bio | website | |
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location | Pasadena | |
age | 26 | |
visits | member for | 5 years, 3 months |
seen | 12 mins ago | |
stats | profile views | 22,997 |
You can contact me at gjergjiz at gmail.com
Mar 12 |
comment |
What's the analogue of a Young symmetrizer in the Brauer algebra?
There is an explicit formula in Nazarov's 2002 ICM address arxiv.org/abs/math/0209129 or his previous paper with a similar title. His setting is slightly more general but it's not hard to specialize to the Brauer algebra. |
Mar 10 |
awarded | Good Answer |
Mar 9 |
comment |
A set of integers whose factorial can be written as a product of two factorials
Some equations of the form $k!=P(n)$ ca be solved, but the case $P(n)=n(n-1)$ is still out of reach, I believe. mathoverflow.net/questions/39210/… |
Mar 3 |
awarded | polynomials |
Mar 2 |
comment |
On a reciprocal of Ostrowski theorem on Newton polytopes and factorization
@Bruno, the stronger question has a negative answer, too. I added an example to the answer. |
Mar 2 |
revised |
On a reciprocal of Ostrowski theorem on Newton polytopes and factorization
added 451 characters in body |
Mar 2 |
answered | On a reciprocal of Ostrowski theorem on Newton polytopes and factorization |
Mar 2 |
awarded | Popular Question |
Feb 28 |
awarded | Enlightened |
Feb 28 |
awarded | Nice Answer |
Feb 28 |
awarded | Nice Answer |
Feb 23 |
revised |
A variant to the Hadwiger-Nelson problem
added 494 characters in body |
Feb 23 |
answered | A variant to the Hadwiger-Nelson problem |
Feb 22 |
comment |
Fixed points of self maps
Do you want assumptions on the underlying field? Are you working over $\mathbb C$? |
Feb 22 |
comment |
Square-free grows as $6n/\pi^2$: $k$-th free?
Yes, your guess of $1/\zeta(k)$ is correct :) en.wikipedia.org/wiki/… |
Feb 17 |
comment |
Which power series have bounded integral coefficients and have an inverse given by a series having bounded integral coefficients
In some sense these are also "products of cyclotomic polynomials", you just need to allow for infinite products. See my answer here mathoverflow.net/questions/50798/… |
Feb 12 |
awarded | Enlightened |
Feb 12 |
awarded | Nice Answer |
Feb 12 |
comment |
Number of semi-standard tableau
@Harry, almost! It's the coefficient of $t^{kn}$ in $(1-t)(1+t+\cdots+t^{2k})^n$. |
Feb 12 |
comment |
Number of semi-standard tableau
I'm not sure what type of calculation you have in mind. A way of doing this without using symmetric functions is to notice that your tableaux correspond to Dyck paths of length $2n$ with no peaks at odd level. You can work out the recurrence relations straight from this, however some amount of work is unavoidable given that the final formula is "not too simple". |