Pascal's square

Mpmath · 2012-11-11 09:50:06

Hi bobbym;

Why do we need two formulas to obtain the table?

Last edited by Mpmath (2012-11-11 09:50:28)

bobbym · 2012-11-11 10:50:15

Hi;

We shouldn't and that is what I am trying to avoid. The formula in post #95 is one formula but it is too difficult. I am still working on something better.

Mpmath · 2012-11-11 17:55:25

Hi;

Ok, but why this formula 2^(n-1)*n + 2^(n+1) is incorrect?

Last edited by Mpmath (2012-11-11 18:47:10)

anonimnystefy · 2012-11-11 21:02:33

If you can prove what I said in post #99 then the proof is not far from it.

Mpmath · 2012-11-11 21:35:55

Hi;

If we take a diagonal of the square:
8 4 4 4 8
Then we trasform the numbers in exponents of 2
2^3 2^2 2^2 2^2 2^3
The exponent of the first term (in this case 3) is the number of the exponents of 2 between the first and the last term. So we can say that 2^n*2 for the first and the last term, then, since that the exponents of 2 between the first and the last term are the half of the first term, so (2^n/2) multiplied by the exponent of the frist term (in this case n), so (2^n/2)*n. Then we add (2^n/2)*n to 2^n*2. The result is the sum of the numbers of each diagonal. The complete formula is 2^n*2 + (2^n/2)*n that we can write it also like 2^(n+1)*n+2^(n-1). This formula is valid for all the diagonals, except for the first one.

bobbym · 2012-11-12 01:07:36

Hi;

Ok, but why this formula 2^(n-1)*n + 2^(n+1) is incorrect?

No one said it was incorrect. I have been trying to come at the problem from another side that gets that formula.

This formula is valid for all the diagonals, except for the first one.

Isn't that what we are trying to prove.

Mpmath · 2012-11-12 01:52:57

Hi bobbym;

I wanted to ask only if my formula was correct or incorrect. Then I tried to prove a different thing. I'm still thinking about what you're trying to prove. My formula was a separate thing.

bobbym · 2012-11-12 02:14:43

Hi;

I think it is. Also I think the part about the compositions is correct too.

Mpmath · 2012-11-12 03:18:08

Hi;

I agree with you, but I've got some difficults regarding what you're try to prove. The formula is hard to find.

bobbym · 2012-11-12 04:11:15

Hi Mpmath;

If we define:

this is anonimnystefy's formula with a index adjustment by me. Then the sum of every diagonal including the first one is:

this can be summed to be

Where dn is the nth diagonal. If we now take the formula for the compositions of n+1 (Paul Barry's) we can see that the (n+1)th diagonal is equal to the (n+1)th composition.

This provides a proof of your statement. anonimnystefy's formula can be proven using induction

Mpmath · 2012-11-12 05:05:56

Hi bobbym;

Excellent! After a lot of work we arrived to a conclusion. Good job!

bobbym · 2012-11-12 10:42:20

Hi;

Math is hard work. Sort of like dragging an ox up a hill.

Mpmath · 2012-11-12 20:33:37

Hi;

You're absolutely right.

Math Is Fun Forum

#101 2012-11-11 09:50:06

Re: Pascal's square

#102 2012-11-11 10:50:15

Re: Pascal's square

#103 2012-11-11 17:55:25

Re: Pascal's square

#104 2012-11-11 21:02:33

Re: Pascal's square

#105 2012-11-11 21:35:55

Re: Pascal's square

#106 2012-11-12 01:07:36

Re: Pascal's square

#107 2012-11-12 01:52:57

Re: Pascal's square

#108 2012-11-12 02:14:43

Re: Pascal's square

#109 2012-11-12 03:18:08

Re: Pascal's square

#110 2012-11-12 04:11:15

Re: Pascal's square

#111 2012-11-12 05:05:56

Re: Pascal's square

#112 2012-11-12 10:42:20

Re: Pascal's square

#113 2012-11-12 20:33:37

Re: Pascal's square

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