Fibred category: Difference between revisions

From formulasearchengine
Jump to navigation Jump to search
→‎Cartesian morphisms and functors: being cartesian is with respect to a functor, not its codomain
No edit summary
 
(One intermediate revision by one other user not shown)
Line 1: Line 1:
In [[number theory]], '''de Polignac's formula''', named after [[Alphonse de Polignac]], gives the [[prime decomposition]] of the [[factorial]] ''n''<nowiki>!</nowiki>, where ''n''&nbsp;≥&nbsp;1 is an [[integer]].  [[Leonard Eugene Dickson|L. E. Dickson]] attributes the formula to [[Adrien-Marie Legendre|Legendre]].<ref>[[Leonard Eugene Dickson]], ''[[History of the Theory of Numbers]]'', Volume 1, Carnegie Institution of Washington, 1919, page 263.</ref>
There is nothing to tell about myself I think.<br>Finally a part of wmflabs.org.<br>I really wish I'm useful at all<br><br>Also visit my web page; [http://muslimloveforum.com/member.php?action=profile&uid=13733 FIFA coin generator]
 
==The formula==
Let ''n''&nbsp;≥&nbsp;1 be an integer. The prime decomposition of ''n''! is given by
 
:<math>n! = \prod_{\text{prime }p\le n} p^{s_p(n)}, </math>
 
where
 
:<math>s_p(n) = \sum_{j = 1}^\infty \left\lfloor\frac{n}{p^j}\right\rfloor, </math>
 
and the brackets represent the [[floor function]]. Note that the former product can equally well be taken only over primes less than or equal to ''n'', and the latter sum can equally well be taken for ''j'' ranging from ''1'' to log<sub>''p''</sub>(''n''), i.e :
 
:<math>s_p(n) = \sum_{j = 1}^{\lfloor \log_p(n) \rfloor} \left\lfloor\frac{n}{p^j}\right\rfloor </math>
 
Note that, for any [[real number]] ''x'', and any integer ''n'', we have:
 
:<math>\left\lfloor\frac{x}{n}\right\rfloor = \left\lfloor\frac{\lfloor x \rfloor}{n}\right\rfloor</math>
 
which allows one to more easily compute the terms ''s''<sub>''p''</sub>(''n'').
 
The small disadvantage of the De Polignac's formula is that '''we need to know all the primes up to ''n'''''.
In fact,
:<math>\displaystyle n! = \prod_{i=1}^{\pi(n)} p_{i}^{s_{p_{i}}(n)} =  \prod_{i=1}^{\pi(n)} p_i^{ \sum_{j = 1}^{\lfloor \log_{p_i}(n) \rfloor} \left\lfloor\frac{n}{{p_i}^j}\right\rfloor  } </math>
 
where <math>\pi(n)</math> is a [[prime-counting function]] counting the number of prime numbers less than or equal to ''n''
 
== Notes and references ==
{{reflist}}
 
{{DEFAULTSORT:De Polignac's Formula}}
[[Category:Number theory]]
[[Category:Factorial and binomial topics]]

Latest revision as of 21:58, 15 November 2014

There is nothing to tell about myself I think.
Finally a part of wmflabs.org.
I really wish I'm useful at all

Also visit my web page; FIFA coin generator