\\
\end{align}
$$
I KNEW I'D SEEN THIS BEFORE!!!!!!!!! The function \(\Pi(m)\) is actually pretty well understood, so no problem there. I believe it grows about \(\sqrt{m}\log(m)\) or something like that. So now when we write:
$$
g(m) = (1)^{m+1}\sum_{n\le m} \frac{\log(n)^{\frac{\log m}{\log n}}}{\frac{\log m}{\log n}!} \chi_m(n)\\
$$
For :
$$
T(x,2) = \sum_{m=1}^\infty g(m) m^{x}\\
$$
We can bound this much better. I just have to go over my Abel summation bound technique notes I have. I'm lazy, but we should expect something like \(g(m) \le \log(m) \Pi(m)\) right off the bat!!!!!! Which is a much better bound than I got initially. It's probably the best naive bound we can get.......
Abel's tricks with arithmetic functions/integrals/zeta functions will probably afford us much more. If you're anxious, I suggest looking for expressions of the form:
$$
A(m) = \sum_{n\le m} f(n) g(m/n)\\
$$
which is Abel's convolution of \(f\) and \(g\). Gottfried's weighted counting function is ALMOST of this form. I'm trying to reduce it to this form to ignore problems. The way I have reduced it is by creating:
$$
\Pi(m) = \prod_{j=1}^\infty \sigma(\nu_{p_j}(m))\\
$$
And showing this is a pretty fucking tight bound on the growth of \(q(m)\). It just ignores the Iverson bracket part of my deduction...
The cleaner version of a bound, which I can call off my head is a bit more difficult to describe. But we can get a better bound than \(g(m) = O(\Pi(m)\log(m))\). We can probably bound a good amount tighter..... To do this, we would have to look at:
$$
y(m) = \sum_{d \mid m} \frac{\log(d)^{m/d}}{m/d!}\\
$$
And guess how this grows. Which is just guessing how fast Gottfried's weights grow.... This should lower the \(\log(m)\) though, this definitely grows slower than \(\log(m)\)........ I think an ambitious guess would be something like \(\sqrt{\log(m)}\) or something like that...
To relate this to Gottfried's \(T(x,2)\) is actually pretty standard. If you can control the growth of the zeta function coefficients, you can control the growth of the zeta function itself. They are both related through a Fourier inversion....
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01/31/2023, 11:34 PM
(This post was last modified: 01/31/2023, 11:39 PM by MphLee.)
(01/31/2023, 10:09 PM)JmsNxn Wrote: So let \(\chi_m(n) = 1\) if there exists a \(k\) where \(m = n^k\), and zero other wise.
Given the notation and this definition this is literally a characteristic function.
I leave this in the thread only as a terminological addendum. Let \(A\subseteq X\) then it's characteristic function \(\chi_A:X\to 2\) is defined as follows: \(\chi_A(x) = 1\) if \(x\in A\), and zero other wise. The concept is deeply categorical... and also an ancient one.
In your case just define for every \(m\in\mathbb N\) the set \(\sqrt m:=\{n\in \mathbb N\, \,\exists k.\, n^k=m\}\), then its characteristic function is the function you've defined, i.e. \(\chi_m(n) := \chi_{\sqrt m}(n)\), this and old old concept.
Now you are moving at the level of numbers... categorifying means switching pojnt of view and instead of studying numbers directly... that are the shadows of some higher level combinatorics business, you directly aim your attention at the set theoretic properties of the mapping \(m\to \sqrt m\) as \(m\) varies. This is basically what seems you are doing.. but implicitly. It turns out, I've studied this under the name of intrinsic iteration ( here and here I hinted at it), that this construction is functorial... and is deeply related to rational iteration (how? by considering exactly the same sets but instead of over the multiplicative monoid of natural numbers you do it over arbitrary monoids, e.g. monoids of functions under composition).
PS: I apologize for not being able to make say something more interesting but I'm lacking mental and physical vigor to do any better.... I'm sorry because I think this is really fascinating.
MSE MphLee
Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)
S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
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01/31/2023, 11:46 PM
(This post was last modified: 01/31/2023, 11:55 PM by JmsNxn.)
(01/31/2023, 11:34 PM)MphLee Wrote: (01/31/2023, 10:09 PM)JmsNxn Wrote: So let \(\chi_m(n) = 1\) if there exists a \(k\) where \(m = n^k\), and zero other wise.
Given the notation and this definition this is literally a characteristic function.
I leave this in the thread only as a terminological addendum. Let \(A\subseteq X\) then it's characteristic function \(\chi_A:X\to 2\) is defined as follows: \(\chi_A(x) = 1\) if \(x\in A\), and zero other wise. The concept is deeply categorical... and also an ancient one.
In your case just define for every \(m\in\mathbb N\) the set \(\sqrt m:=\{n\in \mathbb N\, \,\exists k.\, n^k=m\}\), then its characteristic function is the function you've defined, i.e. \(\chi_m(n) := \chi_{\sqrt m}(n)\), this and old old concept.
Now you are moving at the level of numbers... categorifying means switching pojnt of view and instead of studying numbers directly... that are the shadows of some higher level combinatorics business, you directly aim your attention at the set theoretic properties of the mapping \(m\to \sqrt m\) as \(m\) varies. This is basically what seems you are doing.. but implicitly. It turns out, I've studied this under the name of intrinsic iteration (here and here I hinted at it), that this construction is functorial... and is deeply related to rational iteration (how? by considering exactly the same sets but instead of over the multiplicative monoid of natural numbers you do it over arbitrary monoids, e.g. monoids of functions under composition).
PS: I apologize for not being able to make say something more interesting but I'm lacking mental and physical vigor to do any better.... I'm sorry because I think this is really fascinating.
Lmao!
Hey, Mphlee!
This is standard Analytic Number Theory. This is how Analytic Number Theory works. So if you are finding some relation to Category theorythat is the relation that Analytic Number Theory has to Category Theory. Honestly, I think that's why I understand you better than most people here. I have a strong background in Number Theory/Analytic Number Theoryand these are discrete modes of research. And there's a fairly strong overlap in many ways. I'm pretty sure I just rediscovered things I have read before, I'm just too dumb to remember where I read it. I'm willing to bet the past 3 posts were 90% already proved by greater mathematicians than me
I'll peruse your links though, always love your comments
EDIT: Just want to say I know what a characteristic function is. And I know what a radical is. But I didn't think this was a radical, so that's super cool, that this is a radical... lol I only know radicals from Analytic Number Theory, and I def wouldn't of called my function that. Still super cool! Thanks for your comments!!!!!!!
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02/01/2023, 07:12 AM
(This post was last modified: 02/01/2023, 07:16 AM by JmsNxn.)
Let's have some fun with Mphlee's introduction of radicals. The radical we care about is:
$$
\sqrt{\mathcal{I}_m} = \{n \in \mathbb{N}\,\mid\, m = n^k\,\,k \in \mathbb{N}\}
$$
We are then writing:
$$
\sum_{n \le m} \chi_m(n) = \sum_{n \in \sqrt{\mathcal{I}_m}} 1\\
$$
This becomes the quicker statement of:
$$
\sum_{n \le m} \chi_m(n) = \left \sqrt{\mathcal{I}_m}\right\\
$$
Now this is all super cool and all. But it's just a rewritea translation from the language I was using. But here, I can better explain my solution:
$$
\sqrt{\mathcal{I}_m} \le \Pi(m)\\
$$
Where \(\Pi(m)\) is a much more manageable quantity... If:
$$
\begin{align}
m &= p_1^{r_1} p_2^{r_2} \cdots p_c^{r_c}\\
\Pi(m) &= \sigma(r_1) \sigma (r_2) \cdots \sigma(r_c)\\
\end{align}
$$
This does not prove the size of \(\sqrt{\mathcal{I}_m}\), or even really discuss the radical. All we know, is that there's a pretty good bound of \(\Pi(m)\)...
I'd love to do this radical talk, Mphlee. But I don't see anything obvious that would aid in the questions at hand. It's just a different way of saying the problem (which is still cool), but nothing novel to the problem.
Unless this radical produces a brand new multplication formula or something, I don't give a fuck......
Nothing but love, Vittorio. Just talking straight
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02/01/2023, 07:44 AM
(This post was last modified: 02/01/2023, 08:01 AM by JmsNxn.)
Okay, let's run some radical translations of shit I've already written:
$$
\sqrt{\mathcal{I}_{p^k}} = \sigma(k)\\
$$
for any prime \(p\). And we can write a straight bound:
$$
\sqrt{\mathcal{I}_{p^kq^u}} \le \sigma(k) \sigma(u)\\
$$
While additionally we have:
$$
\sqrt{\mathcal{I}_{ab}} \le \sqrt{\mathcal{I}_{a}}\sqrt{\mathcal{I}_{b}}
$$
so long as \((a,b)=1\), that \(a\) and \(b\) are co prime..... I think the word for this is "sub" multiplicative, or some bs like that...
Fuck, Vittorio I think you are a couple steps away from a hard result........
I should add that I have ultimately proven the sub radical formula:
$$
\sqrt{\mathcal{I}_{ab}} \le \sqrt{\mathcal{I}_{a}}\sqrt{\mathcal{I}_{b}}
$$
by just noting that the left hand adds a bunch of Iverson \([a=b]\), and the right hand ignores that. Where, all we do is add a couple \(1)'s on the right... nothing more.
Once you add this, you have created a weird ring effect, and a better zeta function. We can write \(\sqrt{\mathcal{I}_m} = q(m)\). And this is not multiplicative, but \(q(mn) \le q(m)q(n)\) when \((m,n)=\) are coprime......
