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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

Hi bobbym

The formula I got is (n+2)*2^(n-3) for n>=3. The formula there is (n+3)*2^(n-2). We can see that the difference is only in indexing.

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Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,309

Hi;

You got that formula how, by curve fitting? That only proves for the values you fit for. It does not mean that formula continues for the next diagonal and the one after that.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

We can prove by induction that a(i,j)=2^(i+j-2) for i,j>1. From there, it is easy proving the formula...

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,309

Hi;

What does a(i,j)=2^(i+j-2) for i,j>1 generate?

I would have set it up as

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

Sorry, I got 0 starting arrays in my head.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,309

Hi;

I will leave the inductive proof to you.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

Either way, I think we can be certaing that is how the sequence can be generated...

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,309

Hi;

We can think it all we want. Until we have some proof we ain't gonna convince anybody else of it.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Mpmath****Member**- Registered: 2012-10-11
- Posts: 216

Hi everyone;

I tried to find a formula to obtain the sum of the numbers of each diagonal and this is the result:

(2^n)*2+(2^n)2*n, then I simplified it and I obtained 2^(n-1)*n + 2^(n+1). The result is Number of 1's in all compositions of n+1 (A045623 of OEIS), because 2^(n-1)*n + 2^(n+1. Generate the same terms of (n+3)*2^(n-2), the formula of A045623, proposed by anonimnystefy.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,309

Hi;

To refresh my memory, this square?

1 1 2 4 8

1 1 2 4 8

2 2 4 8 16

4 4 8 16 32

8 8 16 32 64

16 16 32 64 128

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Mpmath****Member**- Registered: 2012-10-11
- Posts: 216

Yes, this one.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,309

Hi;

What is holding up some sort of proof for the problem is that my expression given in post #79 does not cover the first row or the first column.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Mpmath****Member**- Registered: 2012-10-11
- Posts: 216

Hi;

So what do you suggest?

*Last edited by Mpmath (2012-11-10 00:11:42)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,309

Hi;

I am working feverishly on an expression that actually generates that table. Then it should be easier to prove the relation is true.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Mpmath****Member**- Registered: 2012-10-11
- Posts: 216

Hi;

Ok. Thanks.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,309

Hi;

Nothing yet.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Mpmath****Member**- Registered: 2012-10-11
- Posts: 216

Hi;

Ok. Keep me informed, thanks.

*Last edited by Mpmath (2012-11-10 11:11:39)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,309

I finally got an expression that will generate the table but it is too complicated for me to understand. At least we can see more of the table.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Mpmath****Member**- Registered: 2012-10-11
- Posts: 216

Hi bobbym;

Good job! Can I see the expression?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,309

Hi;

I did not post it because it is virtually useless.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

That is equal to 2^(i-2)*2^(j-2).

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,309

Aha! You went for the trap. It is incorrect for

It is also incorrect for the whole first column and first row.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Mpmath****Member**- Registered: 2012-10-11
- Posts: 216

Hi;

Well, good job!

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

bobbym wrote:

Aha! You went for the trap. It is incorrect for

It is also incorrect for the whole first column and first row.

I still think we could use 2^(i-1) and 2^(j-1) for the first column and row respectively, and 2^(i-1) for the rest.

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,309

The only problem is that Mathematica gagged on both those series.

I want to sum along the diagonals but if it takes two functions for every diagonal that is going to make the proof much harder or impossible.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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