Timeline for Can $b^4+1$ be a pseudoprime to base 2 (except for Fermat numbers)?
Current License: CC BY-SA 3.0
3 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 29, 2016 at 21:30 | vote | accept | Jeppe Stig Nielsen | ||
| Mar 26, 2016 at 23:40 | comment | added | Jeppe Stig Nielsen | This was the kind of answer I was hoping for. For comparison, with $b^2+1$, looking at the subsets of $\{1,\ldots,2^{64}\}$, this time they have sizes $118968378$ and $2^{32}$, and the intersection turns out (long search completed) to have cardinality $31$ (excluding one Fermat number, $65536^2+1$). And with $b^3+1$ (Jeremy Rouse's comment to the question) subsets of sizes $118968378$ and $2^\frac{64}{3}$ (quick search!) give an intersection of size $11$. Some statistician may tell us if this is compatible (at some level of confidence) with the idea of stochastic independence of these sets. | |
| Mar 24, 2016 at 14:38 | history | answered | Stefan Kohl♦ | CC BY-SA 3.0 |