Re: Re: Finding the Number of Pythagorean Triples below a bound

*To*: mathgroup at smc.vnet.net*Subject*: [mg68478] Re: [mg68345] Re: Finding the Number of Pythagorean Triples below a bound*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Mon, 7 Aug 2006 01:40:35 -0400 (EDT)*References*: <20060806080123.15221.qmail@web50704.mail.yahoo.com>*Sender*: owner-wri-mathgroup at wolfram.com

On 6 Aug 2006, at 10:01, Titus Piezas wrote: > Hello Dan and Andrzej, > > >I'd say up to 10^6 or maybe 10^7 the way to go is by looking at the > >factorization of c2=c^2+k. This way you only need iterate over c. > > Yes, turns out the problem was equivalent to counting the number of > representations of N as > > a^2+b^2 = N > > where 0<a<=b (I should have phrased it that way, but that is the > benefit of high-sight) which I assume can be given by the function > OrderedSumOfSquaresRepresentations[2,N]? Hence the code you have > given me, in essence, is, say for k = 5, > > Sum[OrderedSumOfSquaresRepresentations[2,c^2+5],{c,1,10^m-1}] > > roughly speaking? (With finer details of removing from > consideration those c^2+5 that are of from 4m+3, etc.) Well, no, it is rather: Sum[Length[OrderedSumOfSquaresRepresentations[2,c^2+5]],{c,1,10^m-1}] In fact I forgot that there was OrderedSumOfSquaresRepresentations as well as SumOfSquaresRepresentations, which is the main reason why I did not think it would be very useful. Another reason is that, I expected, computing Length of many lists and then adding these Lengths up should be slower than just counting the total number of elements. > > >Here are some timings. > > > >In[54]:=Timing[countTriples[10^4,3]] > >Out[54]={0.591 Second,1154} > > > >In[93]:=Timing[countTriples[10^5,3]] > >Out[93]={8.562 Second,12115} > > > >In[94]:=Timing[countTriples[10^6,3]] > >Out[94]={131.91 Second,121054} > There is something odd with the first value. James' count (up to > m=5) vs this gives the sequences, > > {1219, 12115} > {1154, 12115, 121054} > > Why the disparity? If the ratio S(N)/N would converge, it should > start as 0.121... (Note however, that James' table has bound 10^m > *not* 10^m-1.} My code also gives: countTriplesP[4,3] 1219 Changing 10^m-1 to 10^m should give a value at least as large, and in fact gives the same answer. which suggests that something still needs to be looked carefully at in Daniel's code. I can't do it now but will think about it when I can. Andrzej

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**Re: Re: Finding the Number of Pythagorean Triples below a bound**

**Re: Re: Finding the Number of Pythagorean Triples below a bound**