Reputation
36,592
Next tag badge:
115/100 score
17/20 answers
Badges
4 91 143
Newest
 Nice Answer
Impact
~549k people reached

Apr
29
revised Find a subset such that its sum is divisible by $n$
edited tags; edited title
Apr
28
revised Modified Jacobi’s theta function
edited tags
Apr
28
revised Is this group given presentation isomorphic to $\mathbb{Z}_2$, and why?
deleted 2 characters in body
Apr
26
comment Generalizing Ramanujan's “1729 story”
It is classical that $r_2(k)$ and $r_3(k)$ can be arbitrary large. It is conjectured (but it is also widely open) that for $n\geq 5$ we have $r_n(k)\leq 1$. I don't know the status of $n=4$ from the top off my head.
Apr
25
revised Checking Mertens and the like in less than linear time or less than $\sqrt{x}$ space
edited tags
Apr
25
comment Function that gives 1 only when an integer is divisible by another integer
@AndrejBauer: See my "Added" section. (P.S. Trust analytic number theory.)
Apr
25
revised Function that gives 1 only when an integer is divisible by another integer
added 450 characters in body
Apr
25
answered Function that gives 1 only when an integer is divisible by another integer
Apr
24
comment Precise asymptotic of diophantine approximation
@DouglasZare: We talk about different things. Of course the same limsup (or sup), when it is at most $1/3$, can be attained by continuum many (pairwise inequivalent) $\xi$'s. This harmonizes with my response. If the limsup (or sup) exceeds $1/3$, then $\xi$ is equivalent (in a certain sense) to $\sqrt{9-4m^{-2}}$ with $m$ a Markov number, and the limsup (or sup) is given in terms of this $m$, and there are countably many such $\xi$'s.
Apr
24
revised On cluster points of a particular sequence
edited tags
Apr
24
comment Precise asymptotic of diophantine approximation
@DouglasZare: Note also that by an element in the Markov spectrum I meant the quadratic irrational $\sqrt{9-4m^{-2}}$, where $m$ is a Markov number.
Apr
24
comment Precise asymptotic of diophantine approximation
@DouglasZare: I think no, since $\mathrm{SL}_2(\mathbb{Z})$ is countable. Two $\xi$'s are equivalent if and only if their continued fraction expansions are the same apart from finitely many coefficients and a shift of the coefficients.
Apr
23
comment neat identities just for operators?
This is an intense area of reseach. I think a full description of polynomial identities is known for $2\times 2$ matrices, but for higher ranks such a description is out of reach. Check out the papers quoting the last paper, and the literature on "polynomial identities for rings".
Apr
23
revised neat identities just for operators?
edited tags
Apr
23
comment neat identities just for operators?
See also sciencedirect.com/science/article/pii/S002186938371207X
Apr
23
comment neat identities just for operators?
Yes, see en.wikipedia.org/wiki/Polynomial_identity_ring
Apr
23
revised Precise asymptotic of diophantine approximation
added 270 characters in body
Apr
23
comment Precise asymptotic of diophantine approximation
@Wojowu: Thanks, I will clarify my response.
Apr
23
revised Precise asymptotic of diophantine approximation
added 136 characters in body
Apr
23
revised Precise asymptotic of diophantine approximation
edited tags