Re: A special kind of partitions of an integer

*To*: mathgroup at smc.vnet.net*Subject*: [mg48162] Re: A special kind of partitions of an integer*From*: drbob at bigfoot.com (Bobby R. Treat)*Date*: Fri, 14 May 2004 20:59:32 -0400 (EDT)*References*: <200405131305.JAA24791@smc.vnet.net> <c81hq6$4uc$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

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