[seqfan] Re: possible relation in A005132 Recaman's sequence
Georgi Guninski
guninski at guninski.com
Wed Jun 3 16:30:51 CEST 2009
On Wed, Jun 03, 2009 at 10:19:15AM -0400, David Wilson wrote:
> a(n) certainly has to satisfy some such recurrence.
>
> By the definition of Recaman, it is easy to show
>
> [1] |a(n+1) - a(n)| = n+1
>
> hence
>
> [2] (a(n+1) - a(n))^2 = (n+1)^2 = f(n)
>
> But f(n) = (n+1)^2 satisfies the recurrence
>
> [2] f(n+2) - 2f(n+1) + f(n) - 2 = 0
>
> I highly suspect that plugging f(n) = (a(n+1) - a(n))^2 into [2] and
> properly translating the index results in your recurrence.
>
> So your recurrence is likely true, but it is in reality an arcane way of
> saying [1]. Any sequence satisfying [1], e.g, A000217 or A008344(n+1),
> would also satisfy your recurrence.
>
>
yes, it seems true as per the previous post.
another lame way to prove is to bruteforce the 2^3 possible choices of sign
for the next 3 terms and plug in the formula.
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