Timeline for Representation of a number as a sum of co-prime numbers
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Feb 11, 2017 at 5:59 | comment | added | მამუკა ჯიბლაძე | Multiplicativity almost never holds already for $k=3$. E. g. $\beth_3(3)=1$, $\beth_3(4)=3$, $\beth_3(5)=3$, $\beth_3(12)=39$, $\beth_3(20)=93$. | |
| Feb 10, 2017 at 10:08 | comment | added | LRDPRDX | @js21, I agree. What an obvious fact that I did not note. Thank you. | |
| Feb 10, 2017 at 10:04 | history | edited | LRDPRDX | CC BY-SA 3.0 |
added 130 characters in body
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| Feb 10, 2017 at 10:01 | comment | added | js21 | $\beth_2$ is just Euler's totient function. The case $k=2$ is a very special case. | |
| Feb 10, 2017 at 9:56 | comment | added | LRDPRDX | @js21, Of course $\beth_k(a)$ is define for only $a\geq k$. I will add your remark to the description. P.S, I have tested $\beth_2(n)$ for multiplicity for $n$ up to 100. | |
| Feb 10, 2017 at 9:38 | comment | added | js21 | One has $\beth_{pq}(pq) = 1 \neq 0 = \beth_{pq}(p) \beth_{pq}(q)$ for distinct primes $p,q$. Thus $\beth_{k}$ is not multiplicative in general. | |
| Feb 10, 2017 at 8:51 | comment | added | მამუკა ჯიბლაძე | (contd.)$$\beth_k(k+10)=k^2$$$$\beth_k(k+11)=2k^3-k^2$$$$\beth_k(k+12)=k^3-k^2+k$$$$\beth_k(k+13)=k^4-2k^3+2k^2$$$$\beth_k(k+14)=3k^2-2k$$ | |
| Feb 10, 2017 at 8:33 | comment | added | მამუკა ჯიბლაძე | Some empirical observations which might or might not be helpful or deceptive$$\beth_k(k)=1$$ $$\beth_k(k+1)=\beth_k(k+2)=\beth_k(k+4)=k$$ $$\beth_k(k+3)=\beth_k(k+6)=\beth_k(k+8)=k^2$$ $$\beth_k(k+5)=2k^2-k$$ $$\beth_k(k+7)=k^3-k^2+k$$ $$\beth_k(k+9)=2k^3-2k^2+k$$ | |
| Feb 10, 2017 at 7:30 | history | asked | LRDPRDX | CC BY-SA 3.0 |