4
$\begingroup$

I think I stumbled on something and I just wanted to ask why this pattern occurs. I’m not a mathematician

I put the sequence in excel (if anyone wants it here’s a link https://oeis.org/A000045/b000045.txt ) and from 20th number onwards a pattern emerges with the largest digit.

6765 so the 6

10946 the 1

17711 the 1

28657 the 2

46368 the 4

75025 the 7

121393 the 1

196418 the 1

317811 the 3

514229 th 5

832040 the 8

1346269 the 1

2178309 the 2

3524578 the 3

5702887 the 5

9227465 the 9

14930352 the 1

24157817 the 2

39088169 the 3

The number 6 then begins again on the 39th number,

, and the 40th begins with 1

, and the 41st begins with 1

, and the 42nd begins with 2

, and the 43rd begins with 4

, and the 44th begins with 7

, and the 45th begins with 1

so again the 6112471.... begins to appear.

I thought I'd write out everytime I see this 6 in the sequence begins again.

List of line where the 6 sequence repeats. 20 (as shown with first example), 39, 63, 87, 106, 130, 154, 173, 197, 221, 240, 264, 288, 307, 331, 355, 374, 398, 422, 441, 465, 484, 508 ,532 , 551, 575, 599, 618, 642, 666, 685, 709, 733, 752, 776, 800, 819, 843, 867, 886, 910, 934, 953, 977, 996, 1020, 1044, 1063, 1087, 1111, 1130, 1154, 1178, 1197, 1221, 1245, 1264, 1288, 1312, 1331, 1355, 1379, 1398, 1422, 1441, 1465

As you will see below this pattern repeats (0r the 6 1 1 2 4 7 1) begins again every 19, 24, 24 times respectively 6 times.

39 - 20 = 19

63 - 39 = 24

87 - 63 = 24

106 - 87 = 19

130 - 106 = 24

154 - 130 = 24

173 - 154 = 19

197 - 173 = 24

221 - 197 = 24

240 - 221 = 19

264 - 240 = 24

288 - 264 = 24

307 - 288 = 19

331 - 307 = 24

355 - 331 = 24

374 - 355 = 19

398 - 374 = 24

422 - 398 = 24

441 - 422 = 19

465 - 441 = 24

484 - 465 = 19

508 - 484 = 24

532 - 508 = 24

551 - 532 = 19

575 - 551 = 24

599 - 575 = 24

618 - 599 = 19

642 - 618 = 24

666 - 642 = 24

685 - 666 = 19

709 - 685 = 24

733 - 709 = 24

752 - 733 = 19

776 - 752 = 24

800 - 776 = 24

819 - 800 = 19

843 - 819 = 24

867 - 843 = 24

886 - 867 = 19

910 - 886 = 24

934 - 910 = 24

953 - 934 = 19

977 - 953 = 24

996 - 977 = 19

1020 - 996 = 24

1044 - 1020 = 24

1063 - 1044 = 19

1087 - 1063 = 24

1111 - 1087 = 24

1130 - 1111 = 19

1154 - 1130 = 24

1178 - 1154 = 24

1197 - 1178 = 19

1221 - 1197 = 24

1245 - 1221 = 24

1264 - 1245 = 19

1288 - 1264 = 24

1312 - 1288 = 24

1331 - 1312 = 19

1355 - 1331 = 24

1379 - 1355 = 24

1398 - 1379 = 19

1422 - 1398 = 24

1441 - 1422 = 19

1465 - 1441 = 24 (excel reached limit after this, could not continue) The 6 sequence repeats 19, 24, 24 interval 6 times respectively for 19 cosecutive times.

Does anyone have an explanation?

Also I know numbers shouldn’t be looked at from it’s largest number alone but as a whole, but it’s just something I noticed

Thanks in advance

$\endgroup$
6
  • 4
    $\begingroup$ The Fibonacci series, like any similarly structured additive series, approximates in the limit a geometric series with a ratio of $\varphi = \frac{1+\sqrt5}{2} \approx 1.618$. So if you start with any element that begins with a $6$, the next one will be about $1.618$ times larger and therefore be very likely to begin with $1$ (the only other possibility is a $9$). And if that second element begins with a $1$, it will begin with either $10$ or $11$, and if you multiply that by $1.618$, you'll get something that begins $16$, $17$, or $18$. And so on. $\endgroup$
    – Brian Tung
    Oct 20 '18 at 1:06
  • 2
    $\begingroup$ The recurrence at intervals of $19$ or $24$ is related to the fact that $\varphi^{19}$ and $\varphi^{24}$ are reasonably close to powers of $10$: the former is a bit lower than $10000$, and the latter is a bit higher than $100000$. So they trade off, first raising the element (or rather its mantissa) starting with $6$, then lowering it, one after the other. $\endgroup$
    – Brian Tung
    Oct 20 '18 at 1:07
  • $\begingroup$ OP, you can use pastebin next time. But not a bad question for your first one on MSE. Welcome! $\endgroup$ Oct 20 '18 at 1:12
  • $\begingroup$ Reasonably close but not exact. So, if you go far enough, the pattern will break. It's too late here now to calculate when that will be. $\endgroup$
    – badjohn
    Oct 20 '18 at 1:13
  • $\begingroup$ @BrianTung Those comments look like the basis of an answer to me. $\endgroup$
    – David K
    Oct 20 '18 at 13:23
3
$\begingroup$

The Fibonacci sequence is the canonical example of a general class of (usually integer) sequences in which each element is the sum of the previous two elements. The Fibonacci series is the one that begins $(0,) 1, 1$. The Lucas series begins $1, 3$, and so on.

It is an interesting fact (not too difficult to prove) that in all such sequences, the ratio between consecutive elements approaches a limiting value—actually, the same limiting value for all such sequences. That limiting value is

$$ \lim_{n\to\infty} \frac{F_{n+1}}{F_n} = \varphi = \frac{1+\sqrt{5}}{2} \approx 1.618 $$

where $F_n$ is the $n$th Fibonacci number (but again, this would work for Lucas numbers too). This convergence happens fairly swiftly:

$$ \begin{array}{|c|c|c|c|} \hline n & F_n & F_{n+1} & F_{n+1}/F_n \\ \hline 2 & 1 & 2 & 2 \\ \hline 3 & 2 & 3 & 1.5 \\ \hline 4 & 3 & 5 & 1.667 \\ \hline 5 & 5 & 8 & 1.6 \\ \hline 6 & 8 & 13 & 1.625 \\ \hline 7 & 13 & 21 & 1.615 \\ \hline 8 & 21 & 34 & 1.619 \\ \hline 9 & 34 & 55 & 1.618 \\ \hline 10 & 55 & 89 & 1.618 \\ \hline \end{array} $$

So before we've left double digits, we've already approximated $\varphi$ to four significant digits. What significance does this have for the present question? The leading digit is a coarse representation of the mantissa or significand (the former is an ambiguous term) of a number's value. The significand, as the name suggests, gives the significant digits of a value. For instance, the significand of a number in scientific notation—say, $6.022 \times 10^{23}$—is that $6.022$ (or perhaps $6022$).

Well, if we multiply a number that begins with $6$ by $\varphi \approx 1.618$, the result will be a value whose significand is at least

$$ 6\varphi \approx 9.708 $$

but less than

$$ 7\varphi \approx 11.326 $$

If we were to imagine that our starting value is evenly distributed across all numbers beginning with $6$ (in any reasonable sense of "evenly distributed"), about $80$ percent of the results will start with a $1$.

So most of the time that you have a Fibonacci number that begins with $6$, the next value will begin with $1$. What's more, $100$ percent of the time, the next next number will begin with $1$ also, because that number's significand must be at least

$$ 6\varphi^2 \approx 15.708 $$

but less than

$$ 7\varphi^2 \approx 18.326 $$

And $100$ percent of the time, the next number will begin with $2$, and so on. In fact, the full sequence you cited—$6, 1, 1, 2, 4, 7, 1$—will occur starting with any Fibonacci number whose significand is at least $70/\varphi^5 \approx 6.312$ but less than $7$. So it's not surprising that once you see a Fibonacci number beginning with $6$, you'll very frequently see that sequence of initial digits.


Now, why should this pattern repeat after intervals of length $19$ and $24$? We noticed that in order to observe the sequence of initial digits $6, 1, 1, 2, 4, 7, 1$, we need only a Fibonacci number whose significand is greater than about $6.312$ and less than $7$. If you have such a number—say, $F_{20} = 6765$, when would you expect the next such number to show up in the Fibonacci sequence?

We know from the foregoing discussion that $F_{21}$ will be very nearly $\varphi F_{20}$, that $F_{22}$ will be very nearly $\varphi^2 F_{20}$, and in general that $F_{20+k}$ will be very nearly $\varphi^k F_{20}$. So the next Fibonacci number to begin with a significand in our target range will involve an integer power of $\varphi$ that is also close to an integer power of $10$. And sure enough, $\varphi^{19} \approx 9349$, and

$$ \varphi^{19} F_{20} \approx F_{39} = 63245986 $$

whose significand is again (just barely) within our range.

You'll see that we started with a number $F_{20}$ whose significand was smack dab in the middle of our target range, and ended up with a number $F_{39}$ whose significand is just barely above our lower limit. That's because $\varphi^{19}$ isn't quite a power of $10$, but somewhat lower than it. So occasionally, we'll need a power of $\varphi$ that is somewhat above a power of $10$, to nudge that significand back up. And $\varphi^{24} \approx 103682$ obliges quite nicely, so that shows up as an interval too. What's more, because $103682$ is closer to its power of $10$ than $9349$ is to its power of $10$, we would expect $24$ to show up more often than $19$—about twice as often, because $10000-9349 = 651$ and $103682-100000 = 3682$. And that's what you in fact found.


Incidentally, if you actually calculate $\varphi^{19}$ and $\varphi^{24}$, you'll see that they are very close to $9349$ and $103682$; the differences are about $0.0001$ and $0.00001$, respectively. This isn't a coincidence! But that's a question for another time.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.