Let be the sequence $b_n=50+n^2$, for $n=1,2,3...$ . Which is the sequence $a_n=\gcd(b_n,b_{n+1})$?
The first values are $3,1,1,3,1,1,3...$, so I think that is the series. I've tried writing $n$ as $3k+1, 3k+2, 3k$ but I'm not able to prove it.
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1$\gcd(50+33^2,50+34^2)=67$ – J. W. Tanner Dec 02 '22 at 12:33
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1I found this counterexample almost at the same time. $201$ can also occur , for example for $n=100$ – Peter Dec 02 '22 at 12:33
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Interesting is whether those $4$ values are all that can occur. I remember a similar gcd-expression which is $1$ upto a huge counterexample. – Peter Dec 02 '22 at 12:37
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2Note that $50+n^2=(2n+1)(\frac12n-\frac14)+\frac{\color{red}{201}}4$, and the factors of $201$ are $1, 3, 67$, and $201$ – J. W. Tanner Dec 02 '22 at 12:46
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2Did you mean $b_n$ where you wrote $b_1$? – J. W. Tanner Dec 02 '22 at 12:48
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@J.W.Tanner Wow, that's it since the gcd must divide $2n+1$. Nice ! – Peter Dec 02 '22 at 12:51
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1$a_n = \gcd(n-100,201)$ by the formula here in the linked dupe. The rest is easy. – Bill Dubuque Dec 02 '22 at 13:09
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the difference between an index where b is 67 or 201 and the previous index, where b is 67 or 201 seems to be always 67 (at least until n<10^6) – miracle173 Dec 02 '22 at 13:11