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MathGroup Archive 2006

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Re: Mathematica:recursion with 2 arguments?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg64422] Re: Mathematica:recursion with 2 arguments?
  • From: Renan <renan.birck at gmail.com>
  • Date: Thu, 16 Feb 2006 03:05:38 -0500 (EST)
  • References: <200602130815.DAA12589@smc.vnet.net> <dsuqik$70n$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Pratik Desai escreveu no grupo comp.soft-sys.math.mathematica:
> ddjjjkkkk wrote:
> 
> >Hi,everybody:
> >My English is a little poor.But I have a very flustered problem to consult you.Who can help me? Here is the prob:
> >IF function f(m,n)(m,n are non-negative integers)satisfying 3 conditions:
> > f(0,n)=n+1,f(m+1,0)=f(m,1) and f(m+1,n+1)=f[m,f(m+1,n)],then how can 
> > we use Wolfram's Mathematica to solve f(m,1),f(m,2),f(m,3) and so on.
> > I can get f(1,n)=n+2,f(2,n)=n+3,f(3,n)=2^(n+3)-3 by hand and pen(but how by Mathematica?).
> > Surely,f(4,n) is very complicated.
> >Who can use Mathematica's language (recursion or iteration etc.) to solve them out? Friends,help me--a poor person,please!!!!!!
> >  
> >
> Have you tried RSolve???

RSolve doesn't seem to like it:

In[2] := RSolve[{f[0, n] == n + 1, f[m + 1, 0] == f[m, 1], 
    f[m + 1, n + 1] == f[m, f[m + 1, n]]}, f, {m, n}]

RSolve::nestdv : The expression f[1+m,n] has nested dependent variables.

$Version is "5.2 for Microsoft Windows (June 20, 2005)"


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