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	<title>Comments on: Enigma 1436 - One More Step</title>
	<atom:link href="http://www.bottledcity.com/2007/04/04/enigma-1436-one-more-step-2/feed/" rel="self" type="application/rss+xml" />
	<link>http://www.bottledcity.com/2007/04/04/enigma-1436-one-more-step-2/</link>
	<description>Musings and wastes of time, all to put off the inevitable work</description>
	<pubDate>Tue, 06 Jan 2009 20:30:18 +0000</pubDate>
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		<title>By: Doug Davidson</title>
		<link>http://www.bottledcity.com/2007/04/04/enigma-1436-one-more-step-2/#comment-294</link>
		<dc:creator>Doug Davidson</dc:creator>
		<pubDate>Sun, 27 Jan 2008 05:34:44 +0000</pubDate>
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		<description>11011 Twenty seven in Bianary, Three cubed, Simplified trigonometry math of the heavens. Number of books in the Book. It's always much more simple than you think it is.</description>
		<content:encoded><![CDATA[<p>11011 Twenty seven in Bianary, Three cubed, Simplified trigonometry math of the heavens. Number of books in the Book. It&#8217;s always much more simple than you think it is.</p>
]]></content:encoded>
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		<title>By: Alexis</title>
		<link>http://www.bottledcity.com/2007/04/04/enigma-1436-one-more-step-2/#comment-153</link>
		<dc:creator>Alexis</dc:creator>
		<pubDate>Sat, 28 Apr 2007 22:18:21 +0000</pubDate>
		<guid isPermaLink="false">http://www.bottledcity.com/2007/04/04/enigma-1436-one-more-step-2/#comment-153</guid>
		<description>You do know that 1 is a perfect cube too. 


I might be wrong but I am almost certain that 1 is a perfect cube.</description>
		<content:encoded><![CDATA[<p>You do know that 1 is a perfect cube too. </p>
<p>I might be wrong but I am almost certain that 1 is a perfect cube.</p>
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		<title>By: joelbradbury</title>
		<link>http://www.bottledcity.com/2007/04/04/enigma-1436-one-more-step-2/#comment-149</link>
		<dc:creator>joelbradbury</dc:creator>
		<pubDate>Mon, 16 Apr 2007 14:41:13 +0000</pubDate>
		<guid isPermaLink="false">http://www.bottledcity.com/2007/04/04/enigma-1436-one-more-step-2/#comment-149</guid>
		<description>You are totally right! How did that slip by. 
I originally wrote out the whole thing on paper, realized it was wrong and redid it. 

Doh! Must of copied the wrong thing when I was typing it up. 
Never mind.

--I've changed it now</description>
		<content:encoded><![CDATA[<p>You are totally right! How did that slip by.<br />
I originally wrote out the whole thing on paper, realized it was wrong and redid it. </p>
<p>Doh! Must of copied the wrong thing when I was typing it up.<br />
Never mind.</p>
<p>&#8211;I&#8217;ve changed it now</p>
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		<title>By: werd</title>
		<link>http://www.bottledcity.com/2007/04/04/enigma-1436-one-more-step-2/#comment-148</link>
		<dc:creator>werd</dc:creator>
		<pubDate>Mon, 16 Apr 2007 10:29:48 +0000</pubDate>
		<guid isPermaLink="false">http://www.bottledcity.com/2007/04/04/enigma-1436-one-more-step-2/#comment-148</guid>
		<description>the answers are correct but i don't think n^2*(n-1) is the pattern simply because:
if n=1 then 1^2*(1-1)=0
   n=2 then 2^2*(2-1)=4 and
   n=600 then 600^2*(600-1)=215640000

the pattern I have is An=(n-1)n+1
where 
if n=1 then (1-1)*1+1=1
   n=2 then (2-1)*2+1=3
   n=3 then (3-1)*3+1=7     etc

Therefore to get the 600th term, you do (600-1)*600+1=359401

I hope I got my point across.</description>
		<content:encoded><![CDATA[<p>the answers are correct but i don&#8217;t think n^2*(n-1) is the pattern simply because:<br />
if n=1 then 1^2*(1-1)=0<br />
   n=2 then 2^2*(2-1)=4 and<br />
   n=600 then 600^2*(600-1)=215640000</p>
<p>the pattern I have is An=(n-1)n+1<br />
where<br />
if n=1 then (1-1)*1+1=1<br />
   n=2 then (2-1)*2+1=3<br />
   n=3 then (3-1)*3+1=7     etc</p>
<p>Therefore to get the 600th term, you do (600-1)*600+1=359401</p>
<p>I hope I got my point across.</p>
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