# What is the pattern or relation in this table?

Here is the table:

$$\begin{array}{c} 0\\ 1\\ 1& 1\\ 3& 2& 3\\ 5& 3& 3& 5\\ 11 &8 &10 &8 &11\\ 21 &13 &14 &14 &13 &21\\ 43 &30 &37 &36 &37 &30 &43\\ 85 &55 &61 &55 &55 &61 &55 &85\\ 171 &116 &140 &140 &146 &140 &140 &116 &171 \end{array}$$

In this triangle like table, I am unable to find the relation between columnwise. Say for 1st column, I have found the relation $A(n)= A(n-1)+2A(n-2)$.

For the 2nd column, I have the relation $A(n)= A(n-1)+2A(n-2) + x$ [ where $x$ is $-1$ for even row number and $+1$ for odd].

But for other columns, I have not found the relations yet. Can anyone help me find a generalized equation to generate the table?

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Where did you get this table? –  Matthew Conroy Oct 8 '12 at 19:55
in the second last row, shouldn't there be a 85 instead of the 65? :) –  Alex Oct 8 '12 at 19:57
@Alex , yes, fixed. –  Ahmad Faiyaz Oct 8 '12 at 20:04
Just out of interest, what is this table for? –  Alex Oct 8 '12 at 20:51
The first column and the diagonal appear to be A001045, $J_n = \frac{2^n-(-1)^n}{3}$. Wikipedia has an article, too. –  Snowball Oct 8 '12 at 22:16

Updated: let $A(n,k)$ be the number in the table with row $n$ and column $k$, where $0\le n$ and $0\le k\le n$.

$$A(n,k)=A(n,n-k)$$ $$A(n,0)=A(n,n)=0$$

The table would look like this: (top left is row $0$ and column $0$)

$$\begin{array}{c} 0\\ 0& 0\\ 0& 1& 0\\ 0& 1& 1& 0\\ 0& 3& 2& 3& 0\\ 0& 5& 3& 3& 5& 0\\ 0& 11 &8 &10 &8 &11& 0\\ 0& 21 &13 &14 &14 &13 &21& 0\\ 0& 43 &30 &37 &36 &37 &30 &43& 0\\ 0& 85 &55 &61 &55 &55 &61 &55 &85& 0\\ 0& 171 &116 &140 &140 &146 &140 &140 &116 &171 & 0 \end{array}$$

Then, we have: $$A(n+1,k)+A(n,k)+A(n,k-1)=2^{n-1}$$

Example $$A(8,3)+A(7,3)+A(7,2)=37+14+13=64=2^6$$

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how 4C2 gives 8 ? i am unable to understand your example. –  Ahmad Faiyaz Oct 9 '12 at 0:28
@HamadZaraf yeah sorry, that was the wrong row. it should have been 5. –  Alex Oct 9 '12 at 7:10
The choice of binomial coefficient is strange and a bit confusing. –  Emmad Kareem Oct 9 '12 at 9:53
@EmmadKareem Is it better now? –  Alex Oct 9 '12 at 13:10
+1, thank you. It is much better. –  Emmad Kareem Oct 9 '12 at 14:24