Weregoose Authentic INTJ
Super Elite (Last Title)
Joined: 25 Nov 2004 Posts: 3976

Posted: 23 Jun 2010 09:24:59 am Post subject: 


Carrying over from this thread, I decided to refine the search by gauging which of ten dissections of the natural numbers holds all of the square roots matching the initial digit of any given fraction. When the numbers are adjacent to each other within a group (for instance, [font=times new roman][size="3"]√61[/size] and [font=times new roman][size="3"]√62[/size] both "begin" with [font=times new roman][size="3"].8[/size]), only the foremost is considered. Fortunately, every case adopts the same linear recurrence, but with a different series of initial values.
The differences between these sequences and the central polygonal numbers [font=times new roman][size="3"](n–1)n+1[/size] (the "[font=times new roman][size="3"].5[/size]" set) can be described quite easily: I'll leave it up to you to extract it from the following code, if you think you're clever enough.
With D an integer representing the desired digit, seq(I[font=verdana]²I+1iPart(I.2DI+.15D(D<6)),I,0,6 initiates the equation...
[font=times new roman][size="3"]a_{n} = a_{n–7} – 2a_{n–6} + a_{n–5} – a_{n–2} + 2a_{n–1}[/size]
...which is progressed by ΔList(cumSum(augment(Ans,{Ans(1)2Ans(2)+Ans(3)Ans(6)+2Ans(7.
Candidates (those outside of parentheses) are generated thus:
[font=courier new]D  [font=times new roman][size="3"]a_{n}[/size]
[font=courier new]0  1,0,1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324,...
[font=courier new]1  (1,1,2,5,)10,17,27,38,51,66,83,103,124,147,172,199,229,260,293,328,...
[font=courier new]2  (1,1,2,)5,(11,)18,28,39,52,68,85,105,126,149,175,202,232,263,296,332,...
[font=courier new]3  (1,1,2,6,)11,19,29,40,54,69,87,107,128,152,177,205,235,266,300,335,...
[font=courier new]4  (1,1,)2,6,12,20,30,41,55,71,89,109,130,154,180,208,238,269,303,339,...
[font=courier new]5  (1,1,3,7,13,)21,31,43,57,73,91,111,133,157,183,211,241,273,307,343,...
[font=courier new]6  (1,1,3,)7,13,22,32,44,58,74,93,113,135,159,185,214,244,276,310,346,...
[font=courier new]7  (1,1,)3,(8,)14,23,33,45,60,76,95,115,137,162,188,217,247,279,314,350,...
[font=courier new]8  (1,1,4,)8,15,24,34,47,61,78,97,117,140,164,191,220,250,283,317,354,...
[font=courier new]9  (1,1,4,9,16,25,)35,48,63,80,99,119,142,167,194,223,253,286,321,358,...
And it goes without saying that for each "hit," reiterate with one higher until D is no longer matched.
Not necessarily for my purposes is the observation that they all converge on similar curves:
[font=courier new]D  [font=times new roman][size="3"]~a_{n}[/size]
[font=courier new]0  [font=times new roman][size="3"](5x^{2} – 20x + 20)/5[/size]
[font=courier new]1  [font=times new roman][size="3"](5x^{2} – 19x + 21)/5[/size]
[font=courier new]2  [font=times new roman][size="3"](5x^{2} – 18x + 19)/5[/size]
[font=courier new]3  [font=times new roman][size="3"](5x^{2} – 17x + 17)/5[/size]
[font=courier new]4  [font=times new roman][size="3"](5x^{2} – 16x + 15)/5[/size]
[font=courier new]5  [font=times new roman][size="3"](5x^{2} – 15x + 15)/5[/size]
[font=courier new]6  [font=times new roman][size="3"](5x^{2} – 14x + 12)/5[/size]
[font=courier new]7  [font=times new roman][size="3"](5x^{2} – 13x + 11)/5[/size]
[font=courier new]8  [font=times new roman][size="3"](5x^{2} – 12x + 10)/5[/size]
[font=courier new]9  [font=times new roman][size="3"](5x^{2} – 11x + 9)/5[/size]
I'll lastly note that their expansions bear a slight resemblance to the aforesaid differences.
*looks at clock* Yup, that's another allnighter. Heaven.
Last edited by Guest on 24 Jun 2010 12:55:27 am; edited 1 time in total 
