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# Next terms of two-dimensional sequences

## Question

Suppose I conjure a function f(x,y).

I give you a matrix of f values for x = {0 1 2 3 4 5 6} and y = {1 2 3 4 5 6}.
It looks like this:

x\y| 1 2 3 4 5 6
---+------------------
0 | 3 6 9 12 15 18
1 | 1 4 7 10 13 16
2 | -1 2 5 8 11 14
3 | -3 0 3 6 9 12
4 | -5 -2 1 4 7 10
5 | -7 -4 -1 2 5 8
6 | -9 -6 -3 0 3 6

From the first row you can see that f(0,y) = 3y
From the first column you can see that f(x,1) = 3-2x
From the second column it's clear that f(x,2) = 6-2x, and so forth.

At some point it's evident, and the other values so confirm, that f(x,y) = 3y-2x.

OK that's the idea. A 2-diminsional "what's the next term in this sequence" puzzle.
Only we don't ask the next term, we ask for the simplest function that generates the given terms.

See if this one is too easy (there are plenty of clues in the table):

Same range of values for x and y as above.

x\y| 1 2 3 4 5 6
---+--------------------------
0 | 1 1 1 1 1 1
1 | 0 1 2 3 4 5
2 | -1 0 1 0 -7 -28
3 | -2 -1 0 -17 -118 -513
4 | -3 0 17 0 -399 -2800
5 | -4 7 118 399 0 -7849
6 | -5 28 513 2800 7849 0

What is f(x,y)?

If you like this puzzle type, then let the solver make one, and we'll keep the thread going.

Aligning a table like this is easy: just use Courier font.

## Recommended Posts

• 0

Very cool puzzle, I'd love to see more (some harder, but some at this level as well).

The first row really gives the first term away--only y^x could produce 1s across the board at x=0 and non-trivial values after. Then the fact that it's and odd function (or whatever you'd call a multivariable function with rotational symmetry over y=x) gives

f(x,y)=y^x-x^y

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I'll add my own now:

x\y 1 2 3 4 5 6

0 -2 -2 -2 -2 -2 -2

1 -1 0 1 -2 -1 0

2 0 -2 0 -2 0 -2

3 1 0 -1 -2 1 0

4 -2 -2 -2 -2 -2 -2

5 -1 0 1 -2 -1 0

6 0 -2 0 -2 0 -2

Edited by jamieg
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• 0

f(x,y)=(xy mod 4)-2

x\y 1 2 3 4 5 6

0 -2 -2 -2 -2 -2

1 -1 0 1 -2 -1 0

2 0 -2 0 -2 0 -2

3 1 0 -1 -2 1 0

4 -2 -2 -2 -2 -2 -2

5 -1 0 1 -2 -1 0

6 0 -2 0 -2 0 -2

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sorry apparently pressing space 6 times posts it 6 times.

1 2 3 4 5 6

0 0 0 0 0 0 0

1 2 6 12 20 30 42

2 6 16 30 48 70 96

3 12 30 54 84 120 162

4 20 48 84 128 180 240

5 30 70 120 180 250 330

6 42 96 162 240 330 432

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• 0

f(x,y) = xy(x+y)

x\y| 1 2 3 4 5 6
---+---------------------
0 | 1 0 -3 -8 -15 -24
1 | 3 3 1 -3 -9 -17
2 | 7 8 7 4 -1 -8
3 | 13 15 15 13 9 3
4 | 21 24 25 24 21 16
5 | 31 35 37 37 35 31
6 | 43 48 51 52 51 48

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• 0

Nicely done on mine, googon97. Bonanova, I've solved yours:

Next:

f(x,y)=(x+1)

2 - (y-1)2 + x(y-2)

x\y| 1 2 3 4 5 6
---+---------------------
0 | 4 10 28 82 244 730
1 | 1 7 25 79 241 727
2 | 7 13 31 85 247 733
3 | -5 1 19 73 235 721
4 | 19 25 43 97 259 745
5 |-29 -23 -5 49 211 697
6 | 67 73 91 145 307 793

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