Timeline for Does this sequence always give an integer?

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Jul 28, 2017 at 1:04	history	edited	Alexey Ustinov	CC BY-SA 3.0	added 165 characters in body
Apr 9, 2015 at 3:25	comment	added	Alexey Ustinov		@David Speyer I've corrected my answer using your approach.
Apr 9, 2015 at 3:24	history	edited	Alexey Ustinov	CC BY-SA 3.0	solution has been corrected
Apr 8, 2015 at 11:35	history	edited	Alexey Ustinov	CC BY-SA 3.0	added 117 characters in body
Apr 8, 2015 at 10:47	comment	added	David E Speyer		Isn't this circular? If we know that all the $\tau$'s in the matrix except for $\tau_{m+n}$ are integers, then we can deduce that $18 \tau_{m+n} \tau_{m-n} + (\tau_{m+2} \tau_{m-2} + \tau_{m+4} \tau_{m-4})(\tau_{4+n} \tau_{4-n} + \tau_{5+n} \tau_{5-n})$ is an even integer. But how do you rule out that $\tau_{m+n}$ is a half integer and $(\tau_{m+2} \tau_{m-2} + \tau_{m+4} \tau_{m-4})(\tau_{4+n} \tau_{4-n} + \tau_{5+n} \tau_{5-n})$ is odd?
Apr 8, 2015 at 10:38	comment	added	Alexey Ustinov		Yes, because this product $\equiv\Delta\equiv0\pmod 2$. From you claculations probably follows that each multiple here is even.
Apr 8, 2015 at 10:31	comment	added	David E Speyer		But I am missing something. Are you claiming to have a proof, without relying on the computations I already did, that $(\tau_{m+2} \tau_{m-2} + \tau_{m+4} \tau_{m-4})(\tau_{4+n} \tau_{4-n} + \tau_{5+n} \tau_{5-n})$ is always even?
Apr 8, 2015 at 10:28	comment	added	David E Speyer		Ah, nice. And this explains why I could always find a relation of the form $a \tau_n \tau_{n+8k} + b \tau_{n+k} \tau_{n+7k} + c \tau_{n+2k} \tau_{n+5k} + d \tau_{n+3k} \tau_{n+5k} + e \tau_{n+4k}^2$. I knew what I was doing was computing the kernel of a $5 \times \infty$ matrix, but I didn't know why it always had rank $4$.
Apr 8, 2015 at 9:39	history	answered	Alexey Ustinov	CC BY-SA 3.0