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 Dec18 awarded Caucus Mar2 revised How can I solve equations like $A(z)=1+z+(z+z^2)A(z)$? Removed thanks, reworded for introduction of the context before the question, instead of after the question. Mar2 suggested approved edit on How can I solve equations like $A(z)=1+z+(z+z^2)A(z)$? Dec22 accepted Graph on the cover of Bollobás's “Combinatorics” Dec21 comment Graph on the cover of Bollobás's “Combinatorics” BTW, all of the answers are superb and I don't know what to choose as "the" answer to this question... Dec21 comment Graph on the cover of Bollobás's “Combinatorics” @ThomasAndrews, Great. Translating from your description to wikipedia's notation: for any two adjacent vertices, there is exactly 1 path of length 1 between then, which means that any 2 adjacent vertices have exactly 1 neighbor in common (in wikipedia's notation, $\lambda = 1$). For any 2 non-adjacent vertices, there is exactly 2 (vertex disjoint) paths of length 2 between them, which means that 2 non-adjacent vertices have 2 neighbors in common (in wikipedia's notation, $\mu = 2$). The graph is regular of degree $k = 4$ with $v = 9$ vertices and is, therefore, $sgr(9, 4, 1, 2)$. Dec20 asked Graph on the cover of Bollobás's “Combinatorics” Dec15 accepted Showing that the infinite grid is Eulerian Dec15 comment Showing that the infinite grid is Eulerian Your explanation of what you meant with the ellipsis is helpful in understanding what you meant. But there is one problem, though, if I were to transform this into a proof by induction: after I traversed the whole plane, how do you define "the last 3 or 4 or 5" last steps? Perhaps I misunderstood you? Dec15 comment Showing that the infinite grid is Eulerian The definition of Eulerian given in the book for infinite graphs is that you simply have a path that extends from its two end vertices indefinitely, is allowed to pass through any vertex any number of times, but each edge only a finite number of times. Dec15 comment Showing that the infinite grid is Eulerian @GerryMyerson, I didn't get what Eppstein said about an even sided diamond shaped region. That would include the particular case of squares, but I can't even see the case for squares! :) OTOH, I am following (again) his hint of walking the edges of the graph such that the endpoints of a visited region of the graph always lie on its border and it has been working so far for a drawing... I am distilling some observations, but I don't know if I will be able to transform them into an algorithm to cover the whole graph. Let's see how things progress now. :) Dec15 comment Showing that the infinite grid is Eulerian How do you get from the 5th step to the 6th step? In the 5th step, let's call the two extremes $u$ (at the left) and $v$ (at the right). The only way that I see that jump is to get $v$, visit 3 sides of the upper square and then finish there at the top. Is this right? You are basically just walking with one of the endpoints, right? But in your last figure, how do you get of the dead-end? Dec15 awarded Scholar Dec15 awarded Student Dec15 asked Showing that the infinite grid is Eulerian Nov16 suggested rejected edit on Burnside's Lemma, applications Nov14 awarded Editor Nov14 comment How to show that $1 \over \sqrt{1 - 4x}$ generate $\sum_{n=0}^\infty \binom{2n}{n}x^n$ It would be a good thing to note that, while we are using the same variable names (namely $n$) both on the left and on the right side of the equations, for one to derive that $na = 2$ and $n(n-1)a^2 = 12$, as you did, you have to consider that, on the RHS, when setting $x = 0$, you need the coefficients to be already known to be constant---not involving the variable $n$ of the LHS. Nov14 revised Too old to start math Fixed leftover typo for consistency. (As edits can't be this small, I made some other cosmetic changes.) Nov14 suggested approved edit on Too old to start math