Reputation
3,305
Top tag
Next privilege 5,000 Rep.
Approve tag wiki edits
Badges
1 9 23
Newest
 Yearling
Impact
~36k people reached

Aug
7
answered Finding real money on an even stranger weighing device
Aug
6
comment Finding real money on an even stranger weighing device
At each step there is the choice of including coin 0 or not. This decision can be made based on the weights of previous weighings. As far as I can see, none of the discussion takes this possibility (which makes the analysis considerably harder) into account. Perhaps there's some good reason to believe that this extra freedom makes no difference to the answer to the question.
Jun
30
awarded  Yearling
Jan
9
revised Chi-square or chi-squared?
added tag
Jan
9
revised Chi-square or chi-squared?
improved punctuation
Jan
9
asked Chi-square or chi-squared?
Oct
20
comment A counting problem on the integer lattice
@Shibi, I'd misread the question. I've now corrected my answer. It's not just 4-fold symmetry you have to worry about. For example, $(1,7)$ and $(5,5)$ are at the same distance from the origin.
Oct
20
revised A counting problem on the integer lattice
corrected due to misunderstanding of question
Sep
30
awarded  Explainer
Sep
29
answered A counting problem on the integer lattice
Sep
29
comment Countable or uncountable set 8 signs
I stand corrected.
Sep
29
comment Sequence of numbers with prime factorization $pq^2$
@Nishant, the only numbers of the form $pq^2$ having $6$ as a factor are $12$ and $18$.
Jul
2
awarded  Curious
Jun
30
awarded  Yearling
Oct
21
comment How many fours are needed to represent numbers up to $N$?
103 = 44 / .4. + 4
Sep
23
answered Would this be bounded
Jul
25
comment Generating Function for edge-rooted labelled trees
Yes, of course; the labels induce an orientation (of every edge). You didn't state that you were considering labelled trees, but you did state that $T_v(z)$ was the exponential generating function, so that should have been evident.
Jul
25
comment Generating Function for edge-rooted labelled trees
We've both assumed that the edge-root is oriented (i.e. the two ends are distinguishable). If it isn't, then we need to discard one of each asymmetric pair, which gives $T_e(z)=\frac{1}{2}(T_v(z)^2+T_v(z^2))$.
Jul
25
comment Generating Function for edge-rooted labelled trees
This doesn't look correct; you can add a forest of trees to each end of the designated edge (since the two new vertices are external to what is added). It's probably simpler just to take two vertex-rooted trees and identify the roots as the ends of the designated edge, giving $T_e(z)=T_v(z)^2$.
Jun
30
awarded  Yearling