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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

...n<>7

when I tryed n=8 my vb compiler crushed because I had very much AND-s in one row. Tomorrow I can improve my program and I can write a program that calculates by my algoritm.

IPBLE: Increasing Performance By Lowering Expectations.

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**seerj****Member**- Registered: 2005-12-21
- Posts: 42

No no n is the set of integer numbers from n>1

n could be whatever integer number! It's not the lenght of the circular sequence.

The problem is find for which integers n u can arrange all the natural numbers [from 1 to N eg. 1,2,3...N] in a circular sequence.

In the last right example this set of integer numbers was : 1, 24,12,4,5 etc..

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

For the ricky's comments. Ijust have seen them:

1. "The sum of all perfect squares is 2n(n+1)/2=n(n+1)"

I wanted to say that the sum of all the squares that are sum of 2 numbers in the table is exactly the doubled sum of the numbers, because each number takes part to two squares.

2. " The the remainder of the sum of every two consecutive numbers when it's divided by 4 must be 0 or 1."- I don't wanted to say that the two numbers are consecutive at all. I just wanted to say that the sum of two consecutive at the table numbers is square so it's remainder when it's divided by 4 is 0 or 1. This property of the succesive at the table numbers may reduce the computation very much.

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

Seerj, I understood now. n may be greater than N.

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

But wait a minute...

If i've understood correctly

N cant be greater than n so some numbers may be repeated.

IPBLE: Increasing Performance By Lowering Expectations.

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**seerj****Member**- Registered: 2005-12-21
- Posts: 42

sorry!! I wrong in the first post

a)"FOR WHICH ENTIRES N>1 is possible ARRANGE IN A CERTAIN WAY ALL THE ENTIRE NUMBERS 1,2,3...,n AROUND A CIRCLE (CIRCULAR SEQUENCE),THAT THE SUM OF 2 CONSECUTIVE NUMBERS IS ALWAYS A PERFECT SQUARE."?

b) and then WHICH IS THE SMALLEST ENTIRE NUMBER N>1 FOR WHICH IS POSSIBLE TO FIND THIS CIRCULAR SEQUENCE?

But your last example was ok!

EXAMPLE: 1 24 12 4 5 11 14 2 7 9 27 22 3 .........

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**seerj****Member**- Registered: 2005-12-21
- Posts: 42

krassi_holmz wrote:

But wait a minute...

If i've understood correctly

N cant be greater than n so some numbers may be repeated.

No no. No numbers can be repeated!

I read that this N>1 is less than 50

So you have to combine all the numbers from 1 to 50 without repeat a same number twice.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

seerj, only use the report button to report abusive or very spammy posts, not because someone else made a mistake or did something wrong.

I thought the first time might have been an accident, but you've done that three times now.

Why did the vector cross the road?

It wanted to be normal.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

FOR WHICH ENTIRES N>1 is possible ARRANGE IN A CERTAIN WAY ALL THE ENTIRE NUMBERS 1,2,3...,n AROUND A CIRCLE

I'm still getting confused with this "all" thing. When you say all numbers 1, 2, 3... N, I think you need:

1, 2, 3, .... N-2, N-1, N all in the list.

Apparently, that is not the case (from your examples). So I take it that N is just the highest number you use. Is that right?

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**seerj****Member**- Registered: 2005-12-21
- Posts: 42

Sorry mathsyperson! Yesterday it was too late and I was so confused

About the problem..I making some mistakes ...You started well

The only 2 things to do are:

1)Choose a n (entire). Try to arrange in a circulare sequence all the numbers from 1 to n. (1,2,3,4,...,n)

Check if the sum of 2 consecutive numbers is always a perfect square. Which is the smallest n in which u can do that?

the tip is: this n is less than 50..so n<50.

2) Write a sequence (like the example)

Sorry for my mistakes. I think now it's ok.

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**Ricky****Moderator**- Registered: 2005-12-04
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Try to arrange in a circulare sequence all the numbers from 1 to n. (1,2,3,4,...,n)

When you say this, you are saying you need to include every single integer from 1 to n. For example, a list of size 5 would look like:

1 3 4 2 5

Of course, that isn't an answer because they don't all add up to perfect squares. Is this what you mean? Examples in previous posts have been different:

1 24 12 4 5 11 14 2 7 9 27 22 3

It's a list of size 13, but it includes number not within the range 1 to 13.

Which of these are right?

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**seerj****Member**- Registered: 2005-12-21
- Posts: 42

I'm a fu**in stupid!! I didn't see well the list written in old posts..Sorry me. These holidays are making me stupid!

1 24 is right coz 1+24 is 25 ok..

1 24 12 isn't right coz 24+12=36 ok perfect square but 12+1=13 isn't a perfect square!!!

1 24 12 4 isn't right. 12+4 is perfect square but 4+1=5 isn't a perfect square.

All have to work in a circular sequence too. The last number of the list + the first number = perfect square too!!

a little right sequence is:

1 15 10 6 3 1+15 15+10 10+6 6+3 AND 3+1 all perfect squares!

But it isn't enough to explain which is the minimun n to arrange all the entire numbers from 1 to this n in a circular sequence. [as first question]

As second question is to write a sequence longer than that above one.

Sorry me again!

PS: THE TIP is that this minimum n is less than 50. So we need to find a circular sequence made with all number from 1 to this minimum n. (n<50)

HAPPY XMAS TO ALL

*Last edited by seerj (2005-12-24 23:03:12)*

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

So my program did some work. This was the question I wanted to solve in the begining. n>7. And you don't have to check form 1 to 100. It's enough ot check from 1 to n.

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

n<>8

Now I'm waiting the case n=9.

IPBLE: Increasing Performance By Lowering Expectations.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

And you don't have to check form 1 to 100. It's enough ot check from 1 to n.

Krassi, you keep saying that, and I believe this is the third time I'm asking, what is n? The examples in this thread contradict the notion that it is the number of elements in the circle. So what is n?

*Last edited by Ricky (2005-12-26 09:22:03)*

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

Here's what I think. In the question we haven't N and n. We have only n. So we take all the numbers between 1 and n and put them on a sircular table.

Example:n=3 we should find is there an order of the numbers 1,2,3 on a sircular table so the rules to be observed.

Seerj must answer,too. Because we may solving a quite different quest then the original.

IPBLE: Increasing Performance By Lowering Expectations.

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**seerj****Member**- Registered: 2005-12-21
- Posts: 42

ricky I tried to write the correct version of the question. I think that krass is doing well.

This n is the minimum integer tat allow u to arrange all the numbers [1.2.4..till this n] in a circular sequence and the sum of 2 consecutive numbers is always a perfect square , but please read my last post.

And the tip was tat this minimum integer n is less than 50. so we need to look for this n as krassi is doing.

I think u wrote n<>9 coz u trying to arrange all the numbers till 9 right? so u wrote <> coz 9 isn't the minimum numbeR? good!

There is only n.

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**seerj****Member**- Registered: 2005-12-21
- Posts: 42

1 15 10 6 3

This is the correct sequence written by my teacher!

SO we need to look for n from 15 !! so we can think tat this n is 15<n<50....mumble...

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

1. Yes <> means, that it can't be.

2. No, I haven't computed it for n=9 yes so i'm not sure. But i'm sure for n<9.

IPBLE: Increasing Performance By Lowering Expectations.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Here's what I think. In the question we haven't N and n. We have only n. So we take all the numbers between 1 and n and put them on a sircular table.

Example:n=3 we should find is there an order of the numbers 1,2,3 on a sircular table so the rules to be observed.

krassi, that's what I initally thought as well. And on that note, it doesn't work for any list up to size 33 (currently testing 34). But then seerj says:

1 15 10 6 3

This is the correct sequence written by my teacher!

So it's not only 1 to n which can be included, since 15 wouldn't be in a list of size 5, only up to 5 would.

Now, to avoid confusion, I think we should define some terms and stick to them:

n: this is the number of elements in the circular sequence.

k: this the the highest number in the circular sequence.

Now I believe we are supposed to find the smallest k which exists on a circular sequence. So for the above example (the quote from seerj), k=15. The question then becomes, can we find one smaller than k=15? So we need to test n∈[1,50] for which k < 15.

Is this correct?

*Last edited by Ricky (2005-12-26 12:50:26)*

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

I don't think it's a coincidence that that sequence is of triangular numbers.

I played around with it for a bit and found a few other sequences that follow that pattern.

There's (1, 3) or any other combination of numbers that add to give a square.

There's also (1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120), which is a chain 15 numbers long.

There are more chains got by following the same pattern that are 32 long and 90 long. I'm not sure if that's helpful to what you're doing, but it's certainly quite interesting.

Why did the vector cross the road?

It wanted to be normal.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Going back on what I wrote before...

Seerj, you keep saying:

This n is the minimum integer tat allow u to arrange all the numbers [1.2.4..till this n]

Which means that you need a list which contains every number from 1 to n. If you make a list like this, it has to have size n. Ignoring the "perfect square" part:

1, 4, 6, 3, 2, 5

That would be a correct list. However:

1, 4, 2

Would not be a correct list because it doesn't include 3, which is a number from 1 to n, n being 4.

Is this correct? And if it is, how is:

1 15 10 6 3

This is the correct sequence written by my teacher!

A correct sequence since it doesn't contain 2, 4, 5, 7.. etc?

*Last edited by Ricky (2005-12-26 13:16:38)*

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**seerj****Member**- Registered: 2005-12-21
- Posts: 42

O H M Y G O D.

I think tat Santa Claus has stolen my brain!!!!!!! For SURE!

ok ok ok ok... ONLY THE LAST NUMBER OF THE CHAIN OR OF THE SEQUENCE MUST BE SUM WITH THE FIRST NUMBER AND THEM HAVE TO BE A PERFECT SQUARE!!! [sorry again..]

so any lists it's ok,

LAST RIGHT EXAMPLE:

Let a..z be numbers:

a b c d e f g h i

a+b=perfect square b+c=perfect square c+d=perfect square d+e=perfect square e+f=perfect square

f+g=perfect square g+h=perfect square h+i perfect square ANDD i+a= perfect square!!

A THING LIKE THIS. SORRY FOR ALL MY MISTAKES!!

So it's important tat the sum of 2 consecutive numbers it's always a perfect square and the last with the first have to be a perfect square yet! [to complete the circular sequence]

-And it's not important tat all numbers are present.

THE ONLY 2 THINGS ARE:

- Which is this n which u can arrange a..b..z..k...to n in a circular sequence

-The longest list that u can make with this n.

TIP:

This minimum n is less than 50. I think we need to make a circular sequence with all the combinations from 1 since 50 [except 0...in fact n>1]

Thank u again.

Forgive me!

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

So then seerj, why is:

1 3

Not the smallest solution?

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**seerj****Member**- Registered: 2005-12-21
- Posts: 42

no..Not the smallest solution..the smallest n

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