[seqfan] Re: Any other thoughts, opinions on (in)finiteness of A020995?

[Sven Simon]

Hello,

these numbers were checked up to large indices, and no further numbers with
the property were found. They were checked by David Terr for his article
(Link in A020995). He made a statistical analysis and predicted 684+-26
numbers for the same property in base 11. In fact I found 710 of these
numbers in base 11 (A025490). Lucas numbers have the same size, but there
were only 284 having the property in base 11 (A025491). So you see, that
Terr's analysis was quite accurate. This difference is explained a little -
but not in detail - in Terr's article. With Lucas numbers the sequence ends
948539,973261,983101,2528952, so there happened a statistical runaway (right
English?). But these numbers were checked up to indices 7500000, so there is
nearly no chance for more numbers even in base 11. And the same is true for
Fibonacci in base 10, where T.D. Noe checked up to large indices (A004090).
There is no prove for it, but you won't find a new one in base 10 - I would
bet. No risk no fun. 
Sven

-----Ursprüngliche Nachricht-----
Von: seqfan-bounces at list.seqfan.eu [mailto:seqfan-bounces at list.seqfan.eu] Im
Auftrag von Alonso Del Arte
Gesendet: Montag, 19. März 2012 18:26
An: Sequence Fanatics Discussion list
Betreff: [seqfan] Any other thoughts, opinions on (in)finiteness of A020995?

Long ago, Robert G. Wilson v suggested that the sequence of n such that the
sum of the base 10 digits of Fibonacci(n) is equal to n (A020995) might be
infinite, though only twenty terms are known and the largest known term is
quite small. In 2006, Stefan quoted an argument from Robert Dawson that not
only is the sequence likely finite, we might already know all the terms.

Most recently, Charles posted the following argument in a comment for the
Sequence of the Day for April 20, saying that it needs to be checked:

"The number of digits in the n-th Fibonacci number is
<math>n\log_{10}\varphi+o(1)</math>, so the expected digit sum is about
0.94n.  Modeling the result as a normal distribution, the variance is about
<math>82.5\log_{10}\varphi</math> and so the heuristic probability that the
digits of <math>F_n</math> are large enough to be in the sequence is about
<math>\operatorname{erfc}\left(n\frac{1-4.5\log_{10}\varphi}{\sqrt{165}}\rig
ht)/2.</math>
This decays rapidly: by n = 10,000 it is below <math>10^{-935}.</math>"

Have you any other thoughts, opinions, those of you who have pondered this
or similar questions?

Al

--
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Author at
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