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Nov
4
comment Does $A193201$ count the partitions of $n$ of arbitrary dimension?
I get 14 for n=6. I might have not understood what exactly we are counting. This?: A119268
Nov
3
comment Examples of mathematical discoveries which were kept as a secret
@DavidRicherby I think you could add it as a separate answer. While belonging to the same area (security/cryptography), it is a different discovery.
Nov
3
comment How many possible color combinations?
I don't think there is a closed formula for T(n,m).
Nov
3
revised How many possible color combinations?
deleted 1 character in body
Nov
3
answered How many possible color combinations?
Nov
3
comment How many possible color combinations?
I think the answer is 17952, let me check a few things.
Nov
2
comment Combinatorial prime problem
The smallest I found to need a power of 10 are: 799 = 2^10 - 3^2 * 5^2 and 851 = 3 * 5^4 - 2^10
Nov
2
comment Combinatorial prime problem
797 = 2^5 * 5*2 - 3
Nov
2
comment Prove $\left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}$ for two complex numbers
Where did that square root come from? Can't you use that |z| = r when z = r(cosθ+isinθ)?
Nov
2
comment Proving two graphs are isomorphic in polynomial time - Bondy/Murty - Graph Theory Page 6
It works without a hitch because the graph has a lot of symmetries (self isomorphisms). I think you could identify any 5-cycle with any other (as a start) and it will work.
Oct
29
comment A geometric assembly: Triangle, circle, square, pentagon.
I think I've seen somewhere the second problem about the inverse and I vaguely remember the value 12 as limit.
Oct
24
comment $27 | (2x+1)^2 \implies 2x$ is a multiple of 9?
@RyanYu Then edit the title of your question.
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.