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Hi;
L'Hopital's rule is the best way.
Hi;
You're absolutely right.
Hi bobbym;
Excellent! After a lot of work we arrived to a conclusion. Good job!
Hi;
I agree with you, but I've got some difficults regarding what you're try to prove. The formula is hard to find.
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.
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.
Hi;
Ok, but why this formula 2^(n-1)*n + 2^(n+1) is incorrect?
Hi bobbym;
Why do we need two formulas to obtain the table?
Hi;
Well, good job!
Hi bobbym;
Good job! Can I see the expression?
Hi;
Ok. Keep me informed, thanks.
Hi;
Ok. Thanks.
Hi;
So what do you suggest?
Yes, this one.
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.
Hi anonmystefy;
I agree with you.
Hi bobbym;
Thanks for compliment.
Hi;
Now I understand. Thanks!
Hi;
Sorry, but I don't understand what you mean when you say "Does seem to be generating the compositions.
Yes...?
Hi bobbym;
So what are your conclusions?
Ok. Let me know if you find something interesting.
At least we find an alternative system to obtain square array of Delannoy numbers!