Timeline for Proving that every term of the sequence is an integer
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 25, 2013 at 3:02 | review | First posts | |||
| Jun 26, 2013 at 2:36 | |||||
| May 30, 2013 at 14:44 | vote | accept | mathlove | ||
| May 29, 2013 at 16:21 | answer | added | Barry Cipra | timeline score: 1 | |
| May 29, 2013 at 15:32 | answer | added | Joe Silverman | timeline score: 8 | |
| May 29, 2013 at 14:43 | comment | added | mathlove | @J.J. Thank you for your comment. It's easy to prove that $a_{m,4}$ is an integer for any $m$ because of the following. $$a_{m,4}=(m+1)(m+2)-1+\frac{(m-1)m(m+1)}3+\frac{(m-1)m{(m+1)^2}(m+2)(m+3)}{45}$$ I have tried this approach, but faced difficulty. | |
| May 29, 2013 at 14:38 | answer | added | Timothy Chow | timeline score: 23 | |
| May 29, 2013 at 12:50 | comment | added | j.c. | In the comments of the MSE question, mathlove states: @J.J. One of my friends made this. None of us can prove this and no one can get a counterexample even by using a computer. He, who made this, whose major is math, was interested in 'Somos sequence'. That's why the idea I showed on May 23th came from the proof of this sequence. – mathlove 8 hours ago | |
| May 29, 2013 at 11:46 | answer | added | PRobinson | timeline score: 0 | |
| May 29, 2013 at 11:27 | comment | added | Gerry Myerson | Where does the problem come from? | |
| May 29, 2013 at 9:17 | comment | added | J. J. | Adding to the previous comment, a polynomial interpolation suggests that $a_{m,4} = \frac{x^6 + 6x^5 + 10 x^4 + 15 x^3 + 34 x^2 + 114 x + 45}{45}$. | |
| May 29, 2013 at 9:08 | comment | added | J. J. | We also have $a_{m,2} = m+1$, $a_{m,3} = \frac{m(m+1)(m+2)}{3} + 1$ and $a_{m,4} = n+1 + \sum_{j=1}^n \frac{(j^3 - j + 3)(j^3 + 3j^2 + 2j + 3)(n+1)}{9j(j+1)}$. One might be able to simplify the formula for $a_{m,4}$. | |
| May 29, 2013 at 8:39 | answer | added | MHMertens | timeline score: 2 | |
| May 29, 2013 at 8:13 | comment | added | Dietrich Burde | I write $a(m,n)$ for $a_{m,n}$. We have $a(1,n+1)=a(1,n)+a(1,n-1)$ with $a(1.0)=a(1,1)=1$. This is the usual Fibonacci sequence. Now try induction over $m$. | |
| May 29, 2013 at 7:13 | comment | added | J. J. | I think it's worth commenting that this question has been asked previously on math.SE without receiving any answers: math.stackexchange.com/questions/399337/… | |
| May 29, 2013 at 7:06 | history | asked | mathlove | CC BY-SA 3.0 |