Pascal's square

anonimnystefy · 2012-11-05 02:47:15

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.

bobbym · 2012-11-05 02:50:24

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.

anonimnystefy · 2012-11-05 03:10:45

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...

bobbym · 2012-11-05 03:15:32

Hi;

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

I would have set it up as

anonimnystefy · 2012-11-05 03:32:16

Sorry, I got 0 starting arrays in my head.

bobbym · 2012-11-05 03:38:28

Hi;

I will leave the inductive proof to you.

anonimnystefy · 2012-11-05 03:50:35

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

bobbym · 2012-11-05 03:59:30

Hi;

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

Mpmath · 2012-11-09 21:05:55

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.

bobbym · 2012-11-09 21:21:26

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

Mpmath · 2012-11-09 21:50:20

Yes, this one.

bobbym · 2012-11-09 23:37:09

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.

Mpmath · 2012-11-10 00:00:11

Hi;

So what do you suggest?

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

bobbym · 2012-11-10 00:17:01

Hi;

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

Mpmath · 2012-11-10 00:45:59

Hi;

Ok. Thanks.

bobbym · 2012-11-10 10:38:24

Hi;

Nothing yet.

Mpmath · 2012-11-10 11:10:57

Hi;

Ok. Keep me informed, thanks.

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

bobbym · 2012-11-10 13:17:22

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.

Mpmath · 2012-11-10 22:20:35

Hi bobbym;

Good job! Can I see the expression?

bobbym · 2012-11-10 22:34:03

Hi;

I did not post it because it is virtually useless.

anonimnystefy · 2012-11-10 22:48:12

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

bobbym · 2012-11-10 22:51:54

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

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

Mpmath · 2012-11-10 22:57:01

Hi;

Well, good job!

anonimnystefy · 2012-11-10 23:02:46

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.

bobbym · 2012-11-10 23:20:05

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.

Math Is Fun Forum

#76 2012-11-05 02:47:15

Re: Pascal's square

#77 2012-11-05 02:50:24

Re: Pascal's square

#78 2012-11-05 03:10:45

Re: Pascal's square

#79 2012-11-05 03:15:32

Re: Pascal's square

#80 2012-11-05 03:32:16

Re: Pascal's square

#81 2012-11-05 03:38:28

Re: Pascal's square

#82 2012-11-05 03:50:35

Re: Pascal's square

#83 2012-11-05 03:59:30

Re: Pascal's square

#84 2012-11-09 21:05:55

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#85 2012-11-09 21:21:26

Re: Pascal's square

#86 2012-11-09 21:50:20

Re: Pascal's square

#87 2012-11-09 23:37:09

Re: Pascal's square

#88 2012-11-10 00:00:11

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#89 2012-11-10 00:17:01

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#90 2012-11-10 00:45:59

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#91 2012-11-10 10:38:24

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#92 2012-11-10 11:10:57

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#93 2012-11-10 13:17:22

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#94 2012-11-10 22:20:35

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#95 2012-11-10 22:34:03

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#96 2012-11-10 22:48:12

Re: Pascal's square

#97 2012-11-10 22:51:54

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#98 2012-11-10 22:57:01

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#99 2012-11-10 23:02:46

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#100 2012-11-10 23:20:05

Re: Pascal's square

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