Re: Pretty output

*To*: mathgroup at smc.vnet.net*Subject*: [mg35156] Re: Pretty output*From*: "Allan Hayes" <hay at haystack.demon.co.uk>*Date*: Thu, 27 Jun 2002 00:23:20 -0400 (EDT)*References*: <afbih4$54r$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Robert, Using ToString seems to be causing the printing to be in OutputForm which is clearly not very good. The following will print in readable StandardForm Do[s = Sum[x^i, {i, 0, 2^n}]; Print[Factor[s, Modulus -> 2]], {n, 1, 10}]. However, since you want to find the number of factors it is best not to use Print and Do but to make a list of the factorized expressions. fcts=Table[Factor[Sum[x^i,{i,0,2^n}], Modulus->2],{n,1,10}]; The output is not shown because of the final semicolon. If you want to look at the list it might be better to use TableForm[fcts] We can extract individual expressions: for example Part[fcts, 9] But if we only need the number of factors in each expression then the following works Map[Length, a fcts]-1 {1,1,2,2,4,6,10,16,30,52} Here I used the trick of multiplying by a dummy expression a and later subtracting one because the first two expressions are 1 + x + x^2, 1 + x + x^2 + x^3 + x^4 and, for example, Length[1 + x + x^2] would give 3, the number of summands, whereas Length[a*(1 + x + x^2)] gives the number of multiplicands, 2; and on subtracting 1 we get 1, the answer that we want. The trick is not necessary with the third entry, (1 + x + x^2)*(1 + x^3 + x^6), but it still gives the right answer. -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay at haystack.demon.co.uk Voice: +44 (0)116 271 4198 Fax: +44 (0)870 164 0565 "Robert G. Wilson v" <rgwv at kspaint.com> wrote in message news:afbih4$54r$1 at smc.vnet.net... > Help please. > > I put in the following Mathematica code: Do[s = Sum[x^i, {i, 0, 2^n}]; > Print[ ToString[ Factor[ s, Modulus -> 2]]], {n, 1, 10}] > > The output was fairly straight forward until n=6 and beyond. What I am > seeing is: > > " 2 3 4 4 5 6 7 8 12 2 > 5 6 7 10 12 2 3 4 6 8 9 \ > 10 12 3 5 6 7 9 11 12 2 > 3 4 5 6 7 8 9 10 11 12\n(1 + x \ > + x + x + x ) (1 + x + x + x + x + x + x ) (1 + x + x + x + > x + x + x ) (1 + x + x + x + x + x + x + x + x \ > ) (1 + x + x + x + x + x + x + x + x ) (1 + x + x + x + x + > x + x + x + x + x + x + x + x )" > > What is the deal here? > > Also what is the easiest way to count the number of factors? > > Thank you for any assistance in advance. > > Sincerely yours, > > Robert G. Wilson, V >