Timeline for Is every Fibonacci number "Fibonacci-prime"?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 29, 2012 at 21:36 | vote | accept | David Callan | ||
| Nov 29, 2012 at 21:35 | comment | added | David Callan | No, François, you didn't miss anything. It was me that was missing something. | |
| Nov 29, 2012 at 19:09 | comment | added | Joe Silverman | If you want further info on this sort of behavior, search on "strong divisibility sequence". These are sequences $(A_n)$ with the property that $\gcd(A_m,A_n)=A_{\gcd(m,n)}$. As Greg Martin pointed out, the Fibonacci sequence is a strong divisibility sequence, as are many other divisibility sequences arising from algebraic groups. | |
| Nov 29, 2012 at 9:47 | answer | added | Greg Martin | timeline score: 15 | |
| Nov 29, 2012 at 8:24 | comment | added | François Brunault | I must be missing something. Why doesn't Carmichael's theorem already answer your question 2? What do you mean by the entry point of $p$? Furthermore, note that if $p$ is a prime divisor of $F_n$ such that $p$ does not divide $F_k$ for $k<n$, then $p | F_m$ implies $p | F_{\mathrm{gcd}(m,n)}$, thus $\mathrm{gcd}(m,n)=n$, which means that $n$ divides $m$. | |
| Nov 29, 2012 at 6:40 | history | asked | David Callan | CC BY-SA 3.0 |