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#76 2012-11-06 01:47:15

anonimnystefy
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Re: Pascal's square

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.


The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
 

#77 2012-11-06 01:50:24

bobbym
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Re: Pascal's square

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.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
 

#78 2012-11-06 02:10:45

anonimnystefy
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Re: Pascal's square

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...


The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
 

#79 2012-11-06 02:15:32

bobbym
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Re: Pascal's square

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.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
 

#80 2012-11-06 02:32:16

anonimnystefy
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Re: Pascal's square

Sorry, I got 0 starting arrays in my head.


The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
 

#81 2012-11-06 02:38:28

bobbym
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Re: Pascal's square

Hi;

I will leave the inductive proof to you.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
 

#82 2012-11-06 02:50:35

anonimnystefy
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Re: Pascal's square

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


The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
 

#83 2012-11-06 02:59:30

bobbym
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Re: Pascal's square

Hi;

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


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
 

#84 2012-11-10 20:05:55

Mpmath
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Re: Pascal's square

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.


Winter is coming.
 

#85 2012-11-10 20:21:26

bobbym
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Re: Pascal's square

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


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
 

#86 2012-11-10 20:50:20

Mpmath
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Re: Pascal's square

Yes, this one.


Winter is coming.
 

#87 2012-11-10 22:37:09

bobbym
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Re: Pascal's square

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.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
 

#88 2012-11-10 23:00:11

Mpmath
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Re: Pascal's square

Hi;

So what do you suggest?

Last edited by Mpmath (2012-11-10 23:11:42)


Winter is coming.
 

#89 2012-11-10 23:17:01

bobbym
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Re: Pascal's square

Hi;

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


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
 

#90 2012-11-10 23:45:59

Mpmath
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Re: Pascal's square

Hi;

Ok. Thanks.


Winter is coming.
 

#91 2012-11-11 09:38:24

bobbym
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Re: Pascal's square

Hi;

Nothing yet.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
 

#92 2012-11-11 10:10:57

Mpmath
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Re: Pascal's square

Hi;

Ok. Keep me informed, thanks.

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


Winter is coming.
 

#93 2012-11-11 12:17:22

bobbym
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Re: Pascal's square

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.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
 

#94 2012-11-11 21:20:35

Mpmath
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Re: Pascal's square

Hi bobbym;

Good job! Can I see the expression?


Winter is coming.
 

#95 2012-11-11 21:34:03

bobbym
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Re: Pascal's square

Hi;

I did not post it because it is virtually useless.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
 

#96 2012-11-11 21:48:12

anonimnystefy
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Re: Pascal's square

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


The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
 

#97 2012-11-11 21:51:54

bobbym
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Re: Pascal's square

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



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


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
 

#98 2012-11-11 21:57:01

Mpmath
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Re: Pascal's square

Hi;

Well, good job!


Winter is coming.
 

#99 2012-11-11 22:02:46

anonimnystefy
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Re: Pascal's square

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.


The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
 

#100 2012-11-11 22:20:05

bobbym
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Re: Pascal's square

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.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
 

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