[seqfan] Re: possible relation in A005132 Recaman's sequence

[Georgi Guninski]

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.