Nice Sequence: CR4 Challenge (07/28/09)

GA,

Put another way, if the rows are labeled i and the columns j, N(i,j) = ABS( N(1,j-1)+N(2,j-1)+N(3,j-1) -2*N(i,j-1)). Adding a few more columns I get:

Thanks,

Jim

Using Fz solution, if two or three numbers in the first column are equal, the rows following them are the same. If there are three different numbers in the first column, the row with max value and min value switch for successive columns and the row with the initial middle value always has the middle value.

I think the numbers are 7, 13 and 9 from top to bottom

thanks

4,13,6

If each number is designated A(Row, column)

A(0,0) A(1,0) A(2,0)

A(0,1) A(1,1) A(2,1)

A(0,2) A(1,2) A(2,2) etc

Then A((n+1), 0) = Abs( A(n,1) + A(n,2) - A(n,0) )

A(((n+1, 1) = Abs( A(n,0) + A(n,2) - A(n,1) )

A((n+1),2) = Abs( A(n,0) + A(n,1) - A(n,2) )

Giving answers 3, 17, 5

3, 17, 5

To put it yet another way:

Each cell is the difference of the sum of the cells in the other rows in the previous column and the cell in its own row in the previous column.

Let the first 3 vertical numbers be a,b and c

Each column of the sequence is based on the previous one as shown:

a..... |b+c-a| The next to last column is 11

b..... |a+c-b| .....................................1

c..... |a+b-c| .................................... 7

Hence the final column is

|1+7-11| = 3

|11+7-1| = 17

|11+1-7| = 5

It interesting that several other correspondents posted the same result but with somewht different techniques.

If you use the absolute values for all the numbers then it would be 3, 17, 5. which can be shown by mathematical formula.

We seem to have consensus on 3, 17,5.

In case any other sad cases want extensions to the problem:
Suppose now that there is a column to the left.
There are two possible non-negative columns that could occupy this position.
What are they?

Plus a few supplementaries:

a) What is the minimum number of columns that you would need to change from their existing values for this additional column to use positive integers only, and what would the values in the left-most column be?
b) Are there any other positive integer solutions that retain the column 11, 1, 7?
b) What is the furthest you could go to the left while retaining column 11, 1, 7?

There are two possible non-negative columns that could occupy this position.

Oh no there aren't:

For any three natural (reasonable assumption?) numbers there are only four possibilities:

three odds

three evens

two odds and one even

two evens and one odd

In the next column these always become

three odds

three evens

three evens

three odds

There is no combination which becomes two evens and an odd.

I'm anticipating looking very silly here so here's my blushes in advance:.

Read on: that should clarify that when I said "non-negative" I didn't mean integers...
That makes it a warm-up for the next stage, where you need to work backwards from further right.

OK now I've got 4½, 1, 1½ AND 4½, 3, 3½

You only need to change the first column to 1, 5, 3 and the zeroth column is 4, 1, 2 OR 4, 2, 3

I'll think about b.) some more and c.) if I get any time.

the sequence from top to bottom : 3 ,, 17 ,, 5

4, 13, 6

As far as I can determine there are at least 2 valid sequences (in the sense that they follow a set of defined rules which give a number pattern that is repeated at least once) which give different results. The first gives the answer of 4, 13, 6 as first provided by dac1267 although the explanation was a little brief so here is my interpretation. Lets call the three boxes in each column top, middle and bottom and number the columns sequentially from left to right starting at 1. The sequence works as follows:

Every even column has 1 in the middle box, 2 + [the value of the middle box in previous column] in the top box, and [the value of the middle box in the previous column] – 2 in the bottom box. Every odd column has 1 + [value of top box in previous odd column] in top box, 2 + [value of top box in previous column] in middle box, and [value of bottom box in previous odd column] + 1 in bottom box.

The second valid sequence gives 3, 17, 5 first provided by Vi Pham but explained fully by Physicist?.

Both of these sequences are valid although the Vi Pham one is repeated 3 times while the dac1267 one is only repeated once.

Either I still fail to follow the DAC1267 sequence, or several of the elements in the matrix are neither involved in the sequences nor predictable from it. If everything is involved/predictable, could you provide more explanation.

Thanks

Fyz

I apologize for the brief and unclear explanation - I have provided the mathematical justification below. Since this solution uses two alternating patterns with differring elements for each row, I think the other solution presented is better because it is more universal. However, this solution is completely independent of forward or reverse column values.

For odd columns (i=1, 3, 5, ...):

A_1i = (i+3)/2

A_2i = (2i+3)

A_3i = (i+7)/2

For even columns (i=2, 4, 6, ...):

A_1i = (2i+3)

A_2i = 1

A_3i = (2i-1)

Thus, for column 5:

A₁₅ = ((5)+3)/2 = 4

A₂₅ = (2(5)+3) = 13

A₃₅ = ((5)+7)/2 = 6

GA,

I suspect there may be other logical patterns. This one certainly is "nice" because the math required is so simple. Good job.

Jim

Thanks, I see what you have. And the picture is a great explanation.
But it gives us three independent sequences of not-closely-related structure (I know it doesn't affect the next term, but the sequence starting (1,1...) could just as well be the start of a Fibonacci sequence) .

This solution would have convinced me (as a great answer to what would become for me a slightly unsatisfactory challenge) if one row above or below had also been left blank...

Actually in my original answer, I saw the 5, 7, 9, 11, 13 & 2, 3, 4 patterns in the top two rows and the 5, 3, 9, 7, 13 (-2, +6, -2, +6) & 4, 5, 6 patterns in the bottom two rows (which was poorly explained), but when I put it in Excel and started coloring the boxes I saw that the patterns actually worked nicely across all three rows.

Thanks for the GA's

Also, this solution could be considered as two independent sequences instead of three if the 1 in the center square is included with the 2, 3, 4, 5 sequence as the increment value (2+1=3, 3+1=4, etc.). It just makes the math a little uglier.

That is horrid, at least to my way of thinking; the problem is that there is no reason for the numbers in those squares to relate to the zigzag groupss rather than to the clockwise groups.

Another way to look at this: if those squares were left blank you could still produce your original sequences - and have no way of knowing what to place in those specific positions.

I agree. That's why I like the simplicity and generality of the other solution. In that case, the 1's are not arbitrary so it seems a better fit.

It will be interesting to see what the posted solution is.

It would be nice if the solution related to this, but so far I've been unable to make any connection.

btw - very good solutions from yourself and others.

I think you are on to something, as the challenge title has hinted. The column sums for the original four columns (11, 11, 17, 18) are all "nice", but the column sum for my solution (4+13+6=23) is not. The column sum for the other solution (3+17+5=25) is "nice", but it basically stands on its own without a requirement for being "nice". Unless perhaps it defines the first column as the lowest combination of three integers that produce a "nice" number which also works with the sequence.

Perhaps the title used the word 'nice' to ensure that only one answer was possible.

I may be misreading, but my interpretation would make this misleading

A substitute first column that gives the same sequence for the remainder is (1,5,3). This gives a sum (9) that is a perfect square and therefore a nice number - and smaller than 11. That substitution would also allow a preceding column (4,2,3) to be appended - also summing to 9. (and of course 9 is the smallest nice number that is the sum of 3 different positive integers...).

Hi Phys,

I have put it into a table to try and explain it better. The explanation by dac1267 in post 22 is different but equivalent.

Just an observation about the top row, probably doesn't mean much relating to the entire problem:

2, 7, 3, and 11 are primes. Starting with 2 and skipping two primes gives 7. Counting back two primes from 7 gives 3. Skipping two primes from 3 gives 11.

Of course this pattern doesn't apply to the other rows, but I can see an up and down "movement."

Irregular - yes, that is to be expected, as the sequence is maintained positive by using moduli.

But the proofs for triplets being co-prime and the sequence never producing equal pairs in a column are refreshingly straightforward - you only need use a single step forward or back to prove each successive feature.

You will need to extend divisibility by two first. That uses my original form for the recurrence relation.

Then I would look at the constraints for not generating equality - that is easiest done using an "a+b-c" format.

The uniformity of the common factor (in this case co-prime) is easiest addressed using the backwards formula. (I would , however, leave whether even sequences can have common factors with increasing powers of 2 - at least for the time-being)

This whole question is a lot more interesting than I thought on first glance. I shall wait and see how informative the official answer is before delving further, but it does tempt me to explore it more.

Why does the question emphasise the sequence "from top to bottom"? It implies that the correct answer will not be valid otherwise.

I think the purpose is only to allow ensure that answers from different contributors are expressed in compatible formats - e.g. either of (3,17,5) or 3, 17 and 5 would satisfy the description.

I forgot to post a "niceness" requirement for a family of sequences: that sequences that start with different members do not coalesce. For this we need only to observe that (r₁,r₂,r₃) can have at most a single precursor that itself has a precursor. Therefore, other than the first member of a sequence, if (r₁.HCF,r₂.HCF,r₃.HCF) does not itself lie on sequence X, then neither will any of its precursors or successors.

Note that, if we require that (r₁,r₂,r₃) be all-odd-and-different, the first-term ambiguity vanishes, and we have a "straightforward" set of orthogonal sequences.

Many thanks for the explanations, Fyz. I don't think it was off-topic in any way at all. .