Re: Help with While Loop Function
- To: mathgroup at smc.vnet.net
- Subject: [mg115960] Re: Help with While Loop Function
- From: Peter Pein <petsie at dordos.net>
- Date: Thu, 27 Jan 2011 03:42:01 -0500 (EST)
- References: <ihorj6$hfh$1@smc.vnet.net>
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