[seqfan] Re: linear recurrence multiples

[Dale Gerdemann]

That’s  an interesting example which shows me that I haven’t made my
assumptions clear.  The greedy algorithm I used would start off with
 4*A015518(n) = A015518(n+1) + 2*A015518(n-1) + ... .  And, perhaps I’m
also assuming that a formula for multiples of 4 shouldn’t include a
multiple of 6 as one of the terms. I never encountered such a thing with
“nice” recurrences where c <= b. Still, even with my hidden assumptions,
this is a near-miss counterexample.  I suspect that such near misses would
be harder to find when the difference between c and b is larger than 1.


El El dom, 21 jul 2019 a las 19:22, Ehren Metcalfe <ehren.m at gmail.com>
escribió:

> You've written: 4*A007482(n) = A007482(n+1) + A007482(n-1) +
> 2*A007482(n-2).
>
> Regarding an analogue for A015518, it looks like the best match is
> 4*A015518(n) = A015518(n+1) + A015518(n-1) + 6*A015518(n-2).
>
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