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)"

**References**:**Mathematica:recursion with 2 arguments?***From:*ddjjjkkkk <noble_c@hotmail.com>