Can you find the solution with this additional property?

[AntonVrba]

$$A[r,c] = \left(\begin{array}{cccccc}<br /> \frac{1}{2}\multsp \left(1+{\sqrt{5}}\right)&-&-&-&-&...\\<br /> \frac{1}{4}\multsp \left(2+{\sqrt{8}}\right)&\frac{1}{4}\multsp <br /> \left(1+{\sqrt{5}}\right)&-&-&-&...\\<br /> \frac{1}{6}\multsp \left(3+{\sqrt{13}}\right)&-&\frac{1}{6}\multsp <br /> \left(1+{\sqrt{5}}\right)&-&-&... \\<br /> \frac{1}{8}\multsp \left(4+{\sqrt{20}}\right)&-&-&\frac{1}{8}\multsp <br /> \left(1+{\sqrt{5}}\right)&-&... \\<br /> \frac{1}{10}\multsp <br /> \left(5+{\sqrt{29}}\right)&-&-&-&\frac{1}{10}\multsp \left(1+{\sqrt{5}}\right)&...\\<br /> ...&...&...&...&...&... <br /> \end{array}\right)$$

How often have we solved linear series, how about a two dimensional series array

Can you complete the matrix?

Once you have it, instead of white just give the answer as a decimal number for say A[9,11] or any other pair

The answer is suprisingly simple


Edit-1
To note is that each induvidual row or column also forms its own logical arithmic series

Edit-2

The formula I got is:
A(r, c) = 1/(2*r)*(r-c+1+sqrt((r-c+2)^2-2(r-c)+1))
That matches the values you've given.

OOPS - redface shame etc

It is not the solution I had in mind - so you are correct. The solution I had in mind has a further property

$$\frac {A[r,c]}{A[n.r,n.c]} = n$$

 

[BicycleTree]

A[9, 11] = .061803? (Numbering from [0,0]; from [1,1] that would be A[10,12] = .061803[/color]

 

[AntonVrba]

BicycleTree - sorry try again

 

[BicycleTree]

Perhaps:
A(9, 11) = 1/18[/color]

 

[AntonVrba]

Perhaps:
A(9, 11) = 1/18[/color]

Sorry - wrong again

 

[BicycleTree]

Maybe
A(9, 11) = .06867[/color]

 

[<<<GUILLE>>>]

$$A[r,c] = \left(\begin{array}{cccccc} \frac{1}{2}\multsp \left(1+{\sqrt{5}}\right)&-&-&-&-&...\\ \frac{1}{4}\multsp \left(2+{\sqrt{8}}\right)&\frac{1}{4}\multsp \left(1+{\sqrt{5}}\right)&-&-&-&...\\ \frac{1}{6}\multsp \left(3+{\sqrt{13}}\right)&-&\frac{1}{6}\multsp \left(1+{\sqrt{5}}\right)&-&-&... \\ \frac{1}{8}\multsp \left(4+{\sqrt{20}}\right)&-&-&\frac{1}{8}\multsp \left(1+{\sqrt{5}}\right)&-&... \\ \frac{1}{10}\multsp \left(5+{\sqrt{29}}\right)&-&-&-&\frac{1}{10}\multsp \left(1+{\sqrt{5}}\right)&...\\...&...&...&...&...&... \end{array}\right)$$

Is it,
A(11, 13)=O.6867?

 

[marcus]

my first guess:

A(3,2) = 0.80473...[/color]

 

[marcus]

my first guess:

A(3,2) = 0.80473...[/color]

Ahh! I see that my proposal is the same as bicycle tree gave in post #6, just two posts back.

I would also say, like bicycle tree, that

A(9,11) = 0.0.06867...[/color]

so he and I must be using the same formula for the general term A(r,c)

 

[AntonVrba]

Folks, none of the proposed answers match my series - either there are two solutions which would surprise me. So let's check marcus' or bicycletree's solution.

For the answers to be correct, each induvidual row or column also forms its own logical arithmic series .

 

[BicycleTree]

The formula I got is:
A(r, c) = 1/(2*r)*(r-c+1+sqrt((r-c+2)^2-2(r-c)+1))[/color]
That matches the values you've given.

 

[AntonVrba]

The formula I got is:
A(r, c) = 1/(2*r)*(r-c+1+sqrt((r-c+2)^2-2(r-c)+1))
That matches the values you've given.

OOPS - redface shame etc

It is not the solution I had in mind - so you are correct. The solution I had in mind has a further property

$$\frac {A[r,c]}{A[n.r,n.c]} = n$$