Spreadsheet Explorations
The
Problem:
I am to place four numbers in the first row on a spreadsheet
as follows. A B C D
For each successive row replace the entries by the absolute
value of the difference of the entry just above and the entry just to the right
in the previous row. In the fourth position use the absolute value of the
difference of the fourth and the first
|A-B| |B-C| |C-D| |D-A|
________________________________________________________________________
Ok, LetÕs start by
substituting arbitrary values in for A, B, C, and D and see what happens. LetÕs
try: 1, 2, 3, 4
|
0 |
1 |
2 |
3 |
4 |
|
1 |
1 |
1 |
1 |
3 |
|
2 |
0 |
0 |
2 |
2 |
|
3 |
0 |
2 |
0 |
2 |
|
4 |
2 |
2 |
2 |
2 |
|
5 |
0 |
0 |
0 |
0 |
The process ends in zeros
after 4 rows. Not bad for the first try. Now letÕs continue, hoping to find
some sort of pattern.
LetÕs try alternating between
zeros: 11, 0, 29, 0.
|
0 |
11 |
0 |
29 |
0 |
|
1 |
11 |
29 |
29 |
11 |
|
2 |
18 |
0 |
18 |
0 |
|
3 |
18 |
18 |
18 |
18 |
|
4 |
0 |
0 |
0 |
0 |
Wow, that obviously is not
the pattern. It now ends in zeros after only 3 rows. LetÕs try some larger
numbers with zeros at the end : 10000, 14, 0, and 0 .
|
0 |
10000 |
14 |
0 |
0 |
|
1 |
9986 |
14 |
0 |
10000 |
|
2 |
9972 |
14 |
10000 |
14 |
|
3 |
9958 |
9986 |
9986 |
9958 |
|
4 |
28 |
0 |
28 |
0 |
|
5 |
28 |
28 |
28 |
28 |
|
6 |
0 |
0 |
0 |
0 |
It now ends in zeros after 5
rows. It does seem to always end in zero.
LetÕs try decimals: 4.2, 5.3, 6.4, 7.5.
|
0 |
4.2 |
5.30 |
6.4 |
7.5 |
|
1 |
1.1 |
1.1 |
1.1 |
3.3 |
|
2 |
8.88E-16 |
8.88E-16 |
2.2 |
2.2 |
|
3 |
0 |
2.2 |
0 |
2.2 |
|
4 |
2.2 |
2.2 |
2.2 |
2.2 |
|
5 |
0 |
0 |
0 |
0 |
Again, the process ends in zeros
after 4 rows.
LetÕs try numbers that have
some sort of relationship. LetÕs try: 3, 5, 7, 9.
|
0 |
3 |
5 |
7 |
9 |
|
1 |
2 |
2 |
2 |
6 |
|
2 |
0 |
0 |
4 |
4 |
|
3 |
0 |
4 |
0 |
4 |
|
4 |
4 |
4 |
4 |
4 |
|
5 |
0 |
0 |
0 |
0 |
Well, that did not improve
things. I am now back to ending in zeros after 4 rows.
LetÕs try: 2, 4, 8, 16.
|
0 |
2 |
4 |
8 |
16 |
|
1 |
2 |
4 |
8 |
14 |
|
2 |
2 |
4 |
6 |
12 |
|
3 |
2 |
2 |
6 |
10 |
|
4 |
0 |
4 |
4 |
8 |
|
5 |
4 |
0 |
4 |
8 |
|
6 |
4 |
4 |
4 |
4 |
|
7 |
0 |
0 |
0 |
0 |
Well, it looks like IÕm
making progress and that some sort of pattern is the key. The process ends in
zeros after 6 rows. It looks like the pattern could be multiplying by two, or
successive powers lets try multiplying by two first.
LetÕs try 6, 12, 24, 48.
|
0 |
6.00E+00 |
1.20E+01 |
2.40E+01 |
4.80E+01 |
|
1 |
6 |
12 |
24 |
42 |
|
2 |
6 |
12 |
18 |
36 |
|
3 |
6 |
6 |
18 |
30 |
|
4 |
0 |
12 |
12 |
24 |
|
5 |
12 |
0 |
12 |
24 |
|
6 |
12 |
12 |
12 |
12 |
|
7 |
0 |
0 |
0 |
0 |
Again, the process ends in
zeros after 6 rows.
LetÕs try decimals: 0.625,
1.25, 2.5, 5
|
0 |
.625 |
1.25 |
2.50 |
5.00 |
|
1 |
0.625 |
1.25 |
2.5 |
4.375 |
|
2 |
0.625 |
1.25 |
1.875 |
3.75 |
|
3 |
0.625 |
0.625 |
1.875 |
3.125 |
|
4 |
0 |
1.25 |
1.25 |
2.5 |
|
5 |
1.25 |
0 |
1.25 |
2.5 |
|
6 |
1.25 |
1.25 |
1.25 |
1.25 |
Again, the process ends in
zeros after 6 rows.
LetÕs try the successive
exponents, which means that B=A^2, C=A^3, D=A^4.
I have created a formula in
excel to perform the exponential calculations, I only need to enter a value for
A. LetÕs try A=7.
|
0 |
7 |
49 |
343 |
2401 |
|
1 |
42 |
294 |
2058 |
2394 |
|
2 |
252 |
1764 |
336 |
2352 |
|
3 |
1512 |
1428 |
2016 |
2100 |
|
4 |
84 |
588 |
84 |
588 |
|
5 |
504 |
504 |
504 |
504 |
Well, my previous conjecture
seems to be a little flawed.
LetÕs try a decimal, A = 3.1
|
0 |
3.10E+00 |
9.61E+00 |
2.98E+01 |
9.24E+01 |
|
1 |
6.51 |
20.181 |
62.5611 |
89.2521 |
|
2 |
13.671 |
42.3801 |
26.691 |
82.7421 |
|
3 |
28.7091 |
15.6891 |
56.0511 |
69.0711 |
|
4 |
13.02 |
40.362 |
13.02 |
40.362 |
|
5 |
27.342 |
27.342 |
27.342 |
27.342 |
|
6 |
3.55E-15 |
0 |
3.55E-15 |
0 |
|
7 |
3.55E-15 |
3.55E-15 |
3.55E-15 |
3.55E-15 |
|
8 |
0 |
0 |
0 |
0 |
Well, at least I am making
progress. Now I have 7 rows before I reach zero. It appears that decimals are
the key. But where do I start? Is there a particular pattern? LetÕs try another
decimal, A= 1.62.
|
0 |
1.62 |
2.62 |
4.25 |
6.89 |
|
1 |
1.00468 |
1.627707 |
2.637089 |
5.269476 |
|
2 |
0.623027 |
1.009382 |
2.632387 |
4.264796 |
|
3 |
0.386355 |
1.623005 |
1.632409 |
3.641769 |
|
4 |
1.236651 |
0.009404 |
2.00936 |
3.255414 |
|
5 |
1.227247 |
1.999956 |
1.246054 |
2.018764 |
|
6 |
0.772709 |
0.753902 |
0.772709 |
0.791517 |
|
7 |
0.018808 |
0.018808 |
0.018808 |
0.018808 |
|
8 |
2.22E-16 |
2.22E-16 |
2.22E-16 |
2.22E-16 |
|
9 |
0 |
0 |
0 |
0 |
Wow, IÕm not up to 8 rows
before zeros, letÕs try A=1.85.
|
0 |
1.85E+00 |
3.43E+00 |
6.36E+00 |
1.18E+01 |
|
1 |
1.578892 |
2.924685 |
5.417586 |
9.921163 |
|
2 |
1.345793 |
2.492901 |
4.503577 |
8.342271 |
|
3 |
1.147107 |
2.010677 |
3.838694 |
6.996478 |
|
4 |
0.863569 |
1.828017 |
3.157784 |
5.84937 |
|
5 |
0.964447 |
1.329768 |
2.691586 |
4.985801 |
|
6 |
0.365321 |
1.361818 |
2.294215 |
4.021354 |
|
7 |
0.996498 |
0.932396 |
1.727139 |
3.656033 |
|
8 |
0.064101 |
0.794742 |
1.928894 |
2.659535 |
|
9 |
0.730641 |
1.134152 |
0.730641 |
2.595434 |
|
10 |
0.403511 |
0.403511 |
1.864793 |
1.864793 |
|
11 |
0 |
1.461282 |
0 |
1.461282 |
|
12 |
1.461282 |
1.461282 |
1.461282 |
1.461282 |
|
13 |
0 |
0 |
0 |
0 |
|
14 |
0 |
0 |
0 |
0 |
Hooray, I now have 12 rows
before I reach zeros. LetÕs keep it going. LetÕs try 1.84.
|
0 |
1.84E+00 |
3.39E+00 |
6.23E+00 |
1.15E+01 |
|
1 |
1.5456 |
2.843904 |
5.232783 |
9.622287 |
|
2 |
1.298304 |
2.388879 |
4.389504 |
8.076687 |
|
3 |
1.090575 |
2.000625 |
3.687183 |
6.778383 |
|
4 |
0.910049 |
1.686559 |
3.0912 |
5.687808 |
|
5 |
0.776509 |
1.404641 |
2.596608 |
4.777759 |
|
6 |
0.628132 |
1.191967 |
2.181151 |
4.001249 |
|
7 |
0.563835 |
0.989184 |
1.820099 |
3.373117 |
|
8 |
0.425349 |
0.830915 |
1.553019 |
2.809283 |
|
9 |
0.405565 |
0.722104 |
1.256264 |
2.383933 |
|
10 |
0.316539 |
0.534159 |
1.12767 |
1.978368 |
|
11 |
0.21762 |
0.59351 |
0.850698 |
1.661829 |
|
12 |
0.37589 |
0.257188 |
0.811131 |
1.444209 |
|
13 |
0.118702 |
0.553943 |
0.633078 |
1.068319 |
|
14 |
0.435241 |
0.079135 |
0.435241 |
0.949617 |
|
15 |
0.356106 |
0.356106 |
0.514376 |
0.514376 |
|
16 |
0 |
0.158269 |
0 |
0.158269 |
|
17 |
0.158269 |
0.158269 |
0.158269 |
0.158269 |
|
18 |
0 |
0 |
0 |
0 |
Can you believe it? I now have a whopping 17 rows before it
reaches zeros.
LetÕs try 1.86
|
0 |
1.86 |
3.46 |
6.43 |
1.20E+01 |
|
1 |
1.5996 |
2.975256 |
5.533976 |
10.10883 |
|
2 |
1.375656 |
2.55872 |
4.574856 |
8.509232 |
|
3 |
1.183064 |
2.016136 |
3.934376 |
7.133576 |
|
4 |
0.833072 |
1.91824 |
3.1992 |
5.950512 |
|
5 |
1.085169 |
1.28096 |
2.751312 |
5.11744 |
|
6 |
0.195791 |
1.470352 |
2.366128 |
4.032272 |
|
7 |
1.274561 |
0.895776 |
1.666143 |
3.836481 |
|
8 |
0.378785 |
0.770367 |
2.170337 |
2.561919 |
|
9 |
0.391582 |
1.39997 |
0.391582 |
2.183134 |
|
10 |
1.008388 |
1.008388 |
1.791552 |
1.791552 |
|
11 |
2.22E-16 |
0.783164 |
2.22E-16 |
0.783164 |
|
12 |
0.783164 |
0.783164 |
0.783164 |
0.783164 |
|
13 |
0 |
0 |
0 |
0 |
Interesting, it decreases to
12 rows before zeros.
LetÕs try going back down to
A=1.83
|
0 |
1.83E+00 |
3.35E+00 |
6.13E+00 |
1.12E+01 |
|
1 |
1.5189 |
2.779587 |
5.086644 |
9.385131 |
|
2 |
1.260687 |
2.307057 |
4.298487 |
7.866231 |
|
3 |
1.04637 |
1.99143 |
3.567744 |
6.605544 |
|
4 |
0.94506 |
1.576314 |
3.0378 |
5.559174 |
|
5 |
0.631255 |
1.461486 |
2.521374 |
4.614114 |
|
6 |
0.830231 |
1.059888 |
2.09274 |
3.98286 |
|
7 |
0.229658 |
1.032852 |
1.890119 |
3.152629 |
|
8 |
0.803194 |
0.857267 |
1.26251 |
2.922971 |
|
9 |
0.054073 |
0.405243 |
1.660461 |
2.119777 |
|
10 |
0.35117 |
1.255219 |
0.459315 |
2.065704 |
|
11 |
0.904049 |
0.795904 |
1.606389 |
1.714534 |
|
12 |
0.108146 |
0.810485 |
0.108146 |
0.810485 |
|
13 |
0.702339 |
0.702339 |
0.702339 |
0.702339 |
|
14 |
0 |
0 |
0 |
0 |
Now, I have 13 rows before it
reaches zero. It appears that it reaches the maximum number of rows when A is
near 1.84. The maximum number of
rows that I discovered before a zero row is generated was 17. But, I think that
this number is much larger. In fact, I think that this number may approach
infinity.