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

Posted July 27, 2009 5:01 PM

This week's Challenge Question:

What is the sequence (from top to bottom) of the three numbers in the fifth column?

To determine the number that goes into the last column's cells, we will determine a general formula for each cell value. Let's index the cells using the row number and the column number, as shown below,

In the above diagram, i represents the row (i=1,2,3), and j is the column (j =1,2 3,4,5)

By careful observation of the pattern given in the problem statement, we can see that, except for the first column, any cell value is determined by the following formula

Based on this formula, the pattern for the fifth column of the original problem is given in the following diagram:

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#1

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

07/28/2009 11:28 AM

My guess is 4, 13 & 6. Here's the pattern I see on the diagonals:

First row: 2+1=3, 3+1=4

Second row: 5+2=7, 7+2=9, 9+2=11, 11+2=13

or 5-2=3, 3+6=9, 9-2=7, 7+6=13

Third row: 4+1=5, 5+1=6

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#2

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

07/28/2009 1:31 PM

My guess is 3, 17, and 5.

Here's why:

2 + the numbers in the second row = the next number in the first row
2+5=7, 2+1=3, 2+9=11, 2+1=3

|The first number in the first row - the first number in the second row + the first number in the third row| = the second number in the second row
|second in the first row - second in the second row + second in the third row| = third in the second row
the pattern continues
|2-5+4|=1, |7-1+3|=9, |3-9+5|=1, |11-1+7|=17

first in the first row + first in the second row - second in third row = first in the third row
second in the frist row + second in the second row - third in the third row = second in the third row
the pattern continues
2+5-3=4, 7+1-5=3, 3+9-7=5, 11+1-5=7

I hope that's correct!

5
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#3

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

07/28/2009 3:50 PM

I agree the numbers in your answer. But I think your method for the first row is pure coincidence.

I think a proper solution should use the same method for each row, and this is possible as follows:

|Column_sum - 2*any_cell_in_column| = Cell in same-row, next_column

Examples:

First column sum = 11.
Subtract 2*2 -> 7
Subtract 2*5 -> 1
Subtract 2*4 -> 3

Second column sum = 11
Subtract 2*7 -> -3 and modulus -> 3
Subtract 2*1 -> 9
Subtract 2*3 -> 5

Etc.

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#4

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

07/28/2009 6:35 PM

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:

 2 7 3 11 3 19 5 33 9 5 1 9 1 17 9 25 7 43 4 3 5 7 5 15 13 17 23

Thanks,

Jim

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#5

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

07/28/2009 7:53 PM

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.

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#6

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

07/29/2009 12:57 AM

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

thanks

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#7

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

07/29/2009 2:22 AM

4,13,6

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#8

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

07/29/2009 2:35 AM

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) )

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#9

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

07/29/2009 3:16 AM

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.

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#10

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

07/29/2009 4:08 AM

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.

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#11

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

07/29/2009 6:46 AM

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

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#12

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

07/29/2009 7:42 AM

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?

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#14

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

07/29/2009 9:47 AM

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:.

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#15

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

07/29/2009 10:24 AM

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.

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#16

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

07/29/2009 1:11 PM

OK now I've got 4Â½, 1, 1Â½ AND 4Â½, 3, 3Â½

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.

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#13

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

07/29/2009 8:30 AM

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

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#17

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

07/29/2009 1:23 PM

4, 13, 6

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#18

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

07/29/2009 7:31 PM

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.

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#19

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

07/30/2009 5:42 AM

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

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#20

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

07/30/2009 12:05 PM

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, ...):

A1i = (i+3)/2

A2i = (2i+3)

A3i = (i+7)/2

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

A1i = (2i+3)

A2i = 1

A3i = (2i-1)

Thus, for column 5:

A15 = ((5)+3)/2 = 4

A25 = (2(5)+3) = 13

A35 = ((5)+7)/2 = 6

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#21

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

07/30/2009 1:41 PM

Sorry - those i's should be j's

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#22

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

07/30/2009 4:32 PM

One other graphical, non-mathematic way to look at it is shown below:

For the left-most 3x3 grid (red box), the green squares contain the numerical sequence 2, 3, 4, 5 starting from the top left square and moving in a z-pattern. The yellow squares contain the numerical odd number sequence 3, 5, 7, 9 starting from the bottom center square and moving clockwise.

Now shift the 3x3 grid so that the right column of the old grid is the left column of the new grid (purple box), and increment the first numerical sequence by 1 and the second by 4. The green squares give 3, 4, 5, 6 in the same pattern and the yellow squares give 7, 9, 11, 13 in the same pattern. The center square always contains 1.

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#23

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

07/30/2009 4:39 PM

GA,

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

Jim

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#25

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

07/31/2009 6:16 AM

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...

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#26

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

07/31/2009 11:05 AM

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

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#27

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

07/31/2009 11:33 AM

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.

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#28

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

07/31/2009 12:07 PM

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.

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#29

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

07/31/2009 1:18 PM

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.

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#30

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

07/31/2009 1:27 PM

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.

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#31

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

07/31/2009 6:53 PM

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.

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#32

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

08/01/2009 2:18 AM

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

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#33

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

08/01/2009 11:02 AM

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...).

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#24

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

07/30/2009 6:29 PM

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.

 Column/Row C1 C2 C3 C4 C5 R1 C1R1 (C1R2)+2 (C1R1)+1 (C3R2)+2 (C3R1)+1 R2 C1R2 1 (C2R1)+2 1 (C4R1)+2 R3 C1R3 (C1R2)-2 (C1R3)+1 (C3R2)-2 (C3R3)+1
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#34

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

08/01/2009 6:38 PM

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."

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#35

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

08/02/2009 11:05 AM

This sequence continues for a very long time without number duplication in a single column, and without a the members of a column sharing a common factor.

I was wondering if it could be proved that these features continued for ever?

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#36

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

08/02/2009 11:50 AM

Unless I goofed, the sequence goes to even numbers - first such column being 800,268,532,2020.

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#39

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

08/02/2009 5:49 PM

One of us "gooofed". Rarely, I'm reasonably sure it wasn't I ("Third Programme", in case you missed the skit), as the method for creating new columns means that the difference between the sum of one column and the next will always be even** - so if the sum of any column is odd then the sums of all columns will be odd.

Being lazy, I used a spreadsheet*. If we use (x,y) notation, I used
Value(m+1,n) = ABS{Value(m,1) + Value(m,2) +Value(m,3) - 2.Value(m,n)

So, in case it helps to resolve the discrepancy, here are some sample column values, together with their sums:
1: (2,5,4) 11
5: (3,17,5) 25
10: (57,11,29) 97
15: (33,219,119) 371
20: (769,179,661) 1609
25: (1551,3041,1905) 6497
30: (10403,1419,3763) 15585
35: (11477,36313,15741) 63531
40: (149945,8961,13497) 172403
45: (97093,334811,189659) 621563

*OpenOffice.org Calc is a highly acceptable free-to-use spread-sheet. As with all these things, there are details that could be made more convenient. One of my principle gripes is in the graphing area, where multiple scales for different variables on one graph are a real hassle - but really that's just being picky.
** Before you change the sign of a term, the sum of a derived column is the same as its parent. Changing the sign gives a change that is 2xelement_value).

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#40

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

08/03/2009 4:14 AM

-must check skit later-

You're quite right (what a surprise ! ). I used Excel and messed up somewhere as I extended the range. Anyway, I've modified my table and it tallies with your results, though I can't discern anything interesting from a quick glance. The sequence is more 'odd' than 'Nice' !

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#41

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

08/03/2009 7:29 AM

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)

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#42

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

08/03/2009 8:05 AM

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.

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#43

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

08/10/2009 3:56 AM

I've probably goofed again, but numbers from the 59th term onward seem to be divisible by 3 ; 1893291, 12830097, 3669973. I haven't checked that set for any other factor.

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#44

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

08/11/2009 5:02 PM

Dividing 3669973 by 3 gives the remainder 1 (I think you misread your 7)

I'll post the proof that no new factors can appear when I have time to write it in a comprehensible form

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#45

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

08/11/2009 5:50 PM

This is a bit odd. On Excel (2003) I had it dividing all the sets of numbers by 3. Excel says that 3669973/3 = 1223324 ! It was churning out precision to 2 decimal places, so I'm not sure how that error occured. I should have spotted that was wrong - it's a bit obviously so on reflection. I'm out of time here, but will look into this a bit more later. It must be the precision/number of digits. All I can say for now is WAHHHHH!

I look forward to your proof about the factors (with fingers crossed that I can understand it !).

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#37

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

08/02/2009 1:05 PM

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

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#38

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

08/02/2009 4:52 PM

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.

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#46

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

08/13/2009 5:02 PM

An implied question is whether this sequence is "nice". We have already established that the sums of the columns are all "nice" numbers.
The other question is whether the sequence is nice in the sense of having some sort of continuing "good" behaviour.

Therefore, in spite of this not being a direct answer to the challenge, I am going to chicken out of marking it "off-topic".

The basic questions in my mind were:
Will the three entries in a column remain for ever different? and
The entries in a column do not so far share a single common factor - does this remain true for ever?

The answer to both these questions is "yes". Here are my 'proofs':

Suppose column number r has elements that can be factorised as r1.HCF, r2.HCF, r3.HCF, where HCF is the highest common factor of the elements as might normally be written.
The elements in the previous column will have values: (Â±r2Â±r3).HCF/2, (Â±r3Â±r1).HCF/2, and (Â±r1Â±r2).HCF/2.

Clearly, other than a possible single factor of two, the previous column has the same common factors as the succeeding column. It follows that, going forward the only additional common factor that can be acquired is the factor 21.

Now we need to know under what conditions the additional factor of two may appear. For this we revert to the forward formula - the elements in the succeeding column will have values:
|(∑-2.r1)|.HCF, |(∑-2.r2)|.HCF and |(∑-2.r2)|.HCF, where ∑=r1+r2+r3.

Given the 2x multiplier in each of the braces, we can see that, if is even all the values in braces must be even; if is odd all the values in braces must be odd. That means that the HCF will be increased by a factor of 2 if (and only if) the sum of the elements is divisible by a higher exponent of 2 than the HCF.

It is also possible to show that for any sequence ∑/HCF will become odd after a finite number of columns (I'll do that another time if I judge there is sufficient demand).

The above shows (amongst other things) that columns with odd sums are succeeded by columns whose elements are all odd. We now show that, if the odd numbers in a column are different, all the numbers in the succeeding column are also different from each other. In order to minimise the number of cases, I shall rename the numbers r1, r2, r3 in descending order as a, b, c. The numbers in the succeeding column may then be written in descending order as: (a+b-c), (a-b+c), and either (c+b-a) or (a-b-c) – clearly all unequal in both cases (this result relies solely on the numbers being non-negative and ordered, and that being odd none of them can be zero).

The final requirement for a "well-behaved" sequence is that the members do not repeat – i.e. for this case that each column is different from all that precede it. We can see that (subject only to the elements all being different and non-negative) this must be the case, as the largest element in a column is always greater than the largest element in the column that preceded it.

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#47

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

08/14/2009 5:46 AM

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.

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#48

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

08/14/2009 3:33 PM

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

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#49

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

08/14/2009 4:13 PM

1) Would you believe that the whole of these started as an abbreviated formal proof that easily fitted on the back of an A5 envelope?

2) I still think the sequence presented should have started with (4,1,2), which is the smallest possible starter for a good sequence. The next two terms would have been (1,5,3) and (7,1,3)

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#50

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

08/15/2009 2:29 AM

I not only believe it, but I'm having a good chuckle about it as well ! It illustrates why maths is fun, and you don't need anything complex to provide a starting point for a mental work-out.

This sort of question probably frustrates the pragmatists, but all the Challenge Questions are a bit like that. I'm sure I'll play further with it at some point in the future. For now I'm diverted (seems to happen quite a lot ) with my recent discovery that 49 circles can be packed in a square measuring slightly less than 7 x 7. Packing theory has been explored pretty thoroughly so I may not take my reading much further, but it could be very useful if I ever find myself working as a Milkman

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#51

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

08/15/2009 7:17 AM

As a unit diameter circle fits inside a square with space to spare, there must inevitably be a size of square at which the gaps can be exploited - I didn't know that 7 was the minimum.

However, assuming for now that space-saving was a significant requirement for the milk-round - wouldn't you be better advised to use a rectangle that allows internal close-packing?

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#52

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

08/15/2009 9:39 AM

I don't have the book I was reading to hand (not that it elaborated much), and my understanding of this is a bit half-baked ! My surprise is that a sort of random packing is most efficient for 49 objects in a square;

That's from a pdf here (http://www.sfu.ca/~ahmolnar/Projects/Circle%20Packing.pdf - sorry, link no longer available). I'm not sure what the side length of that square is, and it doesn't look at all easy to work out. There are plenty of related web-sites, but they all seem to involve a lot of math-speak and no straightforward answer to that question ! Probably just me being dense in the more usual sense.

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#53

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

08/15/2009 9:56 AM

Looks like an exhaustive search job to me - no equal angles, nothing to help "work it out". (And probably not worth the effort unless you just happen to be designing an oil drum to fit into a pre-existing space - and even for that case you'd probably be best advised to use a different shape).

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#54

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

08/15/2009 10:01 AM

Pretty much my thoughts. There's a table here, but I didn't understand it !

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#55

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

08/15/2009 10:31 AM

Let's take the example of packing of 49 circles into a square.

Packing on grid would require the circle's radius to be less than 1/14 (≈0.071429).

The best packing that Dave Boll quotes allows the circle radius to be 0.071693 - only very slightly larger.

If we are looking instead at the size of the square, for a fixed size of sphere that corresponds to reducing the linear dimension of the square by 0.37%, or the area by 0.74%.

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#56

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

08/15/2009 12:57 PM

Ah, I think the circular copper-plated steel thing has dropped; 49 of radius 1 could be packed into a square of side 1/0.071693 = 13.948. I can't imagine this will ever be of practical use to me, but it's fascinating all the same. Many thanks for helping me with that step.

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#57

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

08/17/2009 5:51 AM

I don't get this: his first "better" packing of 49 circles appears to have a top row which would make the square the same size as (or bigger than) the square matrix:-

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#58

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

08/17/2009 7:37 AM

I think there's a gap between the top right circle and the RHS of the square (well, there must be a gap somewhere in that top line of circles ! His use of hexagons doesn't help clarify it). Hard to verify any of them without drawing the things. I haven't drawn any of them myself yet to check, but may do so with some of the better packings.

There's another thread extolling the merit of 7 as a number (!) - I may just have to refer them here to see it's nastiness

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#59

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

08/17/2009 9:58 AM

It's Ye Olde Drawing Artefact. The circles in the top row are not actually colinear, and the gap at the end is tiny. (I'm sure that, if you could be ***ed to read the text, you would find it was explained there).

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