Re: How do quickest

• To: mathgroup at smc.vnet.net
• Subject: [mg72935] Re: How do quickest
• From: dh <dh at metrohm.ch>
• Date: Thu, 25 Jan 2007 07:55:38 -0500 (EST)
• References: <200701231003.FAA06815@smc.vnet.net> <ep7cho\$bdl\$1@smc.vnet.net>

```
Hi Arthur,

Instead of searching for a solution inside the While loop, it is much

faster ()but a bit more cumbersome to use Reduce. E.g.:

a={3};

For[k=1+a[[Length[a]]],Length[a]<10,

s2=Sum[(a[[t]])^2,{t,1,Length[a]}];

t=Reduce[{y^2+s2\[Equal]x^2,x>0,y>=k},{y,x},Integers][[1]];

t=t/.{Equal->Rule,And->List};

k=y/.t;

AppendTo[a,k]];

a

Artur wrote:

> How do quickest following very slowly procedure:

> a = {3}; For[k = 1 + a[[Length[

>      a]]], Length[a] < 13, While[! IntegerQ[Sqrt[(k)^2 + Sum[(a[[t]])^2,

> {t, 1,

>     Length[a]}]]], k++]; AppendTo[a, k]]; a

>

> ARTUR

>

```

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