Re: Mathematica:recursion with 2 arguments?
- To: mathgroup at smc.vnet.net
- Subject: [mg64441] Re: Mathematica:recursion with 2 arguments?
- From: "Scout" <Scout at nodomain.com>
- Date: Fri, 17 Feb 2006 04:11:50 -0500 (EST)
- References: <email@example.com>
- Sender: owner-wri-mathgroup at wolfram.com
"ddjjjkkkk" <noble_c at hotmail.com> > 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!!!!!! > That is an example of a doubly-recursive function that is not reducible to primitive recursion, it's not so reducible. The function grows extremely fast! In fact f(4,1)=65533 while f(4,2) has 19729 digits! More advanced mathematics is needed. This type of function occurs as "Wilhelm Ackermann" 's function. ~Scout~