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Sep
15
reviewed Close What are the root of $x^3 - 2$ $\in \mathbb{R}[x]$?
Sep
13
reviewed Reopen condition for transitivity
Sep
12
revised Partial sums of Nicomachus' Triangle rows produce Stirling numbers of the 2nd kind?
added 1110 characters in body
Sep
12
answered Partial sums of Nicomachus' Triangle rows produce Stirling numbers of the 2nd kind?
Sep
11
revised 4-part partitions of n and 3n
deleted 90 characters in body; edited tags; edited title
Sep
11
answered Maximum value of $a+b$ given that $\frac{1}{a} + \frac{1}{b} = \frac{1}{20}$
Sep
10
reviewed Reject suggested edit on Sign of a series
Sep
5
reviewed Close Series expansion of $\log(1+x^2)$
Sep
2
answered a 2-regular graph is cyclic or not?
Aug
30
reviewed Close Prove a condition for a Banach algebra
Aug
30
answered Pushdown Automata and Challenge in Grammar
Aug
23
reviewed Leave Open Generating function for binomial coefficients $\binom{2n+k}{n}$ with fixed $k$
Aug
20
reviewed Close In how many ways can a number be expressed as a sum of squares of two natural numbers?
Aug
19
comment The Day Camp Stacking Game
Is rule 4 a repetition of rule 2 or is it trying to say something additional? (I assume that "clockwise" and "left" mean the same thing here).
Aug
17
reviewed Leave Closed Probability of ace in each hand
Aug
6
comment Deformable circle from a cubic Bezier approximation
What do you mean by "smoothness"? Do you want C2 continuity? G2? Something else?
Jul
31
comment Lower bound for a relative of the central binomial coeff
@Raphael, can you explain why there's a singularity at $-2+\sqrt{5}$? I see singularities at $-2-\sqrt{5}$ and $\frac{1}{4}$.
Jul
30
comment Lower bound for a relative of the central binomial coeff
@GerryMyerson, the central binomial coefficient has a rather neat product representation which allows an easy ad hoc proof of the bound. The only way I can see to get a similar product representation is to factor out a central binomial coefficient from each of the binomial coefficients in the second sum. Then bounding with $\frac{1}{\sqrt{m-2j-1}}\ge\frac{1}{\sqrt{m}}$ I get a non-Gosper-summable hypergeometric, and crudely bounding the terms to get a summable geometric sequence gives a worse bound than simply taking the first term of the original sum.
Jul
30
comment Prove equivalence of Diffie-Hellman shared secret
@codeomnitrix, $a \equiv b\pmod c$ is a shorthand notation for $\exists k : a - b = ck$
Jul
30
asked Lower bound for a relative of the central binomial coeff