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# The 1/89 lemma

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Gosh darn it I leave the room for two seconds!

I've been thinking about the interesting fact that 1/89 generates the sequence of Fib numbers distributed over negative powers of ten, at least when expressed in BASE ten. I'm wondering whether other sister series, such as the Lucas, can be similarly distributed, say, over negative powers of some other number in some other base (I've tried base 10 without success), and also still be a simple fraction in that system. Would this have any relation to the fact that the classical Pascal Triangle's rows give powers of 11 (though expressed in base 10)- the sister Pascal Triangles also give powers of 11 plus increments defined by the 'seeds' of their sides that begin sister Fib series.

Jess Tauber

[email protected]

the repeating patterns seen in the moduli of fibonacci numbers varies with the divisor, i.e. the base of expression. in one case i found no repeating pattern insofar as i went. i'll have to do some searching here to find my work, but i won't bother unless you ask to see it.

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Gosh darn it I leave the room for two seconds!

I've been thinking about the interesting fact that 1/89 generates the sequence of Fib numbers distributed over negative powers of ten, at least when expressed in BASE ten. I'm wondering whether other sister series, such as the Lucas, can be similarly distributed, say, over negative powers of some other number in some other base (I've tried base 10 without success), and also still be a simple fraction in that system. Would this have any relation to the fact that the classical Pascal Triangle's rows give powers of 11 (though expressed in base 10)- the sister Pascal Triangles also give powers of 11 plus increments defined by the 'seeds' of their sides that begin sister Fib series.

Jess Tauber

[email protected]

the repeating patterns seen in the moduli of fibonacci numbers varies with the divisor, i.e. the base of expression. in one case i found no repeating pattern insofar as i went. i'll have to do some searching here to find my work, but i won't bother unless you ask to see it.

well, i was curious about my list so looked it up without waiting for you to ask. first, an aside on that topic, pascal, of asking. you never miss a chance to take a poke at [ostensibly] me for rousting things up, yet you add to the mix, as in provocation, by pointedly not answering some of my direct questions. like, "why won't you post some drawings of your described geometries and tables?" let me ask it directly so as not to allow that the first time only referred to a question and so was not actually the question. ahhhem... So Jess, why won't/don't you post any drawings of the geometry and tables you have described?

anyway, back to the Fibonacci numbers. so, on my desktop calculator i divided 1 by 89 and get 0.01123595505617977528089887640449. so, the first 6 decimal digits taken individually match the Fibonacci sequence, 0 1 1 2 3 5 8 13... . :shrug: it's an idle coincidence to me, and a short lengthed one at that. using other bases to divide one by eighty-nine would give different looking, if not different results. for example, how would you evaluate this different base decimal expansion of 1/8910 with The Fibonacci numbers in the same base?: .01D3MN0... (Note: i didn't calculate that in a particular base, rather it is a shorthand representation of the problem of comparing bases as pascal proposed. if you don't "know" what the Fibonacci sequence "looks like" in each different base, looking at the decimal expansion has no context for comparison.)

here's a thread that might shed some light on the issue for you. Repeating digits under long division

so thens, to my claim of having done some original "moduli" analysis of the Fibonacci numbers. phhh! whatever turtle. so here's the post with my table, out of its original context and i'll follow up with an explanation. Katabatak Patterns of Fibonacci Numbers Base Two to Twelve

Base___Katabatak Pattern Length

_2____1

_3____3

_4____8

_5____6

_6___12

_7___24

_8___16

_9___12

10___24

11____? (at least >36)

12___10

the first line is a trivial reduction that we can ignore. the second entry, _3____3, means using base 3, i.e. mod 2, the repeating pattern of remainders one gets when successively dividing the sequention Fibonacci numbers has a length of 3 digits.

like this: 0 mod 2=0; 1 mod2=1; 1 mod2=1; 2 mod2=0; 3 mod2=1; 5 mod2=1; 8 mod2=0;9 mod2=1; 13 mod2=1; ...

the repeating pattern of residues is then {0 1 1} and this pattern has 3 elements or a "Katabatak Pattern Length" of 3.

similarly for each following entry in my table, using the divisor that is 1 less than the number listed in the Base column, gives a repeating pattern length given in the Katabatak Pattern Length column. note that it is mod 10 that i found no repeat for out to 36 Fibonacci numbers. i wonder if there ever is a repeat. mmmmm.... curiouser & curiouser.

Lars and Phillip, or any of you lurking math programming junkies, i invite, if not challenge, you all to unleash your enthusiasm & skills 2.0 by checking and then expanding my table to infinity and beyond.

that's all i got,

like it or not. :lol:

Edit: speeling airs & parenthetical qualifications

Edited by Turtle
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Ok, I yield- I apparently don't really DO graphics. People all over the place, including my research partner, keep asking for them, or models. For years now. The only time I ever cut and pasted a graphic image was to my Facebook page, and those were public images of polytopes and THEIR faces. I suppose I'll have to get over this. After all, I had to do drawings for my patent application.

Now, for 1/89. It really does work infinitely, but you have to remember that each Fib number in succession is represented as a successive negative power of ten. For the ealiest Fib numbers this is ok, but as soon as you hit 13 THERE IS OVERLAP OF COVERAGE on the negative powers of ten, and so you start to see summing. As Fib numbers climb to hundreds, thousands, etc. the overlap becomes so extreme that you cannot recognize individual digits in the sequence for what they are. Ok?

Jess Tauber

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Ok, I yield- I apparently don't really DO graphics. People all over the place, including my research partner, keep asking for them, or models. For years now. The only time I ever cut and pasted a graphic image was to my Facebook page, and those were public images of polytopes and THEIR faces. I suppose I'll have to get over this. After all, I had to do drawings for my patent application.

Now, for 1/89. It really does work infinitely, but you have to remember that each Fib number in succession is represented as a successive negative power of ten. For the ealiest Fib numbers this is ok, but as soon as you hit 13 THERE IS OVERLAP OF COVERAGE on the negative powers of ten, and so you start to see summing. As Fib numbers climb to hundreds, thousands, etc. the overlap becomes so extreme that you cannot recognize individual digits in the sequence for what they are. Ok?

Jess Tauber

excellent! :thumbs_up danke. :) as to drawings of close-packing, we can refer to some by buckminster fuller as i read the copyright, as long as the use is not for profit. while he was in many regards not playing with a full deck, he was an excellent draftsman. my "katabataks" in my previous post he called "indigs". but, he was stuck on the idea that base 10 was the "natural" base because we happen to have 10 fingers so he never tried other bases and so never discovered the patterns i have. so, here's a bucky drawing of close packing spheres that you can refer to while describing your schemas. there is text and equations at the link as well as a larger version of the drawing.

Fig. 222.01 Equation for Omnidirectional Closest Packing of Spheres

as to your last paragraph here quoted, i think when the overlap occurs, let alone becomes extreme, it is the end of the coincidence. off by 1 is the number theorists' nightmare. off by 2 is twice as bad, and so on.

Edited by Turtle
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Re 1/89. I never said the decimal was PRETTY! ;-)

In many systems the order is hidden beneath jumble. The nearest analogy I can think of is that of fossilized animal skeletons, all smashed into a plane. The paleontologist, knowing how real animals are constructed, tries to undo the compression into 2 dimensions and reconstruct the third. I guess the number theoretical paleontologist would throw up his hands and declare that all these prehistoric creatures were monstrosities from Flatland.

Currently I'm hoping to figure out if relativistic distortion of the otherwise highly ordered quantum relations I've written so often about here is actually the time dimension in 4D spacetime. If I'm right, then at least in part, the fact that half-lives of elements decrease with increasing atomic number, generally (but with many exceptions and for a variety of reasons) might be re-thought of as showing that the heavier elements, whose inner electrons are whirling about at close to the speed of light, have used up their electron orbits, as it were. A kind of creeping progeria. They AGE more rapidly than lighter elements. Of course we don't think of atoms as aging, but even the Hubble universe is aging. More massive stars burn out faster. Why should atoms be immune? All goes back to my idea that matter shrinking is equivalent to space expanding. And time elapsing to energy ???

Perhaps during this 'shrinkage' the different elements and isotopes don't do so at the same rate, in keeping with how fast their electrons are moving? Why not- in the Standard Model we have the Higgs and differential coupling to it of fundamental particles giving them mass. Maybe this is related?

Jess Tauber

Edited by pascal
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• 2 weeks later...

I've been thinking about the interesting fact that 1/89 generates the sequence of Fib numbers distributed over negative powers of ten, at least when expressed in BASE ten.

Amazing!

I’m curious, pascal – how or where did you discover this?

I checked it for over 1000 base 10 digits, and it seems true.

Generalizing (via some guesses)

$\frac{0}{10} + \frac{1}{100} + \frac{1}{1000} + \frac{2}{10000} \dots = \sum_{n=1}^{\infty} \frac{\mbox{Fib}_n}{10^n} = 0. = \frac{1}{89}$

to

$\sum_{n=1}^{\infty} \frac{\mbox{Fib}_n}{B^n} = \frac{1}{B(B-1)-1}$

also seems to be true.

B=2 is especially interesting, as [imath]\frac{1}{2(2-1)-1}=1= \frac{0}{2} +\frac{1}{4} +\frac{1}{8} +\frac{2}{16} +\frac{3}{32} \dots[/imath]

I expect this won’t be too hard to prove, and that a proof will make intuitive sense of this. I’ll do and post one when I have some free time, unless someone with a more profound proof obsession does first. :)

I'm wondering whether other sister series, such as the Lucas, can be similarly distributed, say, over negative powers of some other number in some other base (I've tried base 10 without success), and also still be a simple fraction in that system.

I’ve a hunch that the “base” B doesn’t much matter in these series.

Like you, pascal, I’m curious to see if something like the 2-fib sequence using formula above can be found for other sequences.

Would this have any relation to the fact that the classical Pascal Triangle's rows give powers of 11 (though expressed in base 10)- the sister Pascal Triangles also give powers of 11 plus increments defined by the 'seeds' of their sides that begin sister Fib series.

I don’t understand this. Can you give an example?

The sums of the rows of a classic Pascal triangle seem to me to be powers of two, eg:

1+1 = 2

1+2+1 = 4

1+3+3+1 = 8

1+4+6+4+1 = 16

1+5+10+10+5+1 = 32

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Ok, deep breath. I didn't discover the 1/89 relation. It's been known for a very long time. The B=2 giving 1 is VERY interesting. I didn't see that. Also, your sums of the numbers in horizontal rows are the denominators for B=2. Is this relatable somehow to the facts for 1/89, numerators vs. denominators?

As for powers of 11, you don't sum the numbers. You just read them off as if powers of 10. That is, 1=11^0, 11=11^1, 121=11^2 and so on.

There are also vertical relations in the Pascal system, though I don't remember offhand how they work, or what they're called. Thus FOUR distinct straight line relations: deep diagonals (the classic triangular, tetrahedral etc.), shallow diagonals sampling and summing across these (giving Fib, Luc, etc.), the horizontals (giving powers of 11, quadratic factors, numbers of features of polytopes (vertices, edges, faces...), and the the verticals. The darned thing is like a multidimensional computer, not a bad analogy at all given the real dimensional associations of diagonals.

We also have the fact that the powers of the more general Metal Means (of which the Golden Mean is only one special case) have equations whose terms, when laid one atop the other, show numerical coefficients and power units that come directly out of the (2,1) sided Pascal Triangle sister, leading me to suspect that other Sisters deal with the even more general PV numbers I mentioned in earlier posts.

I've also now found that the generative motifs used in the deep diagonals of some of the Pascal Sisters (equivalent to summing different multiples of classical Pascal diagonal numbers, have links to the equations for Metal Means.

That is, for a classical tetrahedral number we take 1x itself, plus zero x the next (or last), but for the (2,1) Sister, with tetahedrals replaced by square pyramidal numbers, we now have values in this replacement diagonal which correspond to the sums of contiguous tet numbers from the classical tet diagonal. And in the next Pascal Sister's tetrahedral equivalent, we have 1x a tet number plus 2x the next, and on it goes, if we limit ourselves to Sisters with sides (2,1),(3,2),(4,3) etc. But what about other side values that differ by more than one unit? This is what makes me think that they might use formulae for deep diagonals that come out of broader PV numbers, rather than just Metal Means.

And with this in mind, note that your B(B-1)-1 is a formula in this family, for the Golden Mean, IIRC. The other Metal Means have B(B-1)-N, where N isn't 1. And if you change the internal integer to non-singular values, you get the PV numbers, so B(B-M)-N (and you can also replace minuses with pluses for related values). I would imagine that even the multiples and powers of B can be manipulated meaningfully, but I haven't read about anything like that yet.

I'm only just starting to learn how to channel Ramanujan, and have a ways to go yet.

Jess Tauber

Edited by pascal
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Here's another question. If 1/89 gives the sequence of Fib numbers distributed over negative powers of 10, in base 10, then is it possible that this is not an arbitrary fact? That is, could 1/55, say, also give the sequence of Fib numbers over negative powers of some other number, in the base equal to that number? In other words, is this a general law linking bases in different numbers to the expression of Fib numbers, and in particular to one Fib number intimately connected to that base?

Starting with 1,1,2,3,5,8,13,21,34,55,89, then 89 is the 11th term (interesting in itself given the link to powers of 11 in the classical Pascal Triangle's rows' values read AS powers of 10 in base 10). So is there a connection here as well? What would Pascal sisters look like in other bases? Would such values suggest other ways of constructing them? 89+11=100. Another arbitrary fact, given that 100 is 10 squared? Does our 10D universe (in some models of string theory) relate? If the universe had a different dimensionality would Pascal, Golden, Metal, PV math look different?

Jess Tauber

[email protected]

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Here's another question. If 1/89 gives the sequence of Fib numbers distributed over negative powers of 10, in base 10, then is it possible that this is not an arbitrary fact? That is, could 1/55, say, also give the sequence of Fib numbers over negative powers of some other number, in the base equal to that number? In other words, is this a general law linking bases in different numbers to the expression of Fib numbers, and in particular to one Fib number intimately connected to that base?

Yes. The formula I gave in my previous post,

$\sum_{n=1}^{\infty} \frac{\mbox{Fib}_n}{B^n} = \frac{1}{B(B-1)-1}$

is such a “general law”, of which

$\sum_{n=1}^{\infty} \frac{\mbox{Fib}_n}{10^n} = \frac{1}{10(10-1)-1} = \frac{1}{89}$

is simply a (not especially) special case. Here it is evaluated for B=2 to 100:

 2  1/1
3  1/5
4  1/11
5  1/19
6  1/29
7  1/41
8  1/55
9  1/71
10  1/89
11  1/109
12  1/131
13  1/155
14  1/181
15  1/209
16  1/239
17  1/271
18  1/305
19  1/341
20  1/379
21  1/419
22  1/461
23  1/505
24  1/551
25  1/599
26  1/649
27  1/701
28  1/755
29  1/811
30  1/869
31  1/929
32  1/991
33  1/1055
34  1/1121
35  1/1189
36  1/1259
37  1/1331
38  1/1405
39  1/1481
40  1/1559
41  1/1639
42  1/1721
43  1/1805
44  1/1891
45  1/1979
46  1/2069
47  1/2161
48  1/2255
49  1/2351
50  1/2449
51  1/2549
52  1/2651
53  1/2755
54  1/2861
55  1/2969
56  1/3079
57  1/3191
58  1/3305
59  1/3421
60  1/3539
61  1/3659
62  1/3781
63  1/3905
64  1/4031
65  1/4159
66  1/4289
67  1/4421
68  1/4555
69  1/4691
70  1/4829
71  1/4969
72  1/5111
73  1/5255
74  1/5401
75  1/5549
76  1/5699
77  1/5851
78  1/6005
79  1/6161
80  1/6319
81  1/6479
82  1/6641
83  1/6805
84  1/6971
85  1/7139
86  1/7309
87  1/7481
88  1/7655
89  1/7831
90  1/8009
91  1/8189
92  1/8371
93  1/8555
94  1/8741
95  1/8929
96  1/9119
97  1/9311
98  1/9505
99  1/9701
100  1/9899

Starting with 1,1,2,3,5,8,13,21,34,55,89, then 89 is the 11th term (interesting in itself given the link to powers of 11 in the classical Pascal Triangle's rows' values read AS powers of 10 in base 10). So is there a connection here as well?

I don’t think so. 55, the denominator for B=8, is the 10th Fibonacci sequence term. I’ve not proved it, but just using a fast calculator, 89 is the last denominator that’s also a Fib seq term for at least a few thousand terms.

It’s kinda cool, though trivial given the formula ([imath]B(B-1)-1[/imath]), that, written in the base of B, the B’s complement (the generalization of the ten’s complement in base 10) of all of the denominators greater than 1 are the same. That is

510 = 123

1110 = 234

1910 = 345

2910 = 456

4110 = 567

5510 = 678

7110 = 789

8910 = 8910

etc.

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Interesting stuff! Bit of work on your part- thanks for that.

Curious that the differences in denominators grow by increments of 2:

2 1/1

3 1/5

4 1/11

5 1/19

6 1/29

7 1/41

8 1/55

9 1/71

and that the rightward terms grow by increments of 11 in the lower set: 12, 23, 34, 45... etc. in the different bases.

Jess Tauber

Edited by pascal
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Curious that the differences in denominators grow by increments of 2:

That tells us they’re terms of a degree 2 polynomial.

We could use Newton’s difference interpolation formula (an old, but very useful one :)),

$D(0,n) = \sum_{a=0}^{\infty} D(a,0)\prod_{b=1}^a \frac{n-b-1}{b}$

, to find the polynomial, like this:

Make a difference table:

 B  n D(0,n) D(1,) D(2,) D(3,)
2  0      1     4     2     0
3  1      5     6     2     0
4  2     11     8     2     0
5  3     19    10     2     0
6  4     29    12     2     0
7  5     41    14     2
8  6     55    16
9  7     71

; substitute it into the formula:

$D(0,n) = 1 +\frac{4n}{1} +\frac{2n(n-1)}{2} +0 \dots = n^2 +3n +1$

; then replace n with B-2

$D(0,n) = n^2 +3n +1 = (B-2)^2 +3(B-2) +1 = B^2 -B -1 = B(B-1)-1$

which matches the term for the denominator in the original formula,

$\sum_{n=1}^{\infty} \frac{\mbox{Fib}_n}{B^n} = \frac{1}{B(B-1)-1}$

and that the rightward terms grow by increments of 11 in the lower set: 12, 23, 34, 45... etc. in the different bases.

This too comes from the denominator’s formula,

$B(B -1) -1 = B^2 -B -1 = (B -2)B +(B-1)$

: each consecutive B’s denominator will consist of a 1st and 2nd digit, base B, that’re exactly 1 greater than the previous B-1’s, base B-1.

:smart: :)

Pleasant as this is, I’m just playing around here – what various wise and knowing folk who know me call “mentally masturbating”. I – or someone – need to get to the real mathic business of a elementary proof of the “1/89 lemma” (for lack of a better name).

Still playing, I found truethsome-looking gem: the lemma seems to hold not just for the classical 2-fib sequence (0 1 1 2 3 5 8 13 21 34 55 89 ...) but for the 3-fib (0 0 1 1 2 4 7 13 24 44 81 149 274 ...), or any n-fib (AKA n-nachi) sequence :Exclamati The amazement from this lemma seems unending!

Ok, deep breath. I didn't discover the 1/89 relation. It's been known for a very long time.

Given that the Fibonacci sequence goes back to … well, to Fibonacci, who in wee years of the 13th century, for all intents and purposes brought the Arabic numeral system that made math as we now know it possible to the western world from which our present day one evolved, I was pretty sure this discovery was an old one. I’m curious, though, about its history. Do you know it, Pascal?

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So the 1/89 match to Fib is just an amazing coincidence? What would the denominators be for other n-nachi inverted sequences?

By the way, I was looking into the way that sister Pascal triangles should stack one atop the other in a third dimension. I'd thought that they should be lined up as a column, but now, considering that all the usual straight line relations within any triangle's plane correspond to close packing of spheres, this made me realize that any volume should also be close packed. It would line things up in a more interesting fashion. Then you can get all sorts of triangular relations that are out of plane, but which may be more relevant to actual behaviors of elements in the periodic system, either directly by simply reading off numbers as constructional factors or via operations such as summing (as in diagonals), dropping over powers (as in powers of 11 in the classical Pascal horizontals, etc.).

I wonder whether some of the things you've found here might map to a larger system like this (and it might be higher-D than 3).

Jess Tauber

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Hi Jess,

So the 1/89 match to Fib is just an amazing coincidence?

It might not be a coincidence.

http://en.wikibooks.org/wiki/Algorithm_Implementation/Mathematics/Fibonacci_Number_Program#Logarithmic-time_Version

Logarithmic-time Version

This version squares the Fibonacci transformation, allowing calculations in log2(n) time:

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So the 1/89 match to Fib is just an amazing coincidence?

Stronger than just a coincidence, it’s the evaluation of the equation

$\sum_{n=1}^{\infty} \frac{\mbox{Fib}_n}{B^n} = \frac{1}{B(B-1)-1}$

for an arbitrary value of B=10.

That it appears true, not just for when Fib is the classical 2-fib sequence, but for any n-fib sequence, tickles my sense of amazement and wonder.

What would the denominators be for other n-nachi inverted sequences?

Provided we follow the convention of starting the sequence with n-1 zeros followed by a 1, the equation holds for all integer n>1. Sketching with numbers and [imath]\dots[/imath]s replacing [imath]\infty[/imath]s:

$\frac{0}{2} +\frac{1}{4} +\frac{1}{8} +\frac{2}{16} +\frac{3}{32} +\frac{5}{64} +\frac{8}{128} +\frac{13}{256} +\frac{21}{512} +\frac{34}{1024} +\frac{55}{2048} +\frac{89}{4096} +\frac{144}{8192} +\frac{233}{16384} +\frac{377}{32768} +\frac{610}{65536} + \dots$

$\frac{0}{2} +\frac{0}{4} +\frac{1}{8} +\frac{1}{16} +\frac{2}{32} +\frac{4}{64} +\frac{7}{128} +\frac{13}{256} +\frac{24}{512} +\frac{44}{1024} +\frac{81}{2048} +\frac{149}{4096} +\frac{274}{8192} +\frac{504}{16384} +\frac{927}{32768} +\frac{1705}{65536} +\frac{3136}{131072} + \dots =$

$\frac{0}{2} +\frac{0}{4} +\frac{0}{8} +\frac{1}{16} +\frac{1}{32} +\frac{2}{64} +\frac{4}{128} +\frac{8}{256} +\frac{15}{512} +\frac{29}{1024} +\frac{56}{2048} +\frac{108}{4096} +\frac{208}{8192} +\frac{401}{16384} +\frac{773}{32768} +\frac{1490}{65536} +\frac{2872}{131072} +\frac{5536}{262144} + \dots =$

$\frac{0}{2} +\frac{0}{4} +\frac{0}{8} +\frac{0}{16} +\frac{1}{32} +\frac{1}{64} +\frac{2}{128} +\frac{4}{256} +\frac{8}{512} +\frac{16}{1024} +\frac{31}{2048} +\frac{61}{4096} +\frac{120}{8192} +\frac{236}{16384} +\frac{464}{32768} +\frac{912}{65536} +\frac{1793}{131072} +\frac{3525}{262144} +\frac{6930}{524288} + \dots =$

$\frac{1}{1}$

$\frac{0}{3} +\frac{1}{9} +\frac{1}{27} +\frac{2}{81} +\frac{3}{243} +\frac{5}{729} +\frac{8}{2187} +\frac{13}{6561} +\frac{21}{19683} +\frac{34}{59049} +\frac{55}{177147} +\frac{89}{531441} +\frac{144}{1594323} + \dots =$

$\frac{0}{3} +\frac{0}{9} +\frac{1}{27} +\frac{1}{81} +\frac{2}{243} +\frac{4}{729} +\frac{7}{2187} +\frac{13}{6561} +\frac{24}{19683} +\frac{44}{59049} +\frac{81}{177147} +\frac{149}{531441} +\frac{274}{1594323} +\frac{504}{4782969} + \dots =$

$\frac{0}{3} +\frac{0}{9} +\frac{0}{27} +\frac{1}{81} +\frac{1}{243} +\frac{2}{729} +\frac{4}{2187} +\frac{8}{6561} +\frac{15}{19683} +\frac{29}{59049} +\frac{56}{177147} +\frac{108}{531441} +\frac{208}{1594323} +\frac{401}{4782969} +\frac{773}{14348907} + \dots =$

$\frac{0}{3} +\frac{0}{9} +\frac{0}{27} +\frac{0}{81} +\frac{1}{243} +\frac{1}{729} +\frac{2}{2187} +\frac{4}{6561} +\frac{8}{19683} +\frac{16}{59049} +\frac{31}{177147} +\frac{61}{531441} +\frac{120}{1594323} +\frac{236}{4782969} +\frac{464}{14348907} +\frac{912}{43046721} + \dots =$

$\frac{1}{5}$

$\frac{0}{10} +\frac{1}{100} +\frac{1}{1000} +\frac{2}{10000} +\frac{3}{100000} +\frac{5}{1000000} +\frac{8}{10000000} +\frac{13}{100000000} + \dots =$

$\frac{0}{10} +\frac{0}{100} +\frac{1}{1000} +\frac{1}{10000} +\frac{2}{100000} +\frac{4}{1000000} +\frac{7}{10000000} +\frac{13}{100000000} +\frac{24}{1000000000}+ \dots =$

$\frac{0}{10} +\frac{0}{100} +\frac{0}{1000} +\frac{1}{10000} +\frac{1}{100000} +\frac{2}{1000000} +\frac{4}{10000000} +\frac{8}{100000000} +\frac{15}{1000000000} +\frac{29}{10000000000} + \dots =$

$\frac{0}{10} +\frac{0}{100} +\frac{0}{1000} +\frac{0}{10000} +\frac{1}{100000} +\frac{1}{1000000} +\frac{2}{10000000} +\frac{4}{100000000} +\frac{8}{1000000000} +\frac{16}{10000000000} +\frac{31}{100000000000} + \dots =$

$\frac{1}{89}$

By the way, I was looking into the way that sister Pascal triangles ...

Like Turtle, even searching the internet, I can’t figure out what a sister Pascal triangle is, Jess. Can you explain :QuestionM

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A sister Pascal Triangle is, like the Classical version, created by drawing integers down the sides of an open-bottomed equilateral triangle (though of course any triangle will do), and then summing any two neighboring terms in the same row and putting the result centered beneath them in the next row down. I COINED the term because I can't find any literature on these either, so I don't know what the professional terminology is.

The classical system has sides (1,1). The classical triangle has deep diagonals like the natural numbers, triangulars, tetrahedrals, pentatopes and such. Summing across these along so-called shallow diagonals generates the Fibonacci numbers, in either direction, because both sides are (1,1).

But if you have sides (2,1), then on one side you'll generate the Fib series again, but moved up one move, and on the other you'll get the Lucas series.

----------------------------------------2

--------------------------------------2---1

------------------------------------2---3---1

----------------------------------2---5---4---1

--------------------------------2---7---9---5---1

------------------------------2---9--16--14--6---1

----------------------------2--11--25--30--20--7--1

--------------------------2--13--36--55--50--27--8-1

------------------------2--15--49--91-105--77--35--9-1

Every sister Pascal system will generate similar series, all of which have ratios of neighboring terms converging on Phi.

The (2,1) system also has numbers, in its diagonals, which conform to the numerical coefficients of equations that define the powers of the Metal Means- a fact I discovered by accident, and have still yet to find out if anyone else ever noticed it before (I find it hard to believe that mathematicians wouldn't have seen this some time in the last 120 years or so). But there always has to be a first. Maybe I'll be named for it, or maybe one of the professionals I contacted has already published in his own name and conveniently forgot either to inform me or give me any credit for the discovery. That's already happened in my life before, in linguistics (twice).

So, all you have to do, for yourself, is set up the outline of a Pascal Triangle, blank inside, and then replace one or both outer numbers, and see what drops out. Interesting stuff. For the (2,1) system again, the natural numbers are replaced by the odd integers, the triangular numbers are replaced by square number, the tetrahedrals are replaced by square pyramidals, and so on. The (3,1) system, I think, gives you pentagonal numbers versus square or triangular. So, a lot of this should be intimately related to all the research on polygonal numbers.

Some people have trouble thinking in terms of abstract equations. I'm one of them. But charts and graphs I can understand. And sequences. More intuitive I guess. So its also surprising that I don't like to create depictions.

Jess Tauber

Edited by pascal
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A fun diversion: but beware links to El Naschie! Grandiosity does not make one wrong (or not even wrong...). Who amongst us has not photoshopped ourselves into group shots with the greats? The Nobel Committee be darned!

Jess Tauber

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OOPS! I forgot to include the link, which was the point of my latest post. http://www.mi.sanu.ac.rs/vismath/spinadel/index.html

I'm now in touch with the author, which just goes to prove I really do need professional help! :)

Jess Tauber

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