Talk:Random Fibonacci sequence

I removed the following text from the article:

the ratio of the absolute values of successive terms converges to the value of the constant

If this were true, then f(n-1) would be approximately Vf(n-2), where V denotes Viswanath's constant. Hence f(n) is either f(n-1) + f(n-2) = (V+1) f(n-1) or f(n-1) - f(n-2) = (V-1) f(n-1), so f(n) / f(n-1) is either (V+1)/V or (V-1)/V. These numbers differ, so the ratio f(n) / f(n-1) does not converge.

I replaced the above text with the definition from Viswanath's paper. -- Jitse Niesen 23:17, 28 Apr 2004 (UTC)

Is the definition correct?

Mathworld (see references) defines the random Fibonacci sequence as

$a_{n}=\pm a_{n-1}\pm a_{n-2}$

with +/- sign in front of the two terms. The definition in the main article has only one +/-. TomyDuby (talk) 18:27, 28 August 2008 (UTC)[reply]

Yep, you're right. I checked against Viswanath's article, and he does the same. A bit unfortunate, in my opinion, because it just complicates the definition without making much difference. But we better follow the source, so I changed the article. -- Jitse Niesen (talk) 19:13, 28 August 2008 (UTC)[reply]

My problem with the latest "recursive" definition change, is that it implies that one should drop or change some of the true signs, since the formula of 4, is actually a reduced form of all 8 possible operations. For example when a negative number is subtracted, it can be reduced to merely addition ofcourse. However "recursively" that sequence will be different, if that negative number remains a reduced postive number upon recursion. Anyways, I think it was inappropriate and/or inacurate to change this wikipedia definition from 1/2 probability to 1/4, instead of a more precise 1/8. Primedivine (talk) 23:12, 19 February 2010 (UTC)[reply]