Timeline for Test for pair of odd primes $(p, 2p^2-1)$
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| May 4 at 16:12 | vote | accept | Notamathematician | ||
| May 4 at 0:17 | comment | added | Will Sawin | @mathworker21 The phrase "not clear" often carries an implied "to me". (Of course one could worry that this is not a helpful thing to say, but I was confident enough that a high enough percentage of mathematicians would find this strange and hard to analyze to be worth the effort of mentioning to a question asker who is stated to be not a mathematician.) | |
| May 4 at 0:11 | comment | added | mathworker21 | @WillSawin "This kind of successively-adding-gcds dynamical system is very bizarre and it's not clear what its behavior should be." how can you authoritatively say such things?!? | |
| May 3 at 23:00 | answer | added | Max Alekseyev | timeline score: 6 | |
| May 3 at 20:40 | comment | added | Max Alekseyev | @BenBurns: As Will pointed out, the last value for $A+B$ is $(2k+1)+1+2(2k+1)2k=8k^2+6k+2$. Hence last $B=6k$ iff last $A=8k^2+2$. | |
| May 3 at 20:30 | comment | added | Ben Burns |
Well, after an embarrassing mistake, let me offer a heuristic I noticed that may be helpful. It seems when the above iff condition holds, the last $A$ value in $b(n)$ is equal to $8k^2+2$, and conversely when $A \not= 8k^2+2$ then $k \not\in A106483$.
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| May 3 at 15:31 | history | edited | Notamathematician | CC BY-SA 4.0 |
added 106 characters in body
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| May 3 at 14:31 | comment | added | Will Sawin | Since $\gcd(A,B)= \gcd(A+B,B )$ and each step $A+B$ increase by $2n$ so before the $i$'th step $A+B= (2i-1) n+1$, we can express b(n) more simply as starting with $B=1$ and then for $i$ from $1$ to $n-1$ apply $B:= B + \gcd( (2i-1)n+1, B)$. This kind of successively-adding-gcds dynamical system is very bizarre and it's not clear what its behavior should be. | |
| May 3 at 13:04 | history | asked | Notamathematician | CC BY-SA 4.0 |