Timeline for Conjecture about a sequence of natural numbers, such that, $\forall n : A_n<P_n<A_{n+1}$
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 9, 2015 at 21:37 | vote | accept | barak manos | ||
| Jun 3, 2014 at 5:52 | answer | added | Aaron Meyerowitz | timeline score: 3 | |
| Jun 1, 2014 at 22:38 | answer | added | Aaron Meyerowitz | timeline score: 4 | |
| Jun 1, 2014 at 22:28 | answer | added | Greg Martin | timeline score: 6 | |
| Jun 1, 2014 at 17:54 | comment | added | barak manos | @AnthonyQuas: Not so true. Take $59$ for example. $59-1=58$, so $58 \in A$. Now, $58$ is the product $2$ prime factors, so all the elements in $A$ prior to $58$ must also be the product of $2$ prime factors at most. As in the example I gave, this is impossible if you start the sequence $P$ at a low value. | |
| Jun 1, 2014 at 17:43 | comment | added | Tony Huynh | @AnthonyQuas, I think the OP wants the number of prime factors of $A_n$ (counting multiplicities) to be monotone, so that won't work. | |
| Jun 1, 2014 at 17:36 | comment | added | Anthony Quas | One sequence of natural number with the right property is $A_n=P_n-1$. | |
| Jun 1, 2014 at 17:05 | comment | added | Tony Huynh | This should follow from the (first) Hardy-Littlewood Conjecture. mathworld.wolfram.com/k-TupleConjecture.html I think the distribution of prime 4-tuples should be enough. | |
| Jun 1, 2014 at 15:23 | history | edited | barak manos | CC BY-SA 3.0 |
added 314 characters in body
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| Jun 1, 2014 at 15:14 | history | asked | barak manos | CC BY-SA 3.0 |