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0.999... does equal 1, see www.physicsforums.com/showthread.php?t=507001 [Broken]

I'll try to come up with a better one, mine's terrible. How on Earth did you work out frg(15)?

I calculated the first 8 and put them in to OEIS, and got: oeis.org/A072752. What you're after is not the gaps, but the difference, so it's one more than the terms in the sequence I linked to. I'm not sure about an efficient algorithm, my jumbled together program could only do 8 before taking...

Stop it. You're absolutely unimaginably nowhere near Graham's number, and it's pointless to try and come up with a larger number.

http://en.wikipedia.org/wiki/List_of_unsolved_problems_in_mathematics#Analysis See the fifth one down.

Euler's formula says that e^{ix} = \cos x + i \sin x. Let x = 2 \pi n, where n is any integer. Then e^{ix} = \cos x + i \sin x = 1 + 0i = 1. Therefore, e^{ix} = 1 has an infinite number of solutions, all of the form x = 2 \pi n. The two you brought up, x=0 and x = 2 \pi are just the ones...

http://en.wikipedia.org/wiki/Transcendental_number#Numbers_which_may_or_may_not_be_transcendental

These are the numbers that are the sums of distinct squares: https://oeis.org/A003995 These are the numbers that are not: https://oeis.org/A001422 And these that are the sums of distinct cubes: https://oeis.org/A003997 These are the numbers that are not: https://oeis.org/A001476 From manually...

a(n) = round(\sqrt{n}) seems to work, not that I could prove it. This is where a(1) is the first entry in the sequence.

You say here "assume x^n + y^n = z^n", then later say "x^n + y^n < = > z^n + u^n". As u^n is positive, clearly x^n + y^n < z^n + u^n. This makes nearly all of this case unneeded. And can you explain how Case 1 shows anything here is false? Case 1 only applies when x^2 + y^2 = z^2, which...

The example of x and y you gave doesn't work.

Right, ignore my posts, I've decided they're nonsense.

Despite the fact I got next to nowhere with it, that was a really fun problem.

Using Goldbach's conjecture, any even integer is the sum of two primes (at least up to 1.609 × 10^18). Meaning that (p+3) is the sum of two primes, and 3 can be subtracted to get p. Or more generally (p+q) is the sum of two primes, where q is a prime number, and q can be subtracted to get p...

It's true for all primes between 13 and 9973.

I don't even think it's worth mentioning, but the square root of zero or one factorial is rational. Also, I'm sure it's obvious for most people, but you could explain how you got from: \lim_{n \to \infty } O(\sqrt{n!}) = 1 to \lim_{n \to \infty } \sqrt{n!} = mO(\sqrt{n!})

I'm not sure what you mean by "but you only can use once", so this could be wrong. Would you need to only weigh exactly either only odd or only even weights? For example, if could weight 35 and 37, and the item you were weighing was 36, you could try it on the 35 balance, see it's too heavy...