Re: doing things on a procedural way and doing them on a functional way

*To*: mathgroup at smc.vnet.net*Subject*: [mg47040] Re: doing things on a procedural way and doing them on a functional way*From*: drbob at bigfoot.com (Bobby R. Treat)*Date*: Mon, 22 Mar 2004 05:18:52 -0500 (EST)*References*: <c3h165$b9o$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Oops, I had that wrong a few minutes ago. c[n_] = (2*n)!/(n!*n!*(n + 1)); FullSimplify[c[n + 1]/c[n]] 4 - 6/(2 + n) Bobby "Wolf, Hartmut" <Hartmut.Wolf at t-systems.com> wrote in message news:<c3h165$b9o$1 at smc.vnet.net>... > A remark to add... > > >-----Original Message----- > >From: Wolf, Hartmut To: mathgroup at smc.vnet.net > >Sent: Friday, March 19, 2004 7:36 AM > >To: mathgroup at smc.vnet.net > >Subject: [mg47040] Re: doing things on a procedural way > >and doing them on a functional way > > > > > > > >>-----Original Message----- > >>From: Paul Abbott [mailto:paul at physics.uwa.edu.au] To: mathgroup at smc.vnet.net > To: mathgroup at smc.vnet.net > >>Sent: Thursday, March 18, 2004 10:38 AM > >>To: mathgroup at smc.vnet.net > >>Subject: [mg47040] Re: doing things on a procedural > >>way and doing them on a functional way > >> > [...] > >> > >>I can't help but feel that there should be an elegant way to do this > >>using Distribute ... > >> > >>Cheers, > >>Paul > >> > [...] > > > >You're completely right, > >the solution, I published yesterday, also works with > >distribute (instead of Outer): > > > >In[32]:= nocnac[{arg_}, op_] := {arg} > >In[33]:= > >nocnac[{args__}, op_] := > > Flatten[ReplaceList[{args}, {a__, b__} :> > > Distribute[op[nocnac[{a}, op], nocnac[{b}, op]], List]], 1] > > > > > [...] > > > Effectively both solutions are the same, as for Distribute you always can > find an equivalent expression with Outer, e.g.: > > In[164]:= Distribute[List[{1, 2}, {{3, 4}, 5}, 6], List] > Out[164]= > {{1, {3, 4}, 6}, {1, 5, 6}, {2, {3, 4}, 6}, {2, 5, 6}} > > ...this is by definition the same as > > In[170]:= Distribute[List[{1, 2}, {{3, 4}, 5}, {6}], List] > Out[170]= > {{1, {3, 4}, 6}, {1, 5, 6}, {2, {3, 4}, 6}, {2, 5, 6}} > > ...functionally equivalent to > > In[166]:= Flatten[Outer[List, {1, 2}, {{3, 4}, 5}, {6}, 1], 2] > Out[166]= > {{1, {3, 4}, 6}, {1, 5, 6}, {2, {3, 4}, 6}, {2, 5, 6}} > > > You may have observed that my first solution using Outer doesn't work with > List, whereas that one above with Distribute does. This is not a problem of > Outer, I just have been a bit careless with levels (at Flatten and Outer). > This code also works with List: > > In[143]:= > nocnac[{arg_}, op_] := {arg} > > nocnac[{args__}, op_] := > Flatten[ReplaceList[{args}, {a__, b__} :> > Outer[op, nocnac[{a}, op], nocnac[{b}, op], 1] > ], 2] > > > > > > > > > >In[39]:= Length[nocnac[Range[#], op]] & /@ Range[10] > >Out[39]= > >{1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862} > > > > > [...] > > > > >I don't know, but assume that this can be expressed simply > >through some sort of combinatorical numbers. > > > Yes, indeed, it is given by > > In[171]:= Binomial[2 #, #]/(# + 1) & /@ Range[0, 9] > Out[171]= > {1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862} > > ...and the numbers > > c[n_] := Binomial[2n, n]/(n + 1) > > > are called the Catalan numbers. See e.g. > http://www.research.att.com/~njas/sequences/ > and my observation... > > > > >In[57]:= nc[#]/nc[# - 1] & /@ Range[2, 30] // N > >Out[57]= > >{1., 2., 2.5, 2.8, 3., 3.14286, 3.25, 3.33333, 3.4, > > 3.45455, 3.5, 3.53846, 3.57143, 3.6, 3.625, 3.64706, > > 3.66667, 3.68421, 3.7, 3.71429, 3.72727, 3.73913, > > 3.75, 3.76, 3.76923, 3.77778, 3.78571, 3.7931, 3.8} > > > [...] > > > >The conjecture of course is, that asymtotically nc[n] ~ 4^n. > > > > is nothing else than the simple fact > > In[182]:= Limit[2n (2n - 1)/(n (n + 1)), n -> Infinity] > Out[182]= 4