Jason S
Reputation
1,188
Top tag
Next privilege 2,000 Rep.
 Feb11 asked numerical integration for N datapoints Sep15 awarded Enlightened Sep15 awarded Nice Answer Jul21 comment closed-form expressions for product of 3n+k where k = 1 or 2 yes, I know. Read the Wolfram Alpha link. The gamma function expressions involve gamma(n+4/3) or gamma(n+5/3) which are not integer factorials. Jul21 asked closed-form expressions for product of 3n+k where k = 1 or 2 Jul20 awarded Yearling Apr5 comment constructing “pseudonoise” sequences other than (2^n)-1? (low cyclical autocorrelation) hmm, looks like there's a lot of research on this: signalslab.marstu.net/?page_id=1769 Apr5 asked constructing “pseudonoise” sequences other than (2^n)-1? (low cyclical autocorrelation) Mar7 comment How can adding an infinite number of rationals yield an irrational number? Infinite sums have meaning, sure. But I have a finite lifetime, and each addition step takes a finite time -- therefore I cannot add infinitely many numbers. Neither can a computer, by the same reasoning. However I can use reasoning in some cases to compute the limit of an infinite sum (as could a computer in some other cases, with sufficient programming). The difference may be a bit pedantic, but this is mathematics, after all. Feb24 awarded Taxonomist Dec26 comment how to distribute n red and m blue balls in some containers to maximize probability of random picking a red one from them? off-topic. Best fit is probably math.stackexchange (NOT mathoverflow.net) Oct6 awarded Nice Answer Jul21 awarded Yearling Jun22 accepted order-independent accumulator operations? May2 comment order-independent accumulator operations? That's reasonable. I was going to put only + and XOR in my original post, but the context for this is an "accumulator" where you start with a = 0 and keep applying a = f(a,x) for some function/operation f. The functions f(a,x) = a+x, f(a,x) = a^x, and f(a,x) = a-x were the only ones I could think of where the order didn't matter. May2 comment order-independent accumulator operations? cool! your constructive approach in the first paragraph is interesting, although in a practical sense, you'd have to be mapping back and forth between regular integers and permuted sets of integers. (which I suppose explains your "tongue-in-cheek" comment) May2 comment order-independent accumulator operations? 0 - a - b - c - d = 0 - b - c - a - d May2 asked order-independent accumulator operations? Apr2 comment There are infinitely many primes such that $p^n \equiv 1 \pmod{b^m}$ implies $b^{m-1} \mid n$ @Arturo: OK, I follow. (though your restatement is slightly off: the claim is that for fixed m and n, for any odd b, there are infinitely many primes p such that (if p^n congruent to 1 (mod b^m), then b^(m-1) | n). (my parentheses as per the original problem statement)) Apr2 comment There are infinitely many primes such that $p^n \equiv 1 \pmod{b^m}$ implies $b^{m-1} \mid n$ :confused: b is any given odd positive integer, that's a condition on b... I'm just not following why the contrapositive is false.