Conics from Repeating Decimals, an exploration.

  A short While back, one of John D Cooks tweets directed me to an interesting/surprising conic. 

*Wolfram Mathworld

 

I was surprised , in case you didn't notice, each point is two adjacent digits in the fraction  1/7.  The surprise went on to show that taking the digits in consecutive pairs, (14, 28); (42,85) ... would also form an ellipse.  The short post is here if you want to read before continuing.

Being a curious guy, I started trying this for 1/13 = .076923076923.....

It wasn't an Ellipse, but a hyperbola, but still a conic. The equation is \(141x^2 + 134 xy + 9y^2 - 1872x -684y -4387 = 0\)

 Shock and Awe!  Is this some property of repeating decimals I somehow never learned?

Ok how about 1/17?  Its period was 8 digits, so more points to try.  Turns out, it didn't work.  

1/17

Both the ones that worked had six digit periods, maybe that was a factor.  Couldn't find any more small prime with period six, 14 had period six, but it was the same cyclic pattern as 7.  26 has period 6, repeating 384615 after an initial zero. Even twenty-sixths would be the 1/13 cycle, so they would work. 

3/26 cycled the pattern of 1/26, and produced the same hyperbola, but 5/26 had 923076, producing a different hyperbola.  All the other odd/26 (excepting 13/26) produced some cycle of one of these.  

Some (all?) of this is because the repeat pattern of 1/16, preceded by a zero, is 384615 x two is a cyclic permutation of the repeatng sequence of 1/13.

Deeper into the well, Alice, the rabbit, and me.  1/28 had the same cyclic six digits as seven.  1/35 did also. 1/39 had a new six digit period, 0,2,5,6,4,1.  

Oh, Yes.  Can I get an A-men, children.

So now what.  (right)

I realized that 2/13 had a completely different repeat pattern than 1/13; 1, 5, 3, 8, 4, 6.  So I tried it and it worked also. It was a hyperbola also.(below)

2/39th was different from 1 /39, and 3/39 would revert to the conic for 1/13.  

So on to 2/39; 0, 5, 1, 2, 8, 2. (below left)

And that was totally unexpected, and not a conic.

So! What does that lead toward.  Do only period six reciprocals work? More to try that are not six.  Still haven't found a six other than 7, and multiples of 13 that do work. And are there others that work for 2/p and 3/p... when multiples of 13? What happens to 7x13=91, 010989??? I want to try 1/19, it is the next after 1/7 to have a full p-1 period. Have I focused on individual digits too much and not enough on the fact that pairs of two worked also?

Took time to do 1/19, total failure. First 15 points made two ellipses and a degenerate hyperbola. 

So 1/13 in pairs worked as another hyperbola.

Trying 1/17 next.

1/17 in pairs has all 16 points lying  along two parallel lines of an infinite ellipse, equally divided on both lines.  Maybe I am on to something here.  

 (OK, it looks like two straight lines. After I had written and printed this, some years later I was reading this and wondered if I had assumed it was straight from the graph....posh, I must have checked it.....Ooops, I apparently didn't, because when I "re-checked" it suddenly had a bend.  So in fact it seems to be a hyperbola as well.  (Mia Culpa)

Will try 1/19 next, 18 digits.  Gonna' get crowded on the dance floor tonight. 052631578947368421, wow.

The first six points for 1/19 in pairs fell along a pair of intersecting lines, degenerate hyperbola, and then the next four wander off into space are not part of either.  My silver bullet has turned into pewter.   

Going to look at 1/91 both single digits and in pairs as points.  The digits, 010989, are not like either 1/13 or 1/7. 

The singles worked well

1/91 not 1/19

. The doubles produced a very eccentric ellipse.  

So far of the three I have tried with full periods, 1/7 formed conics as both single digit and paired, 1/17 worked only with paired digits, and 1/19 failed both ways .

Is there a pattern I'm missing. Will try 1/23 next, which will give me two each of full period 6n+1 and 6n-1 primes.

The first 9 points would form either an ellipse or a hyperbola depending on the five points selected, bur none contained more than the minimum five points, with four scattered around.

The pairs attempt also had self destructed by nine points.  

So some full period primes work and some don't with single and/or paired digits .  Some primes and some composite numbers with six digit periods. Need to find more six digit period composites that don't just echo 1/7 or 1/13.  1/63 has fresh six digit period, 015873.  1/77 has 012987. , and 1/84 has 190476. 

1/63 forms an ellipse in both one digit and pairs, 1/77 forms hyperbolas in both methods, and 1/84 failed in both methods.  As the old Soma Cube adds used to say, "Soma Do, Soma Don't"

1/7, 1/13, 1/91, At this Point I made up a fraction with six digit period with only one rule, it obeys the rule that numbers in first half (I chose 123) have their nines compliment in second half (876) and plotted the points with single digits (124/1001 as it turns out) and it was a Hyperbola. When I plotted Pairs they were a hyperbola as well.

Some six digit periods did not have the halves with nine compliments.  All the fractions with full periods have this complimentarity, including 1/19. 1/63 has eight compliments between halves, and 1 /84 has no patterns of compliments in the halves.  1/39 had  compliments of six, 2/39 had no complimentary between halves.  so????  Do failures of six digits period just lack complimentary halves?

Next I tried one with 123 and 8-compliments in second half (765) . Another hyperbola with singles and pairs.

Will 7 compliments work; 123654?  Yes for both.

At this point I tried going to negative compliments. Compliments of -2 would give 1,2,3,-3,-2,-1.  For singles it was another hyperbola, and pairs would not work at all as some digits would be mix of positive and negative digits, numbers that would be really imaginary

No idea what all this suggests, but try your own ideas, and do share what you find! I'm making a spread sheet of all the six digit periods I can find and test noting compliments or not.

One more observation.  The numbers that formed ellipses were 1/7, 1/39, 1/91, and 1/63 .  These all are 4n-1 numbers.  The hyperbolas were formed by 1/13, 1/77 .  These are all 4n+1 numbers.  And 1/17, is a 4n+1 number, but it's repeat period is 16 digits.  Just for a kick I took the first three digits of the repeating digits and appended the first three digits from the second half of the repeating digits, and they formed an ellipse.  

If someone else makes more (any?) sense of this, would love a copy of what you find.