Another Binomial It starts with the first row of n elements as 1  1  0  0  0 ...
For n = 5,
The 1st row is 1  1  0  0  0
The next row is generated as ith element = the sum of (i)th and (i-1)%nth elements of previous row. 
So the generated rows are 
1  2  1  0  0
1  3  3  1  0
1  4  6  4  1
2  5  10 10  5
7  7  15 20  15
....
My question is if exists an expression for the ith element of kth row
 A: This is probably closer to a comment rather then an answer, but it seems that the comment I wrote above needs to be expanded and clarified.
The number in the $n$-th row and $i$-th (where $i=0,1,2,3,4$) column of your table can be written as
$$a_{n,i} = \sum_{k=-\infty}^\infty \binom n{5k+i}.$$
This formally looks as an infinite sum, but notice that there are only finitely many non-zero terms.
Notice that the above definition makes sense for arbitrary $i$ (not only $0\le i \le 4$), but clearly we is periodic:
$$a_{n,i+5}=a_{n,i}.$$
From Pascal's rule we get
$$a_{n,i} = \sum_{k=-\infty}^\infty \binom n{5k+i} =
\sum_{k=-\infty}^\infty \binom {n-1}{5k+i-1} + \sum_{k=-\infty}^\infty \binom {n-1}{5k+i} = a_{n-1,i-1} + a_{n,i}$$
which corresponds to your recurrent definition. (If we consider also $a_{n,i+5}=a_{n,i}$.)

Example
From your definition you get
$$\begin{array}{ccccc}
  1 & 0 & 0 & 0 & 0 \\
  1 & 1 & 0 & 0 & 0 \\
  1 & 2 & 1 & 0 & 0 \\
  1 & 3 & 3 & 1 & 0 \\
  1 & 4 & 6 & 4 & 1 \\
  2 & 5 &10 &10 & 5 \\
  7 & 7 &15 &20 &15 \\
 22 &14 &22 &35 &35 \\
 57 &36 &36 &57 &70 \\
 127&93 &72 &93 &127
\end{array}$$
If we look at the numbers in the $9$-th row of the Pascal's triangle (i.e., $1$, $9$, $36$, $84$, $126$, $126$, $84$, $36$, $9$, $1$) then we can see that
\begin{align*}
127&=1+126\\
93&=9+84\\
72&=36+36\\
93&=84+9\\
127&=126+1
\end{align*}

So your problem is equivalent to determining
$$a_{n,i} = \sum_{k=-\infty}^\infty \binom n{5k+i}.$$
You can probably find more about sums of this type in other questions on this site, or elsewhere. I was able to find this: How to find sums like $\sum_{k=0}^{39} \binom{200}{5k}$
