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November 23rd, 2016, 09:18 AM   #11
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Originally Posted by TobiWan View Post
but for n=3 it's a false, because only (0,1) applies to the second condition. To the first condition applies (2,0), (3,0), (4,0) and so on
Whatever are you talking about. In the case of n = 3, you are dealing with triplets, not pairs, of non-negative integers. And obviously none of the integers can exceed 3. So there are 64 cases. And only three of them are solutions, namely

$1(0) + 2(0) + 3(1) = 3.$ First solution.

$1(1) + 2(1) + 3(0) = 3.$ Second solution.

$1(3) + 2(0) + 3(0) = 3.$ Third solution.

The first solution satisfies the condition that each element of the triplet be either 0 or 1. As I pointed out before, FOR ANY n

$x_k = 0\ if\ 1 \le k < n\ and\ x_n = 1 \implies$

$\displaystyle \sum_{i=1}^n(i *x_i) = \left (\sum_{i=1}^{n-1}(i * 0) \right) + (n * 1) = 0 + n = n.$

The third solution satisfies the condition that each element of the triplet except the last either be positive or be followed by a zero element because the first element is both positive and followed by a zero element and the second element is followed by a zero element. As I pointed out before, FOR ANY n

$x_k = 0\ if\ 1 < k \le n\ and\ x_1 = n \implies$

$\displaystyle \sum_{i=1}^n(i *x_i) = 1(n) + \sum_{i=2}^n(i * 0) + (n * 1) = n + 0 = n.$

The second solution satisfies both conditions.

So two solutions satisfy the first condition, and two solutions satisfy the second condition. So the number of solutions that satisfy the first condition equals the number of solutions that satisfy the second condition.

I admit that checking out n = 4 seems a bit daunting because there are 625 cases, but it is not that hard to do with a computer.
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November 23rd, 2016, 11:32 AM   #12
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yes, thanks, I am trying to prove it now, vainly as yet
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November 24th, 2016, 05:13 AM   #13
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maybe have you found a proof ?
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November 25th, 2016, 05:52 AM   #14
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or a hint?
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