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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)

[/tex]

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

:surprised OOPS - redface shame etc BicycleTree said: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.

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

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

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# Complete the Array Series

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