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MiszaBot I (talk | contribs) m Archiving 2 thread(s) from Talk:Fibonacci number. |
MiszaBot I (talk | contribs) m Archiving 2 thread(s) from Talk:Fibonacci number. |
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:It is already rigorous. If ''x'' is composite, so is ''F<sub>x</sub>'', as the article already states. ''n''!+2, ''n''!+3, ''n''!+4, ..., ''n''!+''n'' are all composite, so ''F''<sub>''n''!+2</sub>, ''F''<sub>''n''!+3</sub>, ''F''<sub>''n''!+4</sub>, ..., ''F''<sub>''n''!+''n''</sub> are also all composite. —[[User:David Eppstein|David Eppstein]] ([[User talk:David Eppstein|talk]]) 22:23, 7 February 2009 (UTC)
== Origins ==
The rabbit population example is inconsistently presented: first the rule is that every pair has two pairs of offspring and then dies, but then the recursive relation is justified in terms of rabbit fertility. Both schemes (two offspring pairs then death vs. offspring every month from the second month on) give rise to the same sequence, but the presentation should not mix both. Should I fix this? [[Special:Contributions/139.19.84.14|139.19.84.14]] ([[User talk:139.19.84.14|talk]]) 17:08, 15 February 2009 (UTC)
== recognizing ==
1st way to recognize it listed in the article makes no sence at all, it is just as good as table look-up or, if no table is available, re-computing all the numbers up to z from scratch. [[Special:Contributions/95.132.178.230|95.132.178.230]] ([[User talk:95.132.178.230|talk]]) 14:04, 24 February 2009 (UTC)
:No, it is useful. If you want to find a Fibonacci number close to 1000, for example, you do:
::<math>\bigg\lfloor\log_\varphi(1000\sqrt{5})+\frac{1}{2}\bigg\rfloor = 16</math>
:then you reverse the process:
::<math>\bigg\lfloor \frac {\varphi^{16}}{\sqrt{5}}+\frac{1}{2}\bigg\rfloor = 987</math>
:which not only tells you that 1000 is ''not'' a Fibonacci number, but also that 987 ''is'' a Fibonacci number. [[User:Gandalf61|Gandalf61]] ([[User talk:Gandalf61|talk]]) 15:43, 24 February 2009 (UTC)
:: but it actually says F(16) in the article, not that you should "reverse the process"; it is not obvious that such reversal will give you F(16) and not F(16) +1 or -1. [[Special:Contributions/95.132.178.230|95.132.178.230]] ([[User talk:95.132.178.230|talk]]) <span style="font-size: smaller;" class="autosigned">—Preceding [[Wikipedia:Signatures|undated]] comment was added on 22:46, 24 February 2009 (UTC).</span><!--Template:Undated--> <!--Autosigned by SineBot-->
:::F(''n'') is always the closest integer to <math>\frac {\varphi^{n}}{\sqrt{5}}</math>. Or, if ''n'' is large and you want to stick to integer operations, then F(''n'') is value of the off-diagonal entries of <math>\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^n</math>, which you can calculate with an [[exponentiation by squaring]] method. So there are definitely methods of calculating F(''n'') that do not require the calculation of all the preceeding Fibonacci numbers. [[User:Gandalf61|Gandalf61]] ([[User talk:Gandalf61|talk]]) 10:50, 25 February 2009 (UTC)
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