Timeline for On Fibonacci numbers that are also highly composite
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 16, 2022 at 13:50 | comment | added | Nandakumar R | Note: There are instances of "highly composites between twin primes" such as 60 which falls between twin primes 59 and 61 and 180 which is between 179 and 181. What is not clear is if there is an upper bound. | |
| Jan 14, 2022 at 12:16 | vote | accept | Nandakumar R | ||
| Nov 25, 2021 at 5:58 | comment | added | Nandakumar R | An observation: none of the first 30 odd Fibonacci numbers is sandwiched between twin primes - maybe none are. | |
| Nov 25, 2021 at 4:27 | comment | added | Nandakumar R | Thanks again Prof Rouse. I guess sequences like "highly composites sandwiched between twin primes" would be a lot harder to decide. | |
| Nov 24, 2021 at 23:29 | comment | added | Jeremy Rouse | Define $a_{n}$ by $a_{0} = 1$, $a_{1} = 1$ and $a_{n} = 9a_{n-1} + 29a_{n-2}$ for $n \geq 2$. It should be possible to show that the largest highly composite number in the sequence is $a_{6} = 166320$. (None of the terms in this sequence are multiples of $29$.) Theorems (like the 2001 result of Bilu, Hanrot and Voutier) put restrictions on how far out in a Lucas sequence one can find a term without a primitive prime divisor. These theorems simultaneously make it possible to prove there are finitely many highly composite numbers, while also putting a limit on where they can appear. | |
| Nov 24, 2021 at 8:51 | comment | added | Nandakumar R | I know this is a bit imprecise but it would be nice to know some series in this ballpark that is both provably finite AND with a large highest number - the 'series' of highly composite fibonacci numbers is finite but has only 1 entry and that too 2. | |
| Nov 14, 2021 at 21:12 | comment | added | Jeremy Rouse | The question of Fibonacci numbers with few prime factors is much more difficult. Given that the number of positive integers with $k$ prime factors $\leq x$ is asymptotic to $\frac{x (\log \log x)^{k-1}}{(k-1)! \log x}$, it's natural to conjecture that for a fixed positive integer $k$, there are infinitely many primes $p$ so that $F_{p}$ is a product of $k$ distinct primes. Proving anything about this is probably not possible with current technology however. | |
| Nov 14, 2021 at 10:23 | comment | added | Nandakumar R | Thank you very much Prof Rouse. Hope you could also clarify how many Fibonacci numbers are semiprimes - or have 3 prime factors. | |
| Nov 13, 2021 at 14:44 | history | answered | Jeremy Rouse | CC BY-SA 4.0 |