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MathGroup Archive 2007

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Re: Re: How do quickest


Comparision 3 procedures Length[a] < 11:

Daniel Huber        4.487 Second
Daniel Lichtblau   13.219 Second Second
Myself             58.744 Second

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Dnia 25-01-2007 o 13:55:38 dh <dh at metrohm.ch> napisa³(a):

> 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



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