You're absolutely right.

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

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

]]>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

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

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

]]>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.

]]>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.

]]>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.

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

]]>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.

]]>Why do we need two formulas to obtain the table?

]]>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.

]]>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.

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