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comment Determine all one to one functions $f:\mathbb{N}^* \rightarrow \mathbb{N}^*$ having the following property:
@barto Ah, so that shows $f(2) = 2$ (since it's the only number occurring in every $4$-tuple, from which follows $f(3) = 3, f(6) = 6$. This does lend weight to the hypothesis that the identity map is the only solution. This would be easy if one could prove something like "if $m < n$ then for some sufficiently large $k$ there are more $k$-tuples containing $m$ than there are containing $n$".
4h
revised Which constellations of primes recur forever?
deleted 784 characters in body
4h
revised Which constellations of primes recur forever?
added a construction, some details missing
5h
answered Which constellations of primes recur forever?
10h
comment Determine all one to one functions $f:\mathbb{N}^* \rightarrow \mathbb{N}^*$ having the following property:
@Dominik See Egyptian fractions. It's trivial to create infinitely many examples by splitting $\frac1n \mapsto \frac1{n+1} + \frac1{n(n+1)}$, but there are many many more than just this.
11h
comment Solving general linear ODE $\sum_{k=0}^n y^{(k)}=0$
Great, then you should be able to recognize that marty's answer is the key to understanding the common pattern to the coefficients in your table above (hint: $e^{(a+bi)x} = e^{ax} (\cos bx + i \sin bx)$). If it's important to you that the basis representatives are real-valued functions, then think about how the roots pair into complex conjugates (this will work out slightly differently depending on whether $n$ is odd or even).
21h
comment Anagrams contained within random strings
The probability is possible to calculate, but it depends very critically on the number of repetitions of each letter in the shorter string and as the answers below indicate, it'll be very messy. On the other hand, it's a lot easier to compute the average number of anagram occurrences. (The only difference is when the anagram occurs multiple times in the same string, including the case where the multiple occurrences overlap each other). When the probability is fairly low, then this will be a good approximation to the probability... usually.
22h
comment Prove that $s(n-1)s(n)s(n+1)$ is always an even number
@You-know-me Have you thought about the conditions under which $\sigma(n)$ is odd? Hint: try to show that when $n$ is odd, then $\sigma(n)$ is odd iff $n$ is a perfect square. What can you say about the case of $n$ even?
22h
comment Prove that $s(n-1)s(n)s(n+1)$ is always an even number
@Moya It probably isn't the restricted divisor function since the first four values of that are $1,1,3,1$.
1d
comment Solving general linear ODE $\sum_{k=0}^n y^{(k)}=0$
How much have you learned about complex roots of unity?
1d
comment Crude verification of Goldbach conjecture
@BeWakePandey It isn't clear what your question is. You give a crude heuristic which could be refined to a sharp heuristic, and though neither is proven. What type of answer are you expecting that you'd accept?? As it stands, there isn't any question in your question.
1d
comment Crude verification of Goldbach conjecture
@BeWakePandey If it had been proven that GC can't be disproven, that would amount to a proof that it's true, since any counterexample would become a trivial disproof.
1d
reviewed Close Primes and even numbers
1d
comment Proof check: $(4n)!$ is divisible by $2^{3n}3^{n}$
It might be, but you didn't really show your induction for $2^{3n}\mid (4n)!$. If anything, the $3^n$ part is the much easier part (and it would be very easy to prove by induction). So I would question why you are uncertain about the $3^n$ part but don't seek verification for the messy induction?
1d
comment For all $n$ there exists $x$ such that $\varphi(x)<\varphi(x+1)<\ldots<\varphi(x+n)$
@PaoloLeonetti You seem to be sending mixed signals: in the question you say you know the answer is positive, and in one comment you assert that the lower asymptotic density is positive. But here you ask how to prove it. Do you already have a reference for the result? If so, then what exactly are you looking for? If not, then how do you know it's true?
2d
revised The longest sequence of numbers with a certain divisibility property
Added OEIS reference
Aug
27
awarded  prime-numbers
Aug
26
comment The longest sequence of numbers with a certain divisibility property
@BradGraham Don't focus so much on the result (it may look complex but that's only because the amount that can be accomplished is so small that a certain level of precision is needed just to distinguish the asymptotic from the trivial bound and earlier results). From what I can tell, the covering question exactly matches your definition of longest denizen. If you can see a specific way it doesn't match, then you need to provide a clear definition (for once).
Aug
26
answered The longest sequence of numbers with a certain divisibility property
Aug
26
comment Non repeatable combinations
@Angie Your question is very poorly worded and I don't think it can be answered in its current state. However, the expression "we're interested only in the number of" indicates that when counting the number of different arrangements, you should consider two arrangements to be the same if just the number of boys is the same in each dance (even if it's a different group of 5 boys, say).