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MathGroup Archive 2004

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Re: A special kind of partitions of an integer

  • To: mathgroup at smc.vnet.net
  • Subject: [mg48181] Re: A special kind of partitions of an integer
  • From: drbob at bigfoot.com (Bobby R. Treat)
  • Date: Mon, 17 May 2004 03:21:49 -0400 (EDT)
  • References: <200405131305.JAA24791@smc.vnet.net> <c81hq6$4uc$1@smc.vnet.net> <c83ssk$m94$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

OOPS! I misunderstood your problem; Rob Pratt has it right.

Bobby

drbob at bigfoot.com (Bobby R. Treat) wrote in message news:<c83ssk$m94$1 at smc.vnet.net>...
> The same problem was solved in another thread, a few weeks ago. Also
> look up CatalanNumber in Help; it counts the number of these
> partitions.
> 
> http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&frame=right&th=ff1de8017f9f5222&seekm=c38uvb%24g39%241%40smc.vnet.net#link2
> 
> wolf[{arg_}, op_] := {arg}
> wolf[{args__}, op_] := Flatten[ReplaceList[{args}, {a__, b__} :>
> Outer[
>     op, wolf[{a}, op], wolf[{b}, op], 1]], 2]
> 
> wolf[Array[1 &, 5], CenterDot]
> 
> Displaying with Plus is a bit tricky:
> 
> stringIt[a_, b_] := "(" <> ToString[a] <> "+" <> ToString[b] <> ")"
> List @@ (2 @@ wolf[Array[1 &, 5], List] /. List -> stringIt)
> 
> DrBob
> 
> Cezar Augusto de Freitas Anselmo <cafa at ime.unicamp.br> wrote in message news:<c81hq6$4uc$1 at smc.vnet.net>...
> > Dear friends,
> >  
> >  I'm looking in literature theory of certain partitions of a integer. 
> >   
> >  Have you worked with partitions of a integer n in n terms where the 
> >  summation is not associative (or know someone or texts about it, or 
> >  other discussion list)? 
> >   
> >  Example: 
> >  I have to count P(n): the number of partitions of n with n positive 
> >  integers (the only so integer is one) terms where the + operator is not 
> >  associative, but is commutative; but the recurrence isn't simple. See 
> >  below 
> >   
> >  2 = (1+1); 
> >  (thus P(2)=1) 
> >  3 = ((1+1)+1); 
> >  (thus P(3)=1) 
> >  4 = (((1+1)+1)+1), ((1+1)+(1+1)); 
> >  (thus P(4)=2) 
> >  = ((((1+1)+1)+1)+1), (((1+1)+(1+1))+1), (((1+1)+1)+(1+1)); 
> >  (thus P(5)=3) 
> >   
> >   
> >  P(6)=6  
> >  P(7)=11 
> >  
> >  Thanks for all help,
> >  
> >  -- 
> >  ========================================
> >  Cézar Freitas (ICQ 109128967)
> >  IMECC - UNICAMP / IME - USP
> >  Campinas / São Paulo, SP - Brasil


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