Recent content by shmoe

How you've picked your b, such a k may not exist. You seem hung up on looking at n mod 10. You want to find a k where 10^k=1 mod n, this should strongly suggest looking at powers of 10 mod n.

You wrote it the other way around. You can't have looked very far if you haven't been able to find a k where 7 divides (10^k)-1, what values of k did you try?

There isn't one, but that's irrelevant to this question... you're mixing up what is supposed to divide what. the log thing won't help, you should be able to say why. You may already have the notion of order that matt is talking about without ever looking at groups. Elementary number...

The average density tending to infinity does not imply the series will diverge. How fast it tends to infinity is important. Here, you only have log(T) times as many zeros up to T as you do integers (ignoring constants and lower order terms), which really isn't that much, and not enough to make...

Hey, I'm in North America and Thanksgiving holidays were several weeks ago :-p I'll up the ante on pet photos to one adult cat and two 3-4 month old kittens. I'm thankfull that the adult has accepted the kittens very quickly (4 days or so after bringing them) and that this acceptance goes...

That's the example I was talking about, I didn't say they showed you what primitive root to use or how to find it. I said you can fill in their steps using the primitive root I suggested, i.e. take the data from (6.10) and you can reproduce all the numbers in their example when squaring...

The abc conjecture is quite well known, google will turn up many, many hits. Mathworld's, http://mathworld.wolfram.com/abcConjecture.html

Of course you want it in the hands of the experts you know, arxiv isn't stopping that. It can make it available to both the experts you want and others. Was her error found by someone who read the paper on arxiv? I don't know. experts certainly had the time to read perelman's work on...

the FFT is a way to compute the DFT with less operations than the 'naive' way of perform the transformation (just computing the terms in the sum as they appear). the DFT is a special case of the DWT, taking the weights to be 1's. For their algorithm in section 6, use a primitive Nth complex...

I'm a little late commenting, but this sort of thing is a bad idea. getting 'results' announced in the popular press before they are 'results' makes mathematicians look like tools to the general public if they have to be retracted. It's not a good idea to have the public thinking mathematicians...

Are you familiar with the usual discrete Fourier transform? This weighted version appears to be a modification the same thing. "appropriate domain" can be whatever ring you are considering your x_i's to be a part of. You could take the complex numbers for example, a primitive nth root of unity...

I think you mean the repeated squaring idea for finding exponents? To find x^n you first find x^1, x^2, x^(2^2), x^(2^3), x^(2^m) where 2^(m+1)>n. this takes some log(n) squaring operations. Then write n in binary and multiply the appropriate numbers from your list to get x^n, another log(n)...

In defense of 1729 it is possible to discuss it's fame in the mathematical folklore without boring the pants off the 'average' person (though boring their pants off may be the goal), the relationship between Hardy and Ramanujan should have appeal to anyone. GautamAishwarya, the advice to talk...

rule b tells you g(4) is either g(3)-1=0 or 4*g(3)-1=3, nether work of course, but 15 isn't an option.

It's not that hard to do by hand, you'll get the sequence 2, 1, 3, 11, 10, ..., 4, 15, 14, ..., 12, 47, 46, ..., 16, 63, 62, ...48, ... just group as: (2, 1), (3), (11, 10, ..., 4), (15, 14, ..., 12), (47, 46, ..., 16), (63, 62, ...48), ... the number of elements in the groups are...