On the decimal expansion part, 1⁄7 has always fascinated me, having something very similar going on. Doubling from 7, you get 14, 28, 56; and 1⁄7 is 0.1̅4̅2̅8̅5̅7̅, 2⁄7 is 0.2̅8̅5̅7̅1̅4̅, 3⁄7 is 0.4̅2̅8̅5̅7̅1̅, <i>&c.</i> (just changing which digit you start the recurring sequence with). <a href="https://en.wikipedia.org/wiki/142,857" rel="nofollow">https://en.wikipedia.org/wiki/142,857</a> talks about it a bit more; the doubling sequence thing is covered in the section <i>1⁄7 as an infinite sequence</i> (including the reason the recurring decimal has 57 instead of 56—that 56 doubled is 112, so the hundred there overlaps with the six, much as the ten of the 13 adds to the 8 in the 1⁄89 expansion of this article).
From an archived talk (2011) on wikipedia [0]:<p>"The linked page misleadingly suggests that a certain Cody Birsner discovered the relationship between the series and the fraction, whereas it had been known for a considerable time before".<p>Günter Köhler, 1983 (published in the The Fibonacci Quarterly, 1985; who cites earlier papers from 1977 and 1981): <a href="https://www.fq.math.ca/Scanned/23-1/kohler.pdf" rel="nofollow">https://www.fq.math.ca/Scanned/23-1/kohler.pdf</a><p>[0] <a href="https://en.wikipedia.org/wiki/Talk%3AFibonacci_number%2FArchive_3#1/89" rel="nofollow">https://en.wikipedia.org/wiki/Talk%3AFibonacci_number%2FArch...</a>
"The successive ratios of the terms, i.e. 1/1, 2/1, 3/2, 5/3 ... tend to a number called the Golden Ratio by the Greeks."<p>Fun fact: take <i>any</i> two numbers (e.g. chosen randomly), and use them as the seeds for a Fibonacci-like sequence by summing the last two terms to generate the next term. The ratio of any two consecutive terms in that series will tend towards the golden ratio.
This is even more clear with larger expansions like `1/99989999 =1.00010002000300050008001300210034005500890144...×10−8`<p>See discussion on math overflow here: <a href="https://math.stackexchange.com/questions/656183/why-does-frac1-99989999-generate-the-fibonacci-sequence" rel="nofollow">https://math.stackexchange.com/questions/656183/why-does-fra...</a>
I went to college in '88 or '89 and one of my teachers showed me the 1/89 trick, along with a few others, e.g. 1/7, and so forth. My teacher claimed to have been shown the various tricks by one of his teachers back in the '60s.<p>In the same period I "discovered" an error detection technique which is commonly known as Hamming codes. I would never dare to claim I invented them or discovered them.<p>In '92 and '93, I was heavily in to the BBS scene, some people had 56K modems, others had 14.4K. With verifiable evidence of written notes and digital artifacts (a BBS door and protocol for AmiBBS and Citadel and a couple of others) , I created a technique whereby multiple peers with low-bandwidth connections could transfer small fragments of a larger dataset to a peer with a lot of bandwidth.<p>If you were heavily in to the warez scene and/or part of Fate, it is probably you made use of this protocol to transfer pirated software between FTP sites. A warehousing server would tell each peer who had what part of a piece of data, and any peer could make requests for any piece of the data from any other peer who happened to have a copy of that data. Today, a very similar protocl is commonly known as BitTorrent.<p>I would not say I "discovered" or "invented" the protocol as my work was based on the various X-, Y- & Z- modem protocols. There was a TCP/IP packet to -Modem packet translator so that a BBS talking over that new fangled internet thing could take advantage of a T1 (1.5Mbps) connection for instance, which really helped with spreading the warez around the various FTP sites by the couriers.<p>I doubt the veracity of the claim by the author to have "discovered it as original" in 1994 before anybody else.
For people who are generally interested in these kind of identities relating a fraction to a recursion, please check out the generatingfunctionology book [1]. The basic idea is to define f(x) = sum_n f_n x^n where f_n satisfies some recursion equation. Very often you can find f(x) as p(x) / q(x) where p and q are polynomials in x. Now you can simply evaluate f(0.1) or more generally f(b^-l) where b is a base and l a positive integer, which gives you on the left-hand side your rational number as a fraction, and on the right-hand side the decimal expansion.<p>[1] <a href="https://www.math.upenn.edu/~wilf/gfologyLinked2.pdf" rel="nofollow">https://www.math.upenn.edu/~wilf/gfologyLinked2.pdf</a>
Okay, as a non-mathematician, I see something like this and I think... “neat coincidence?”<p>But the world of numbers seems to be full of these neat coincidences. So do any of the math folks here have a theory or explanation of <i>why</i>?
Presumably in a different number base (base 12, eg) it would be a different number that had this reciprocal property. Would it still be in the fibonacci series in that base?
I noticed a while ago that the powers of 1001 encode the successive rows of
pascal's triangle/the binomial coefficients. This is not so surprising, since
we are effectively taking powers of the polynomial (1+x), but replacing x with
1000 (clearly it works the same if we use any other power of ten instead). I
wonder if we can find a relationship to that here. We might start by looking at
this relation:<p><pre><code> 1/(1-x) = 1 + x + x^2 + x^3 + ...
</code></pre>
Hopefully this will yield an operator x whose succesive powers are the
fibonacci numbers. Take .01/(1-x) = 1/89, then x = 0.11. Actually, the powers of
x, just like 1001 above, will yield rows of pascal's triangle. So the taylor
expansion above tells us that F(k) = Σ(n=0..k-1) B(n, k), in other words that
each fibonacci number is the sum of a diagonal of pascal's triangle (like here:
<a href="https://cdn1.byjus.com/wp-content/uploads/2018/11/maths/2016/08/01092801/PascalTriangleFibonacci.png" rel="nofollow">https://cdn1.byjus.com/wp-content/uploads/2018/11/maths/2016...</a>)<p>More generally, we can compute numbers with decimal expansions of the fibonacci
numbers with 10^-2n / (1 - 10^-n - 10^-2n). Notice that this is just the
z-transform of the recurrence relation of the fibonacci series, with 10^n
replacing z:<p><pre><code> Z(f(n)) = Z(f(n-1) + f(n-2) + δ(n-2)) (δ is the kronecker delta)
F(z) = z^-2 / (1 - z^-1 - z^-2)
</code></pre>
The taylor expansion of this expression has coefficients equal to the terms of
the fibonacci sequence - which makes sense, because that's the definition of
the z-transform. We can, with a little rearranging, get an explicit formula
for the fibonacci sequence from it too:<p><pre><code> Take φ± = (1 ± √5)/2
Then F(z) = z^2/(z - φ+)/(z - φ-)
= ( φ+ z/(z - φ+) - φ- z/(z - φ-) )/√5 (by partial fraction decomposition)
Z^-1(F(z)) = f(n) = (φ+^(n+1) - φ-^(n+1))/√5</code></pre>
In case anyone is interested, I ran this procedure using integer bases besides 10 and obtained the following sequence<p>1, 5, 11, 19, 29, 41, 55, 71, 89, ...<p>Searching it on the OEIS (a great resource for mathematics) gave<p><a href="http://oeis.org/search?q=1%2C5%2C+11%2C+19%2C+29%2C+41%2C+55%2C+71%2C+89&language=english&go=Search" rel="nofollow">http://oeis.org/search?q=1%2C5%2C+11%2C+19%2C+29%2C+41%2C+55...</a><p>It turns out that these are precisely the first values of the Fibonacci polynomial n^2 - n - 1<p>I haven't verified this fact, but it seems like it comes from an application of the generating function of the Fibonacci numbers.<p>I posted my code in another comment<p><a href="https://news.ycombinator.com/item?id=24933085" rel="nofollow">https://news.ycombinator.com/item?id=24933085</a>
I think I found a typo in the final line (not impactful for the result, however).<p>11/89 should be 11/8900<p>This can be verified by using the following Sage code (simply go to <a href="https://sagecell.sagemath.org/" rel="nofollow">https://sagecell.sagemath.org/</a> to avoid installing sage)<p><pre><code> b = 10
A = Matrix([[0, 1],[1/(b**2), 1/b]])
I = matrix.identity(2)
show((I - A).inverse() * Matrix([[1/b**2],[1/b**3]]))</code></pre>
Question (perhaps naive):<p>Given the decimal expansion of 1/89, is there any way to directly retrieve the Fibonacci sequence from it? (that is, not using any knowledge of the sequence itself)<p>I'm assuming it can't be done because different sets of fractions can sum to 1/89, but maybe I'm missing something.
btw, there's a typo on the page for the value of 1/89, it should be :<p><pre><code> .01123595505617977528
</code></pre>
but the math does work of course, it converges :<p><pre><code> >>> .01 + .001 + .0002 + .00003 + .000005 + .0000008 + .00000013 + .000000021 +
.0000000034 + .00000000055 + .000000000089 + .0000000000144 + .00000000000233 +
.000000000000377
0.011235955056107002</code></pre>