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12:10 AM
What is the Matrix? Control

12:47 AM
@Jakobian @Koro I haven't read all the discussion while I was gone for hours and hours. But I propose the function $f(x) = \begin{cases} n, & x=1/n \\ x, &\text{otherwise}\end{cases}$. Did I misread the given hypothesis? It sure seems to me the limit in the hypothesis is $0$ but $f$ has no limit at $0$.
Hmm, I read the hypothesis as $\lim\limits_{x\to 0} f(x)\big(\frac1x-\left[\frac1x\right]\big) = 0$.
Was it something else?

1:09 AM
Yes, the whole thing is inside of $f(...)$
$f$ evaluated at $x\cdot (\frac{1}{x}-[\frac{1}{x}])$
@Koro sorry, I wasn't here, but looks like you resolved that already

1:33 AM
Ah, thanks. Then it’s pretty easy.

1:44 AM
Any hint to find an upper bound for $\dfrac{x+y}{(y)(y^2-1)(\ln(x+y))}$ for $y \ge 2$ and $x \in \Bbb N$? What I have so far is that $y^2-1 > y$ so $\dfrac{1}{y^2-1}< \dfrac{1}{y}$ and hence $\dfrac{x+y}{(y)(y^2-1)(\ln(x+y))} < \dfrac{x+y}{y^2(\ln(x+y))}$. Also I've noticed that $\dfrac{x+y}{\ln(x+y)}>1$
I suspect an upper bound would be $\dfrac{1}{y^2}$ but $\dfrac{x+y}{\ln(x+y)}>1$ confuses me, unless its false.

Anyone ever participated in crowd math?

1:59 AM
what is that? sounds like some kind of sex thing. count me out.

"participated in," that's what did it for me. made it sound like a whole "scene."

@leslietownes You never fail to disappoint. 0-0

It's a thing on the Art of Problem Solving website. They invite high schoolers and college students to work on open problems as a group. Anyone can join, so I thought it looked fun, but the message boards don't have much activity.
This year's problems are on "Factorizations in Additive Structures".

2:06 AM
sounds like fun, if it weren't ageist and exclusionary of those of us who are not in high school or college. but hey, you do you.
is it open problems? like those polymath things that were popular a while ago?

@leslietownes Yep, that's pretty much what the site description alludes to.

that's cool. i would have been into them if they existed when i was in college.

You finally admit to being a fellow ageist and exclusionist as well.

i love ageism and exclusion i just want to be on the right side of it.

2:10 AM
there's a lot of essentially BS awards and stuff given in the legal profession. "young attorneys to watch" etc. which sounds pervy to me.
i was asked by someone i knew if i was interested in being nominated for it and had to break the news to them that i am not, by their definition, a young attorney to watch.
i just seem young because of my up-to-the-minute cultural references.
dabs

tim robinson's series "detroiters" had a good scene where they film a local ad premised on the idea that tim, who is approximately my age, is popular with high schoolers. they go there.

This is more of a physics question but still
im trying to solve these coupled differential equations - I know $\omega$ is in the answer, but when I calculate it it disappears when taking the determinant. Can anybody see what I’m doing wrong??

2:31 AM
Does this sketch for the proof of Weierstrass' Approximation Proof make sense? BTW, the outline of the proof comes from my textbook:
Let $f(x)$ be a continuous function in $[a, b]$; we want to show for every position $\epsilon$, there exists a polynomial $P(x)$ such that $|f(x) - P(x)| < \epsilon$ for all values of $x \in [a, b]$.

My textbook says to first apply the substitution $t = a + (b-a)x$ to transform the function $f(x)$ into a function $\phi(t)$, where $t \in [0, 1]$. Now, we approximate $\phi(t)$ using a piecewise linear approximation $\psi(t)$ to within $\epsilon/2$. We represent

you'll need both positive and negative coefficients c_i in general.
this does seem like a crap way of proving the theorem.

@leslietownes does that work even if u do?

jake's left-slanted writing is uncommon. i haven't thought through the problem.
david: something like that ought to work although figuring out that it does work seems about as hard as proving the weierstrass approximation theorem.

im pretty sure u can't claim u can represent a piecewise linear approximation in the form $a + bt + \sum c_i |t-t_0|% in general rightttttt, honestly im just gonna skip this one i'm trying to think about it. a is obviously the value at 0, no mystery about that. you can form little hat functions ^ using functions of the given type, and then i guess you need to think about what happens when you add those together. maybe there's some clear way of seeing that you get all piecewise linear functions from this procedure, immediately. it seems icky to me. 2:43 AM Any elegant way to show that$\dfrac{x+y}{y\ln(x+y)} <1$? I proved it with the mean value theorem but was looking for something without involving calculus. @leslietownes what are hat functions? functions whose graphs are straight lines connecting (0,0) to (1,a) to (2,0) and zero elsewhere, and their dilations and translations. my problem with$a + bt + \sum c_i |t-t_0|$is that u can write it as$a + bt + |t-t_0|\sum c_i = a + bt + k|t-t_0|$, where$k = \sum c_i$, and i don't see how u can all the pieces of the linear approximation with this form the general spirit of many proofs of the WAT is to establish it for some class of functions that you 'know' and then show that the properties of this class of functions extends to everything. oh, is it supposed to be the same t_0 for all of them? that doesn't seem right. yea that's what it said on my textbook its weirddd but even if it was |t - t_i| the later part of approximating it to a polynomial doesn't work 2:50 AM it would seem more natural if you tossed a t_i in there and then let things vary at each point that the graph has a corner. well, that's my intuition shot. cuz then u get polynomial approximations at each section of the linear approximation not a single polynomial approximation if u use the power series method yeah, i think my mind is just struggling with this totally not being how i would prove the WAT. yeah its a weird proof - just gonna move on with my life thank so much for ur input as always much love the best proofs of the WAT use ideas from pavel korovkin. you can prove it in one page, simply, with his notions. it is not what most books do. i'll definitely look into that 2:56 AM there are a whole series of results in approximation theory deriving from his work. i don't see them in textbooks, but they are very powerful. maybe in the russian-language literature his work is better known. or maybe i should write a textbook. @leslietownes if u do, i'll read it ahahaha yea most courses seem to use bernstein polynomials, whatever they are 3:13 AM guiness had a crowd math video. i forgot, we are in victorian era land. nsfw. 18+ www.youtube.com/watch?v=km-rN1T_zfI david, the usual route if you do use korovkin is to show that the simple hypotheses of his result apply to the bernstein approximants. the bernstein approach is a good approach. copper this is asking me for a credit card number. why are you hawking pornography on a math chat? encyclopediaofmath.org/wiki/Korovkin_theorems provides background on the korovkin approach. roughly, certain approximation schemes have to converge uniformly for all continuous functions on a bounded interval if they converge uniformly for 1, x, and x^2. it led to a bunch of fairly unimaginative ripoffs of the general theorem but is still surprising to most people. because the bernstein scheme is within the realm of his theorem, you can prove WAT by proving the three bernstein approximants, of 1, x, and x^2, converge to what they're supposed to. which is like two sum formulas. it's nothing. i'd love to know what led korovkin to his result. it's fascinating. not mind blown levels of fascinating, just, what led you there? he is gone now and we cannot ask. as far as i know he did not write about it in english. 3:54 AM @copper.hat grabbed it ;-) :-) 4:10 AM @MethNoob how? when the following matrix drops below the maximal rank. Obviously, its maximum rank is 3 (i.e. pivots' numbers). @Odestheory12 it is not true.${(1+1) \over 1 \ln (1+1)} >1$. @CroCo Since it has a 3x3 identity there, it never drops rank. @copper.hat in the book, "Mathematically, a singular posture is one in which the Jacobian J(θ) fails to be of maximal rank." I am not disagreeing, but the above matrix always has rank 3. The problem I'm trying to solve for this manipulator 4:22 AM it may be that if the expressions in the denominators of that thing are zero, the manipulations used to get that matrix do not make sense. sometimes happens. but i agree with copper with the result. which obviously has singularities when all angles are equal. that's what she said. sorry copper, couldn't resist. with the image... it was all too much. did you watch the guiness ad (not with your daughter please) i didn't grasp your wit, the robot said look at the denominators in the rref. those have to be nonzero for that to be the rref. oh, i still haven't. i will momentarily. daughter is not yet abed. @CroCo i do not understand what a singular posture is? unless it involves ivermectin 4:25 AM so maybe the question is asking you to understand when the denominators of those terms in the 'rref' (which won't be if those things are zero) are zero @copper.hat when the robot loses degree of freedom. you sometimes get nonsense with a symbolic rref. the symbolic calculator has to choose what to divide by, and it can always do that if assumes the things it's dividing by are nonzero. @CroCo i am afraid i do not understand the control conditions under which that happens. the conditions for gimbal lock are not present here, that's about my limit that is probably what is going on here. @leslietownes I'm aware of sin(a_2) != 0. but this doesn't show the rest of singularities. 4:27 AM that, somewhat strangely, is also what she said. there is no$\theta_4$in the matrix in an incident that beggars belief, that too is what she said. so if the arm is completely stretched out it would not be able to reach any further, for example? the book doesn't have examples for this section yet the exercises for this problem are alot. 4:29 AM typical engineer stuff (i am an engineer) (or pretentdto be, at least.) math humour (for me at least): math.stackexchange.com/questions/4364550/… I had a discussion with Prof. Ted yesterday and he told me to check when the rank drops but I don't see any way to get out of this tunnel. almost too much for me " How wild are weakly continuous curves in a Banach space?" @CroCo i can help with mathematical stuff, but my robotic knowledge is limited to helping some relatives & friends who knew the robotics. The above matrix is not defined for$\theta_2 = 0$. @CroCo the matrix above never drops rank, but is undefined for some$\theta_2$. but i can't see how it is not singular when fully stretched out. @copper.hat I think it is a mathematical question here. The book says find the Jacobian matrix and determine when the rank doesn't reach its maximal value. From my calculation, it seems this occurs when sin(a2) != 0. But physically, there is another possibility when the robot is stretched. the potential for double entendre is too much for my simple mind @CroCo only so many ways (& times) i will repeat the same statement. by$a$do you mean$\theta$? it may be that other conditions arise that are not present in the symbolic form of the "rref" if the symbolic package used to define that knows how to simplify expressions, which most do. 4:38 AM @copper.hat yes a2 = \theta_2 sorry for the sloppy notations. other than not being defined for some$\theta_2$it always has rank 3 np, sorry for my lack of knowledge. i find it interesting much more so than reading a dry manifolds book but do not let my curiousity distract @Koro sorry i was a bit grumpy earlier i am still grumpy, but managing myself a little better :-) what's grumpy? were you ever not grumpy? well, we are talking degrees here dammit I heard that term before but don't really know what that means. lol 4:40 AM koro: irish blood boils at low temperatures. we have to tread carefully not to set them off. i had a nice cycle along the water front on a beautiful day, and got to chat with my daughter who is far far away ah, easy to get angry I thought that (the word 'grumpy') had something to do with age, not sure well, not really angry grumpy is more just in a bad mood. yeah, sort of a morning thing 4:41 AM although with copper (but not me) it probably also has something to do with age. age too, i prefer to call it experience girls call it creepy 22 hours ago, by Wolgwang I have proved that$\sin A \sin B =\sin x \sin y $. How to show$\sin x \sin y \leq \sin^2 (\frac C2)$? i had a fun text exchange with one of my older friends today. we were making jokes premised on the human anatomy, just the stupidest stuff we could come up with. we met at age 15 which explains all of this. @CroCo i suspect whatever you did to the matrix above may have eliminated some interesting stuff. @copper.hat not me but Matlab did it. 4:48 AM that's it, blame matlab re-upping chat.stackexchange.com/transcript/message/60235042#60235042 in full agreement with copper. this is the actual matrix @copper.hat ^ i guess we can assume the L's are nonzero? yes for sure this can drop rank 4:50 AM but at which angles? so if we step through this symbolically the first case to worry about is s1 or c1 being 0. you need the 4th & 5th row to lie on the same line that is, one is a multiple of the other but I need all of them. Prof. Ted told me yesterday to work out the rref() lemme think the first two rows will not contribute to the rank, nor the last row. the third row has to contribute to the rank. so it's about this other stuff. 4:52 AM well, ted has more experience than me, but let me think in any event, rref woudl not be my first shot you have rank at least 2 if s1 or c1 is nonzero. the rank can drop by 1 at most. obvious, just recording it but rref allows us to see the rank. I think this is the right start. more joes is always good beyond that i haven't done the mental math. croco the issue is that with symbolic matrices, sometimes the case analysis required to compute the rref is more complicated than finding the rank in some other way. i don't know that we're there here, but that is a background principle to keep in mind. 4:53 AM obviously I can guess some angles but I need to find all of them. ted's suggestion makes sense to me, i just don't want to carry it to completion. i'm lazy. if one of s1 or c1 is nonzero you can turn one of the column 2 entries into a pivot by dividing by the relevant expression. then the column 3 entries become slightly more complicated expressions and you wonder if their numerators are zero. I can keep guessing but for more complicated robots, I may not be able to guess. That is why I'm looking for a systemic way to do it. Notice this manipulator is simple again, word for word, what she said. you say simple. the configuration space of an arbitrary linkage, i think even with just one arm, is enough to generate all compact surfaces. so as simple as the classification of surfaces. which is not simple. one way is by a process of elimination. take the 4:5, 2:3 submatrix. this has full rank as long as$c_1 s_2 -s_1 c_2 \neq 0$. i haven't thought about this in a long time. 5:00 AM this reduces the collection of angles to look for @CroCo are you with me? I'm trying to understand your comment. the 4:5, 2:3 submatrix. what are these numbers? i mean$c_1 = \cos \theta_1$,$c_2 = \cos (\theta_1+\theta_2)$, etc that is lucid. i am not 5:02 AM jk lol in matlab you can select a submatrix, i was using that notation 4:5 you mean rows and 2:3 columns, right? i mean remove the left most column and bottom row and take the bottom left 2x2 do I need to put in rref() for the submatrix? 5:05 AM to start with. if the above is$\neq 0$the rank of the whole matrix is 3. so we only need to deal with the$=0$angles. that is the ted & leslie show teh rref i mean that would be a good show. i think he's the host and i'm ed mcmahon. on our left is our guest trump who is our rref today @CroCo i mean, do you see what i am trying to do, that is more important rn i am trying to eliminate angles that are 'good' to reduce the subsequent search. good meaning the rank of the whole matrix is 3 well, i liked my version better, i took the det of the 2x2 matrix i don't like dividing by 0, that is how we got the earlier version with had rank 3 i could be wrong, keep that in mind i think the 2x2 matrix i am referring to drops rank iff$\theta_2 = 0$. regardless of$\theta_1$. @copper.hat$\theta_1$doesn't impose any problem. 5:14 AM So, if$\theta_2 \neq 0$( or more correctly, if$\sin \theta_2 \neq 0$) then we have full rank. @copper.hat but this is same conclusion reached by @leslietownes So, to continue to find rank dropping angles, we can assume that$\theta_2 \neq 0$. ok i'll stop Ted never said rref He suggested intelligent row ops And then lost interest. but you said see when the rank drops. sounds like a military intervention to me 5:17 AM I asked when one column was a scalar multiple of the other I meant set$\sin \theta_2 =0 $to continue the search, excuse that when they are linearly dependent. i suspect the other condition is (have not worked out the details)$\sin \theta_3 = 0$. I can see for the second column, it is when L1sin(a1) = 0 & -L1cos(a1)=0 this is one possible linearly depedent case. that can't happen 5:22 AM We were looking at the transpose yesterday. Great switching ? true So my comments are for the transpose. But I’m not going any further. @TedShifrin true. my apology. I've missed the forces/Torques when the Jacobian. But I think your suggestion matches the book when the rank drops my current money (not much) is that rank drops when$\sin \theta_2 = \sin \theta_3 = 0$. 5:25 AM we have hit a loop @TedShifrin thank you. I do appreciate your invaluable time and feedback Sir. what am i, chopped liver? @copper.hat @leslietownes you too. accept my apology. :-) jk I said this to Prof. Ted because I made a mistake yesterday. 5:28 AM i think from a physical standpoint, that my above restriction makes sense so that's it then, right but when all angles are equal, the last column is [0,0,1,0,-(L1+L2+L3),0], still not linearly depedent. equal to zero* i goofed.$\sin \theta_2 = 0$does not mean that$\theta_2 = 0$. i think that$\theta_1$is irrelevant here. if the two rows are multiples of each other then the determinant of any 2x2 taken from those rows must be zero that will give$\sin \theta_2 = 0$, if i have computed correctly. this is from the third column this one such possibility. if you simplify that you will get$s_{\theta_2}=0$. yes. and the third column will be linearly dependent with the first column. 5:38 AM In particular, if$s_{\theta_2} \neq 0$it has full rank. So, we can assume$s_{\theta_2} = 0$and continue. now I think I need to check when the last column is linearly dependent I think i need to solve this @copper.hat huh? you are trying to find conditions when one row is a multiple of the other yes Am I correct? no, how would that show that one row is a multiple of the other? literature says that nilpotent groups are those that can be obtained from the commutative group by successive central extension. But I can't see why. 5:49 AM we know that if$s_{\theta_2} = 0$then$4,2:3$is a multiple of$5,2:3$(or the other way around). so the last column would have to have the same ratio to drop rank. yes I can see this now. damnnn not sure what I was thinking. the ratio (if i have nt messed up) is${c_{\theta_1} \over s_{\theta_1}}$(ignoring niceties like zeros) I can see why Prof. Ted lost interest on my attempts. well, calculation is v tedious i am curious if the calculation validates the physical intuition. I'm following the book so I don't see why this is not acceptable. I mean from physical interpretation. 5:56 AM well, if$s_{\theta_2} \neq 0$then you can wiggle$\theta_2$to go in a radial direction. if$=0$then that angle is stretched out. my guess is that the rank drop condition is$s_{\theta_2} = s_{\theta_3} = 0$. but I believe if angle 1 and angle 2 are equal, then this is a possibility of singularity. i am going to have to leave shortly, i have a presentation to a crew in India tomorrow morning so need to get up early zero is a special case. have a goody day and I appreciate your invaluable feedback. oh crap, copper reminded me i also have a work call at 6am. i do love working with people overseas but aaaaaaaaaagghghghghgh good luck @CroCo, sorry to drop & run 5:59 AM no worries I got some directions I can work them out. . and it is obviously the same annoyance for them to work with us. we have a client in taiwan and you can see people checking their watches and falling asleep. it's hard to hide it. there must surely be a quicker way of checking robot configurations. we need a bridge and hamilton around. well, i brought this upon myself. anytime will do. what was i thinking. i'll need a few cans of high strength lager and i'll go under the bridge and explore all the configurations. colt45ernions laughed out loud. reminded me of a time in grad school where we showed up at a BYOB party with 40s to learn that, uh, it wasn't the kind of party where you showed up with something in a brown wrapper. we just committed to it. almost 20 years ago. if they'd just said BYOB means bring someone else's drink, ok, we could have done that. i brought my own. my wife actually knew me then. it's a miracle we're still together. we didn't have google maps. i didn't know i was bringing malt liquor to a house with a nice deck. 6:13 AM :-). i have mellowed so much i cannot believe i am the same person. no drugs involved wither. the equivalent in ireland growing up was to bring a bottle of blue nun white wine. the student's wine. was not that easy to get low quality beer in ireland. that was a felony i believe it was immediately obvious when we arrived it was a grown-ups party. a few professors were there. i was thinking, shit, nobody told me this. wine, well.. we know who drinks that so a nice deck with a view of zinfandel drinking east bay areans and then me with my absurdly alcoholic malt liquor. dang, i must have taken my roommates beer instead, he's got the grolsch it's funny how beer has different reputations in different areas. like stella here has some kind of cachet but it is a code word for lower class in other places. and american budweiser is somehow not trash in some parts of the world. i don't remember what i brought to the party but i do remember paying less than$2.50 for it.

6:20 AM
the funny thing is, bud is not bad at all. just has that connotation

No, it’s plenty awful.

i like it. i drink all of that stuff. it's not trash if it's trying to do exactly what it does.

have done blind taste tests. only one friend passed consistently
he is/was the gm of major asian luxury hotels
can pinpoint many wine origins from taste

ted's here, time to get out the champagna and, i dunno, pheasant appetizers.

i am happy to say that tjs has cotswold crack back again

6:23 AM
the usual criticism of wine is, it's all the same, and that's not true at all. you can definitely tell the difference. i'm not as sure about beer. some of my favorites are very cheap.

and i am about to have some chloe, some italian amusment
i am better at distinguishing reds than i am whites
and even then...
however, my head can easily discriminate in the morning

i have trouble distinguishing whites. i can get oaked vs. non and sweet vs. dry but beyond that it is difficult. it doesn't help that it's served slightly under what we in southern CA would consider room temperature.
and a lot of white wines are just, ick, at room temp.
i don't want to tell them apart if they taste like that.

apparently if you must taste white (as opposed to enjoying it), you should not chill it as that makes it harder to separate the tastes
yep

as they say on the bottle of thunderbird and night train, "SERVE COLD."

:-)

6:28 AM
night train i think says serve VERY cold.
which is a good serving suggestion.
of course it's cold, you're under an overpass when you're drinking it.
but that's another story. just more of my excursions to the quaternion bridge.

:-). champagne if i must drink white
broome bridge is not very impressive other than typical arch bridge of that time
the canal construction is far more interesting (to me)
wow, i was going to sleep back then

6:49 AM
What is the matrix?
Control

7:02 AM
@love_sodam No, Get Smart has Control. The Matrix has The Resistance

1 hour later…
8:10 AM
Can there be a sequence $(x_n)$ whose set of all limit points is $\mathbb N$?

8:23 AM
Let
$$r(k)=\left\lfloor\frac{-1+\sqrt{1+8k}}2\right\rfloor$$
then look at the sequence
$$a_k=k-\frac{r(k)^2+r(k)}2+\frac1{r(k)+2}$$

Thanks a lot @robjohn. I saw the solution that you posted to my other post on mse today. :)
I'm looking at the sequence.

Can anyone suggest to me an informal definition of topology. I'm struggling to understand the topic. Also, any book recommendations for absolute beginners.

8:43 AM
@robjohn I understand that $a_k$ gives a bounded sequence $\frac{\sqrt{1+8k}}{4k}\le a_k\lt \frac{3\sqrt{1+8k}+1}{2k}$

No it doesn't
just to make the output nicer, I am making one more change...
the sequence has each non-negative integer as a limit point

\begin{align*} \frac{1+2k-\sqrt{1+8k}}{2}<\frac{r^2(k)+r(k)}{2}\le k \tag 1\\ \frac{2}{\sqrt{1+8k}+1}\le \frac 1{r(k)+1}\lt \frac 2{\sqrt{1+8k}-1} \end{align*}
@robjohn it looks nice. This graph plotting is the power of mathematica?

The plot was made with Mathematica

But I still don't understand how you created the sequence.
I mean what was the motivation etc.

How would you create the sequence $0,0,1,0,1,2,0,1,2,3,0,1,2,3,4,\dots$ That $a_k$ without the $\frac1{r(k)+2}$

8:53 AM
And I'm still verifying what went wrong in my calculation according to which $(a_n)$ is bounded.

$r(k)$ is the inverse of $\frac{k^2+k}2$

what? inverse?
My brain stopped working now :(

$\frac{4^2+4}2=10$ and $r(10)=4$

I'm trying to figure out the nth term of 0,0,1,0,1,2,0,1,2,3,0,1,2,3,4,...

you know that $\sum\limits_{j=1}^kj=\frac{k^2+k}2$

8:57 AM
yes professor Rob, I do.

The sequence of integers restart every 1,2,3,4,5,...

We want nth term $x_n$ so let it be in $(1,2,...,r)$ for some $r$, if $x_n$ is non zero.

That is where the function $\frac{k^2+k}2$ comes in
so $k-\frac{r(k)^2+r(k)}2$ is $0$ at each $k$ that is in $\{0,0+1,0+1+2,0+1+2+3,\dots\}$
and increases by $1$ in between
So $k-\frac{r(k)^2+r(k)}2$ is $\overbrace{\ \quad0\ \quad}^{r(k)=0},\overbrace{\quad0,1\quad}^{r(k)=1},\overbrace{\ \ 0,1,2\ \ }^{r(k)=2},\overbrace{0,1,2,3}^{r(k)=3},\dots$

I'lll think more on this professor Rob, it still is not clear to me because something is wrong with my head :(.

9:13 AM
....Can anyone suggest to me an informal definition of topology. I'm struggling to understand the topic. Also, any book recommendations for absolute beginners....

$$\begin{array}{c|cc} k&0&1&2&3&4&5&6&7&8&9&10&11&12&13&14\\\hline r(k)&0&1&1&2&2&2&3&3&3&3&4&4&4&4&4\\\hline \frac{r(k)^2+r(k)}2&0&1&1&3&3&3&6&6&6&6&10&10&10&10&10 \end{array}$$
@Koro This might help

@robjohn this without the zeroes does work as an example that I was looking for. I'm still trying to understand how to write its nth term. I'll get back for sure.

@Koro what do you mean "without the zeroes"?

I was starting with basics: finding nth term of 1,2,2,3,3,3,4,4,4,4,....
I think that along with this table will show me the light.
I meant that 1,1,2,1,2,3,1,2,3,4,... is the sequence which has $\mathbb N$ (the set of all naturals) as its set of all limit points.

You are leaving $0$ out of $\mathbb{N}$?

9:24 AM
yes, :D
For me $\mathbb N=\{1,2,3,...\}$

If you don't want $0$ then, starting with $k=1$, use $b_k=a_{k-1}+1$

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