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Jul
13
awarded  Populist
Jul
4
answered Sum and product of linear recurrences
Jun
28
revised $\binom{2p-1}{p-1}\equiv 1\pmod{\! p^2}$ implies $\binom{ap}{bp}\equiv\binom{a}{b}\pmod{\! p^2}$; where $p>3$ is a prime?
edited body
Jun
28
answered $\binom{2p-1}{p-1}\equiv 1\pmod{\! p^2}$ implies $\binom{ap}{bp}\equiv\binom{a}{b}\pmod{\! p^2}$; where $p>3$ is a prime?
Jun
28
comment $\binom{2p-1}{p-1}\equiv 1\pmod{\! p^2}$ implies $\binom{ap}{bp}\equiv\binom{a}{b}\pmod{\! p^2}$; where $p>3$ is a prime?
The right hand side of the congruence in the last line should be ${a-1\choose b-1}$. Also I don't see how the last line follows from what came before.
Jun
26
comment $\binom{2p-1}{p-1}\equiv 1\pmod{\! p^2}$ implies $\binom{ap}{bp}\equiv\binom{a}{b}\pmod{\! p^2}$; where $p>3$ is a prime?
Both congruences actually hold modulo $p^3$. Did you mean to ask just about $p^2$?
May
25
revised Stably-free ideals are free?
edited body
May
25
answered Stably-free ideals are free?
May
25
comment Stably-free ideals are free?
Oh, right. I guess this is only true for finitely generated modules.
May
25
comment Stably-free ideals are free?
You have shown that $I$ is stably free, which in general is much stronger than projective (= locally free).
May
13
comment Evaulate/approximate a series formula $\sum_{i=1}^{n}\left ( \frac{1}{n}\right)^i \left(\frac{n-1}{n}\right)^{n-i}$
The sum isn't a binomial expansion because the binomial coefficients are missing.
Apr
5
awarded  Yearling
Mar
7
awarded  Good Answer
Feb
25
comment A non-nilpotent formal power series with nilpotent coefficients
@Gauloises Yes, it's not hard to check that the Freshman's Dream $(a+b)^2=a^2+b^2$, which holds in characteristic 2, extends to formal power series: $(\sum a_n x^n)^2=\sum a_n^2x^{2n}$
Feb
20
answered Even/Odd Binomial Coefficients
Feb
9
comment Binomial expansion (sort of ) rearrangement
I've added an extra step, hopefully this helps some. At the added step, I'm using the fact that the $k$-th power of a sum is the sum of products of ordered $k$-tuples of summands.
Feb
9
revised Binomial expansion (sort of ) rearrangement
added 141 characters in body
Feb
8
revised Why is the following subset of $\mathbb{C}$ simply connected.
deleted 4 characters in body
Feb
8
answered Why is the following subset of $\mathbb{C}$ simply connected.
Feb
8
answered Binomial expansion (sort of ) rearrangement