11,710 reputation
236107
bio website mit.edu/~darij/www
location Karlsruhe (home), Munich (until summer 2011), Cambridge/MA (2011-)
age 26
visits member for 5 years, 1 month
seen 20 mins ago
I'm just here for asking stupid questions.

19h
comment A possibly surprising appearance of Lucas numbers
My definition of the energy is somewhat wrong, and I am not really sure how to fix it.
1d
comment A possibly surprising appearance of Lucas numbers
By the last $q$, I meant $q+1$.
1d
comment A possibly surprising appearance of Lucas numbers
... adding the entry $u$ in the appropriate place.)
1d
comment A possibly surprising appearance of Lucas numbers
The induction relies on the lemma that for every $q \in \mathbb{N} + \mathbb{N}r$, the energy of $q+1$ is $\leq$ to (the energy of $q$) plus $1$. This is rather annoying to formalize, but I am fairly sure that the "follow your nose" proof goes through. (The idea is: Write $q$ as $r^{a_1}+r^{a_2}+...+r^{a_k}$ as above. Let $u$ be the smallest even integer which does not belong to $\left\{a_1,a_2,...,a_k\right\}$. Then, $q=r^{b_1}+r^{b_2}+...+r^{b_l}$, where $\left(b_1,b_2,...,b_l\right)$ is the result of removing the entries $0,2,...,u-2$ from the sequence $\left(a_1,a_2,...,a_k\right)$ and ...
1d
revised A possibly surprising appearance of Lucas numbers
edited body
1d
comment A possibly surprising appearance of Lucas numbers
... for every $n \in \mathbb{N}$, every element of $\mathbb{N} + \mathbb{N} r$ having energy $n$ belongs to $G(n)$, and that every element of $G(n)$ must have energy $n$ (these inductions need to happen in lockstep). As a consequence, it easily follows that $G(n)$ is the set of all elements of $\mathbb{N} + \mathbb{N} r$ having energy $n$.
1d
comment A possibly surprising appearance of Lucas numbers
I think the main step here is the following observation (for $r = \sqrt 2$): Every element $q$ of $\mathbb{N} + \mathbb{N}r$ can be written uniquely as a finite sum $r^{a_1} + r^{a_2} + ... + r^{a_k}$ for nonnegative integers $a_1,a_2,...,a_k$ satisfying $a_1 < a_2 < ... < a_k$. (Indeed, just write the integral and the $r$-integral parts of $q$ in binary.) We denote the nonnegative integer $k + a_k$ as the energy of $q$ (where $a_k := 0$ if $k = 0$). Now if I am not mistaken, then an inductive argument shows that ...
1d
comment A possibly surprising appearance of Lucas numbers
Do you perhaps mean $g(2) = \{2, x\}$ instead of $\{1, x\}$ ? And $p+1$, not $x+1$, being in $S$?
1d
revised Proof of closed walk generating function identity
URL
1d
comment A possibly surprising appearance of Lucas numbers
Actually, adding $1$ does not give your formulas, so I expect that the condition you omitted is subtler than that.
1d
comment A possibly surprising appearance of Lucas numbers
Welcome to MO! I have edited your post to use LaTeX instead of ascii (which was conflicting with MO's markup syntax). But I think something is missing in your definition of $S$, as multiplying $0$ with $x$ only gives back $0$. Maybe it is also allowed to add $1$?
1d
revised A possibly surprising appearance of Lucas numbers
LaTeX
Jan
27
revised Has Reifegerste's Theorem on RSK and Knuth relations received a slick proof by now?
correcting stupid errors
Jan
26
comment Artin Jacobson-semisimple rings are semisimple. Constructively, too?
Very nice counterexample! So your subring essentially steals its zero-radical property from the $M_2\left(\mathbb F_2\right)$ factors, its lack of von Neumann regularity from the (limit) ring of upper-triangular matrices, and its strong $\pi$-regularity from both.
Jan
26
revised Artin Jacobson-semisimple rings are semisimple. Constructively, too?
added 198 characters in body
Jan
26
revised Artin Jacobson-semisimple rings are semisimple. Constructively, too?
added 745 characters in body
Jan
26
comment Artin Jacobson-semisimple rings are semisimple. Constructively, too?
I'm not really sure about this all. But I don't think that you can constructively prove that the Jacobson radical is zero for a ring which "floats" somewhere between $\left\{0,1,r,1+r\right\}$ and the whole $R$ (e.g., the ring containing $0,1,r,1+r$ and also containing $r^T$ if some given undecidable statement is true). But Artinianity is also far from obvious for these rings. Let me edit my OP actually.
Jan
25
revised What are the reasons for considering rings without identity?
no idea why i had a ² here
Jan
19
comment What is a good introduction to cluster algebras from surfaces?
I know of them, though I'd like something with proofs. Thanks for reminding me of them, though.
Jan
19
comment What is a good introduction to cluster algebras from surfaces?
I am so looking for a nice readable introduction to precisely the other half of cluster algebra theory (the algebraic and combinatorial parts). But on the geometric side, I have been recommended Gekhtman, Shapiro, Vainstein (the AMS book, not the arXiv paper) -- or is it yet another geometric component of cluster algebra theory? (NB: I have no idea about geometry.)