The number of families of subsets of $\{1,2,...,n\}$ whose union is not the whole set. I wrote a code in Mathematica that returns the number of families (collections) of subsets of $\{1,2,...,n\}$ whose union is not the whole set.  The code can only return values for $n = 1,2,3,4$.  The respective values are $2,6,40, 1376$.  These are the first four terms of Sloane's A051185 which counts the number of intersecting families.  Is this a coincidence or is there some reason why these two counts are equal?   
 A: A051185 is the number of (pairwise-) intersecting families.  Two subsets of $\{1,\ldots, n\}$ intersect iff the union of their complements is not $\{1,\ldots,n\}$.  But you want not just the pairwise unions but the union of 
all sets in your family to not be the whole set.  So you should get a different result for $n=3$: the family $\{\{1,2\}, \{1,3\}, \{2,3\}\}$ is pairwise intersecting, so it should be included in A051185, but its complements $\{\{3\},\{2\},\{1\}\}$ should not be included in your count.
In fact, I get $38$, not $40$, for $n=3$ and $942$, not $1376$, for $n=4$. 
A: If I understand correctly what you're trying to count, your code is wrong. I think the sequence you're after is this one: https://oeis.org/A005530
A: The problem is equivalent to count the number of covers of a finite set. 
It is mentioned at the beginning of the paper Minimal Covers of Finite Sets by Hearne and Wagner that the number is given as the following (where "$n=1$" should be a typo):


Alternatively, the number of possible covers for a set of $N$ elements are
$$ C(N)=\frac{1}{2}\sum_{k=0}^N(-1)^k\binom{N}{k}2^{2^{N-k}}, $$
the first few of which are $1, 5, 109, 32297,\cdots$ (OEIS A003465: Number of ways to cover an n-set. ).
A: Here is  an approach using  the Polya Enumeration Theorem  (PET) using
the       same       technique       as       at       this       MSE
link.
We start  with the complementary problem counting  families that cover
all  of $[n].$  We get  the  following generating  function for  these
subsets:
$$\prod_{q=1}^n (1+A_q).$$
Selecting a subset  from these of size $k$ is  done with the unlabeled
set operator $\mathfrak{P}:$
$$Z(P_k)\left(\prod_{q=1}^n (1+A_q)\right).$$
This operator has OGF
$$Z(P_k) = [w^k] 
\exp\left(\sum_{d\ge 1} (-1)^{d+1} a_d \frac{w^d}{d}\right)$$
The substitution into the cycle index uses
$$a_d = \prod_{q=1}^n (1+A_q^d).$$
We use inclusion-exclusion to compute  the terms where not all $A_q$ a
present. These are obtained by  first subtracting those where a single
$A_q$  or  more  are  not  present,  i.e. by  setting  that  $A_q$  to
zero. Then we add  in those where a pair of two  $A_q$ or more are not
present, again by  setting these $A_q$ to zero and  so on. The remaing
$A_q$ are  set to  one. Hence  if $p$ or  more of  the $A_q$  were not
present the substitution becomes
$$a_d = 2^{n-p}.$$
We thus obtain for the substituted cycle index
$$[w^k] \exp\left(\sum_{d\ge 1} 
(-1)^{d+1} 2^{n-p} \frac{w^d}{d}\right)
\\ = [w^k] \exp\left(2^{n-p} \sum_{d\ge 1} 
(-1)^{d+1} \frac{w^d}{d}\right)
\\ = [w^k] \exp\left(2^{n-p} \log(1+w) \right)
= [w^k] (1+w)^{2^{n-p}} = {2^{n-p}\choose k}.$$
This means we  have by inclusion-exclusion for the  problem of the set
being covered
$$\sum_{p=0}^n {n\choose p} (-1)^p {2^{n-p}\choose k}.$$
Summing over all possible $k$ (can be anywhere from zero to $2^n$)
we finally obtain
$$\sum_{p=0}^n {n\choose p} (-1)^p 
\sum_{k=0}^{2^n} {2^{n-p}\choose k}
= \sum_{p=0}^n {n\choose p} (-1)^p 2^{2^{n-p}}.$$
This yields for the original problem whose answer is being sought
$$\bbox[5px,border:2px solid #00A000]{
2^{2^n} - \sum_{p=0}^n {n\choose p} (-1)^p 2^{2^{n-p}}.}$$
which is the sequence OEIS A005530
$$2, 6, 38, 942, 325262, 25768825638, 
\\ 129127208425774833206, \ldots$$
as observed in a companion post.
The Maple code for this  (which will compute the first four values)
was  like this  (shown  here  to clarify  what  interpretation of  the
problem is being used):

with(combinat);

ENUM :=
proc(n)
option remember;
local set, res, covered;

    res := 0;

    for set in powerset(powerset(n)) do
        covered := `union`(op(set));

        if nops(covered) < n then
            res := res + 1;
        fi;
    od;

    res;
end;

X := n-> 2^(2^n)
-add(binomial(n,p)*(-1)^p*2^(2^(n-p)), p=0..n);

Addendum. Some  thought produces the following  simple Perl script
which  adhereres  to first  principles  (problem  definition), yet  is
capable of  computing the first  five values  of the sequence  in less
than three seconds,  which appears to be the natural  boundary of what
is possible. We solve the complementary problem as above.

#! /usr/bin/perl -w
#

sub recurse {
    my ($src, $seen, $pos, $n, $cref) = @_;

    my $chk;
    for($chk = 0; $chk < $n; $chk++){
        last if $seen->[$chk] == 0;
    }

    if($chk == $n){
        $$cref += 1 << ((1<<$n)-$pos);
        return;
    }

    return if $pos >= (1 << $n);

    recurse($src, $seen, $pos+1, $n, $cref);

    foreach my $el (@{ $src->[$pos] }){
        $seen->[$el]++;
    }
    recurse($src, $seen, $pos+1, $n, $cref);
    foreach my $el (@{ $src->[$pos] }){
        $seen->[$el]--;
    }

    1;
}

MAIN : {
    my $n = shift || 4;

    my $srcset = [];

    for(my $idx = 2**$n-1; $idx >= 0; $idx--){
        my @data;

        for(my ($bitpos, $idx2) = (0, $idx); 
            $bitpos < $n; $bitpos++){
            my $bit = $idx2 % 2;

            push @data, $bitpos if $bit == 1;
            $idx2 = ($idx2-$bit)/2;
        }

        push @$srcset, \@data;
    }

    my (@vis) = (0) x $n;

    my $result = 0; 
    recurse($srcset, \@vis, 0, $n, \$result);
    $result = 2**(2**$n) - $result;

    print "$result\n";

    1;
}


