Timeline for A weaker version of the ABC conjecture
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Nov 13, 2012 at 8:16 | comment | added | joro | Re: "tends to 1". I suppose in general it tends to 1. Repeatedly multiplying by 4 is exponential in 2. Unfortunately squaring $2n+1$ gives doubly exponential growth. Doubt the product of the smaller $n$ can compensate the doubly exponential growth. Well might be wrong. | |
| Nov 13, 2012 at 6:16 | comment | added | joro | @Gerhard it is interesting how relatively big the 3-full part of n(n+1) can be. If it is sufficiently big it will disprove both this and the abc conjecture. | |
| Nov 12, 2012 at 22:42 | comment | added | Gerhard Paseman | In fact this sequence deserves more attention. Can one prove that from the sequence (n,n+1) one has infinitely many prime members of (2n+1)? We may have the opportunity of putting two conjectures head-to-head. Gerhard "Ask Me About System Design" Paseman, 2012.11.12 | |
| Nov 12, 2012 at 22:21 | comment | added | Gerhard Paseman | The fact that you keep multiplying by 4 seems promising. Are you sure that sequence tends to 1? Gerhard "Ask Me About System Design" Paseman, 2012.11.12 | |
| Nov 12, 2012 at 15:13 | history | answered | joro | CC BY-SA 3.0 |