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 Oct 3 comment Is this a isomorphism $(\mathbb Z,+) \to (Z,+)$ where $\varphi(n) = 2n$? Why not? You are missing that there is no n such as φ(n)=1. Or φ(n)=3. So, the image of φ is not Z but a subset. You say " for every element in Z in the first group there is a element 2n in the second group." Right. But for every element y in Z in the second group is there is an element n in the first group such as φ(n)=y? Oct 3 comment Solving $p_1^{e_1} p_2^{e_2}…p_k^{e_k}=e_1^{p_1} e_2^{p_2}…e_k^{p_k}$ @Tom-Tom I'm not sure but I see 4 or 5 adjacent related seqeunces, so I guess no. The "last modified" footer seems to in all OEIS sequences. Oct 3 comment Solving $p_1^{e_1} p_2^{e_2}…p_k^{e_k}=e_1^{p_1} e_2^{p_2}…e_k^{p_k}$ @AnalysisIncarnate No, only just now searched for it. It doesn't have a date but several adjacent sequences were created by the same author (Olivier) Oct 3 comment Solving $p_1^{e_1} p_2^{e_2}…p_k^{e_k}=e_1^{p_1} e_2^{p_2}…e_k^{p_k}$ @Tom-Tom and possible combinations of the above 3. (e.g. 2^4 * 3*3 * 5^7 * 7^5). Or what you said with "cycles of length 1 or 2" and the exception for 2^4. Oct 3 comment Solving $p_1^{e_1} p_2^{e_2}…p_k^{e_k}=e_1^{p_1} e_2^{p_2}…e_k^{p_k}$ OEIS sequence A008478 Oct 3 comment Solving $p_1^{e_1} p_2^{e_2}…p_k^{e_k}=e_1^{p_1} e_2^{p_2}…e_k^{p_k}$ There's also 2^3 * 3^2. And (2^17 * 17^2) * (3^31 * 31^3), ... Oct 3 comment Quick and painless definition of the set of real numbers Continued fractions might be a good idea. Aug 26 comment How many ways to generate unique multiplication result from given set? There are 5 elevens, 5 sevens, 4 fives, 3 threes and 3 twos. And there are in total 5 distinct primes in the set. The (-N-1) calculation is needed because you don't include the "one factor" and the "zero factor" multiplications. Aug 26 comment How many ways to generate unique multiplication result from given set? If pi are the distinct prime numbers in the (multi)set and ni are the number of times the respective pi appears, then the result is Product(ni+1) - N - 1 (oh and N is the number of disticnt primes.) So for you last example would be (5+1)*(5+1)*(4+1)*(3+1)*(3+1) - 5 - 1 = 2874 Aug 26 comment How to show the convergence of this infinite series: $\frac{x}{1+x}- \frac{x^2}{1+x^2}+ \frac{x^3}{1+x^3}\dots$ @mike the question has: `Given: 0