Re: Help with While Loop Function

*To*: mathgroup at smc.vnet.net*Subject*: [mg115991] Re: Help with While Loop Function*From*: Daniel Lichtblau <danl at wolfram.com>*Date*: Fri, 28 Jan 2011 06:15:41 -0500 (EST)

----- Original Message ----- > From: "Peter Pein" <petsie at dordos.net> > To: mathgroup at smc.vnet.net > Sent: Thursday, January 27, 2011 2:42:01 AM > Subject: [mg115960] Re: Help with While Loop Function > On 26.01.2011 11:04, KenR wrote: > > Please Help me to Get the following to work as desired > > t is the triangular number. I increment it repeated ly by adding the > > counter x to it. When t*((y^2+1)/2)^2 + y^2 is a perfect square i > > want > > to append t to my list of trangular numbers with the list starting > > with {0,1}, and exit the loop when 3 more triangular numbers have > > been > > added or when t>= 1000000000000. It is not working for me. I am new > > to Mathematica. Thanks > > > > Clearall[x,y,t,w] > > x = 2 > > y = 3 > > t = 1 > > w = 0 > > List1 = {0,1} > > f[t_,y_] ==(Floor(Sqrt ((t/4) (y^2-1)^2 + y^2)))^2 - ((t/4) > > (y^2-1)^2 > > + y^2) > f[t_,y_] = Floor[Sqrt[(t/4) (y^2-1)^2 + y^2]]^2 - ((t/4) (y^2-1)^2 + > y^2) > > While[t<1000000000000,t = t+x;If[f[t,y] = 0, List1 = > > Append[list1,t];w > > = w +1]; > ... , AppendTo[List1,t]; w++ ... > > If [w = 3,t = 1000000000000, x++]]; > .. w== 3 .. > > List1 > > > > If I didn't miss anything, you want the Sophie Germain triangular > numbers (http://oeis.org/A124174). This is a problem for which Reduce > has been made: > > In[1]:= t /. {ToRules@ > Reduce[t == n (n + 1)/2 && 2 t + 1 == m (m + 1)/2 && > Element[{m, n, t}, Integers] && 0 <= t <= 10^12 && 0 <= n && > 0 <= m, t, Backsubstitution -> True]} > > Out[1]= {0, 1, 10, 45, 351, 1540, 11935, 52326, 405450, 1777555, > 13773376, 60384555, 467889345, 2051297326, 15894464365, 69683724540, > 539943899076} > > In[2]:= SophieGermain[n_] = Collect[FindSequenceFunction[%, n] // > RootReduce, _^n, Simplify] > > Out[2]= -(11/32) + 1/64 (-3 - 2 Sqrt[2])^n (1 - Sqrt[2]) + > 5/64 (3 - 2 Sqrt[2])^n (2 + Sqrt[2]) + > 1/64 (1 + Sqrt[2]) (-3 + 2 Sqrt[2])^n - > 5/64 (-2 + Sqrt[2]) (3 + 2 Sqrt[2])^n > > hth, > Peter I thought it would be faster to do a direct computation. Silly of me. First attempt failed miserably. Using Compile with Listable attribute succeeded. pred = Compile[{{m, _Real}}, Module[{n = Floor[Sqrt[2.]*m] + 1.}, n^2 + n == 2.*m^2 + 2.*m + 2.], RuntimeAttributes -> Listable, RuntimeOptions -> "Speed"]; In[62]:= Timing[ With[{rng = Range[0, 10^7]}, Map[(#^2 + #)/2 &, Pick[rng, pred[rng]]]]] Out[62]= {1.656, {0, 1, 10, 45, 351, 1540, 11935, 52326, 405450, 1777555, 13773376, 60384555, 467889345, 2051297326, 15894464365, 69683724540, 539943899076, 2367195337045, 18342198104230}} Without that attribute I can get a Select variant to work fast, but not hugely faster than Reduce. I regard this experiment as a victory for Reduce. If only because I had several hours to think about it, whereas Reduce did both the thinking and computing in seconds. Also my method will fizzle around 10^8, when machine precision becomes no longer sufficient for the task. Daniel Lichtblau Wolfram Research