1. ## Fibonacci?

1. Write down 5 single digits numbers in a row horizontally
2. Pick another single digit and repeat it underneath ( 2 picked below)
3. Add vertically and write down the last digit
4. Then repeat by adding previous two rows..

Example

1 4 5 4 2 3
2 2 2 2 2 2
3 6 7 6 4 5
5 8 9 8 6 7
8 4 6 4 0 2

Something strange happens on the 17th row , regardless of the starting points?
Any ideas what's going on?

2. ## Re: Fibonacci?

Have you tried going out to the 17th row to see what this "something strange" is?

3. ## Re: Fibonacci?

Yes I have. But I didn't want to spoil it for others to try!
Still can't quite see why this emerges? I mean why on the 17th every time?

4. ## Re: Fibonacci?

Happens at 17th row (15 rows after initial 2 rows), then every 15 rows.
For your given example:
17: all 4'4
32: all 8's
47: all 6's
62: all 2'2
77: all 4's ...above repeats

5. ## Re: Fibonacci?

Without having time to work out the problem right now (at work), I would suspect that this is an emergent property of addition mod 10.

To begin, I would write the first row as a,b,c,d,e (none need be distinct)
Next I would choose "a" to be the entries for the second row. (without loss of generality, as I can sort elements such that the one element I choose to be the second row will be a)
After that, I would find out what the entries would be on the 17th row as simple addition of the previous rows (not by writing only the last digit). It would only be after I get the full algebraic sum for each entry on the 17th row that I would apply modulo 10. Modulo is nice to work with because of equivalence that, for example, 10a = 0 mod10

EDIT://
Even easier, express the sums in row form first.
ie. $r_1$ is row 1, $r_2$ is row 2, with $r_i$ being the sum of rows $r_{i-1}, r_{i-2}$. Expressing $r_{17}$ as a sum of $r_1$ and $r_2$ should make finding the algebraic sum for each entry in $r_{17}$ a lot easier, as a final step before applying modulo 10.

6. ## Re: Fibonacci?

Interesting: if 2nd row is all 0's, then same cycle,
however result always all 0's.

Also pattern is identical regardless of number of
digits in first 2 rows; like:
1,2,3,4,5,6,7,8,9
2,2,2,2,2,2,2,2,2

And of course 1st row need not be all 1 digits:
5,87,45,68,32,548 (as example) = same results

7. ## Re: Fibonacci?

Solution (don't open if you want to figure it out on your own!)

Spoiler:

First, express $r_{17}$ as the sum of $r_1$ and $r_2$

$r_{17} = r_{16} + r_{15}$
$r_{17} = 2r_{15} + r_{14}$
$r_{17} = 3r_{14} + 2r_{13}$
$r_{17} = 5r_{13} + 3r_{12}$
$r_{17} = 8r_{12} + 5r_{11}$
.
.
.
$r_{17} = 887r_{2} + 510r_{1}$

From this, I can express $r_{17}$ with the smallest positive integer coefficients modulo 10.
ie.
$r_{17} = 887r_{2} + 510r_{1} \equiv 7r_{2} + 0r_{1} = 7r_{2} \mod 10$

Notes:
So, as an example, if I know that my 2nd row contains elements "2", then I know my entries in my result row are $r_{17} = 7*2 \mod 10 = 14 \mod 10 \equiv 4 \mod 10.$ (It'll be a row of 4's.)

When adding rows, the process can be simpler using coordinate forms instead of writing out r_i's for each row, and if I 'drop the 10's' along the way. For example, $13r_{11}$ has the same remainder as $3r_{11}$ modulo 10. I don't show this here, but it was a good verification step to ensure my last row was correct.

To be aware of the cycle that exists, simply set rows $q_{1},q_{2} = r_{16},r_{17}$

EDIT:// Expressed in a more elementary way that doesn't use modulo but means the same thing:

$r_{17} = 887r_{2} + 510r_{1} = 880r_{2} + 510r_{1} + 7r_{2} = 10 * (88r_{2} + 51r_{1}) + 7r_{2}.$ The last digits of $7r_{2}$ is what we'd see as a result of the operation described in the problem, for each entry.

8. ## Re: Fibonacci?

I think there's a typo here - I believe this should actually be: $r_{17} = 987r_2 + 610r_1$, based on the fact that the 15th term of the Fibonacci series is 610 and the 16th term is 987.

9. ## Re: Fibonacci?

Do'h! You're right. Thanks.