Number of possible access structure with $n$ person I want to count number of possible access structures in closed form(if possible) as a variable of $n$,the number of share holder. I have tried in several ways. Still no good progress. It is very hard to consider all cases and count it and I think there will be work by others on approximating the number. Is there any approximation of that count so that I can get some idea ? 

All possible access contains $(k,n)$ - Secret sharing scheme( general access structures). Can anyone give any idea how to proceed?

Example:  In General access structure, we will call the set of collection of all people who can get the secret by $\Gamma_{qual}$ . Consider a case of $n=5$ , we have to consider the case like $$\Gamma_{qual}= \{\{1,2\},\{2,4\},\{1,3,5\}\}$$ but we can't have an access structure where a shareholder is not present, means we always need $$ \cup_{X\in \Gamma_{qual}}X = \{1,2,\cdots,n\}$$
An access structure is just a function $f:\{0,1\} ^n\to \{0,1\}$. It takes a subset of $\{1,\ldots,n\}$ as input (represented as it's characteristic vector) and returns 1 if authorized and 0 if unauthorized. Additionally, the function must be monotone, meaning that $A\subseteq B \implies f(A) \le f(B)$.

The number of monotone Boolean functions over $n$ variables is given by the Dedekind numbers (OEIS). It is a nontrivial formula!

Edit: You are interested in $f$ which are sensitive to all $n$ inputs. Let's call $f$ insensitive to input $i$ if there is a function $g$ (defined on subsets of $\{1,\ldots, i-1,i+1,\ldots,n\}$) such that $f(A) = g(A \setminus \{i\})$ for all $A$. If this is not the case then we say that $f$ is sensitive to $i$. Now say that $f$ is $k$-sensitive if it is sensitive to $k$ of its inputs. You can count the number of $n$-sensitive monotone functions using the inclusion-exclusion principle.

The idea is that $M(n)$ ($n$th Dedekind number) includes things you don't care about like $(n-1)$-sensitive functions. There are ${n \choose 1} M(n-1)$ Monotone functions of $n$ variables that are (at most) $(n-1)$-sensitive, so subtract them off. But since $M(n-1)$ also "overcounts" all the $(n-2)$-sensitive functions, you've subtracted them off too much and need to add them back. There are ${n \choose 2} M(n-2)$ of these. Etc etc etc. You finally get the thing you want as $\sum_{i=0}^n (-1)^i {n \choose i}M(n-i)$.