## Complete the Array Series

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

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
 Quote by BicycleTree 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

$$\frac {A[r,c]}{A[n.r,n.c]} = n$$
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 A[9, 11] = .061803? (Numbering from [0,0]; from [1,1] that would be A[10,12] = .061803
 BicycleTree - sorry try again

## Complete the Array Series

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

 Quote by BicycleTree Perhaps: A(9, 11) = 1/18
Sorry - wrong again
 Maybe A(9, 11) = .06867
 $$A[r,c] = \left(\begin{array}{cccccc} \frac{1}{2}\multsp \left(1+{\sqrt{5}}\right)&-&-&-&-&....\\ \frac{1}{4}\multsp \left(2+{\sqrt{8}}\right)&\frac{1}{4}\multsp \left(1+{\sqrt{5}}\right)&-&-&-&....\\ \frac{1}{6}\multsp \left(3+{\sqrt{13}}\right)&-&\frac{1}{6}\multsp \left(1+{\sqrt{5}}\right)&-&-&.... \\ \frac{1}{8}\multsp \left(4+{\sqrt{20}}\right)&-&-&\frac{1}{8}\multsp \left(1+{\sqrt{5}}\right)&-&.... \\ \frac{1}{10}\multsp \left(5+{\sqrt{29}}\right)&-&-&-&\frac{1}{10}\multsp \left(1+{\sqrt{5}}\right)&....\\....&....&....&....&....&.... \end{array}\right)$$ Is it, A(11, 13)=O.6867????
 Recognitions: Gold Member Science Advisor my first guess: A(3,2) = 0.80473...

Recognitions:
Gold Member
 Quote by marcus my first guess: A(3,2) = 0.80473...
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...

so he and I must be using the same formula for the general term A(r,c)
 Folks, none of the proposed answers match my series - either there are two solutions which would surprise me. So lets check marcus' or bicycletree's solution. For the answers to be correct, each induvidual row or column also forms its own logical arithmic series .
 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.

 Quote by BicycleTree 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

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

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