Suppose $n$ is the number of elements and $k$ the size of the subset.
The idea here is to construct a generating function by choosing the
first element of each subset and thereafter appending a series of gaps
between adjacent values, marking runs of consecutive elements. We use
$w$ for these runs and $u$ to mark runs of at least three consecutive
elements and $z$ for ordinary gaps of size at least two. We also use
$v$ to count elements. We then have from first principles the OGF
$$f(z,w,u,v) = \frac{vz}{1-z}\sum_{q\ge 0} \frac{v^q z^{2q}}{(1-z)^q}
\\ \times \left(\sum_{p\ge 0} (vw+uv^2w^2+uv^3w^3+uv^4w^4+\cdots)^p
\left(\sum_{q\ge 1} \frac{v^q z^{2q}}{(1-z)^q}\right)^p\right)
\\ \times (1+vw+uv^2w^2+uv^3w^3+uv^4w^4+\cdots) + 1.$$
As a sanity check we should get all subsets from
$$\frac{1}{1-z} f(z,z,1,1)$$
(the multiplier by $1/(1-z)$ sums the contributions as these subsets
are classifified according to the last element). We obtain
$$\frac{z}{1-z} \sum_{q\ge 0} \frac{z^{2q}}{(1-z)^q}
\\ \times \left(\sum_{p\ge 0} (w+w^2+w^3+w^4+\cdots)^p
\left(\sum_{q\ge 1} \frac{z^{2q}}{(1-z)^q}\right)^p\right)
\\ \times (1+w+w^2+w^3+w^4+\cdots) + 1.$$
This simplifies to
$$\frac{z}{1-z} \frac{1}{1-z^2/(1-z)}
\left(\sum_{p\ge 0} \frac{w^p}{(1-w)^p}
\left(\frac{z^2/(1-z)}{1-z^2/(1-z)}\right)^p\right)
\frac{1}{1-w} + 1
\\ = \frac{z}{1-z-z^2}
\frac{1}{1-wz^2/(1-z)/(1-w)/(1-z^2/(1-z))}
\frac{1}{1-w} + 1
\\ = \frac{z}{1-z-z^2}
\frac{1}{1-wz^2/(1-w)/(1-z-z^2)}
\frac{1}{1-w} + 1
\\ = z
\frac{1}{1-z-z^2-wz^2/(1-w)}
\frac{1}{1-w} + 1
= z \frac{1}{(1-z-z^2)(1-w)-wz^2} + 1.$$
To conclude we put $w=z$ and get
$$z \frac{1}{(1-z-z^2)(1-z)-z^3} + 1
= z \frac{1}{1-z-z^2-z+z^2+z^3-z^3} + 1
\\ = z \frac{1}{1-2z} + 1 = \frac{1-z}{1-2z}.$$
We see that on multiplying by $1/(1-z)$ we obtain
$$\bbox[5px,border:2px solid #00A000]{\frac{1}{1-2z}.}$$
There are $2^n$ subsets and the sanity check goes through.
Now to eliminate subsets containing three consecutive elements we
compute
$$\frac{1}{1-z} f(z,z,0,v)$$
which yields
$$\frac{vz}{1-z}\sum_{q\ge 0} \frac{v^q z^{2q}}{(1-z)^q}
\\ \times \left(\sum_{p\ge 0} v^p w^p
\left(\sum_{q\ge 1} \frac{v^q z^{2q}}{(1-z)^q}\right)^p\right)
\\ \times (1+vw) + 1$$
which is
$$\frac{vz}{1-z} \frac{1}{1-vz^2/(1-z)}
\times \left(\sum_{p\ge 0} v^p w^p
\left(\frac{vz^2/(1-z)}{1-vz^2/(1-z)}\right)^p\right)
\times (1+vw) + 1
\\ = \frac{vz}{1-z} \frac{1}{1-vz^2/(1-z)}
\times \left(
\frac{1}{1-v^2wz^2/(1-z)/(1-vz^2/(1-z))}\right)
\\ \times (1+vw) + 1
\\ = \frac{vz(1+vw)}{(1-z)(1-vz^2/(1-z)) - v^2wz^2} + 1
\\ = \frac{vz(1+vw)}{1-z-vz^2 - v^2wz^2} + 1.$$
On replacing $w$ by $z$ and multiplying by $1/(1-z)$we thus have
$$\frac{1}{1-z}\frac{vz+v^2z^2}{1-z-vz^2 - v^2z^3} + \frac{1}{1-z}.$$
The second term only contributes for $k=0$ while the first does not
contribute when $k=0.$ This means there is one subset of zero elements
not containing three consecutive elements, namely the empty set, and
this is obviously true.
Simplifying we find
$$\frac{1}{1-z}\frac{vz+v^2z^2-(1-z)/z}{1-z-vz^2 - v^2z^3}
+ \frac{1}{z} \frac{1}{1-z-vz^2 - v^2z^3}
+ \frac{1}{1-z}
\\ = \left(1-\frac{1}{z}\right)\frac{1}{1-z}
+ \frac{1}{z} \frac{1}{1-z-vz^2 - v^2z^3}
\\ = -\frac{1}{z} + \frac{1}{z}
\frac{1}{1-z-vz^2 - v^2z^3}
\\ = -\frac{1}{z} + \frac{1}{z} \frac{1}{1-z}
\frac{1}{1-vz^2/(1-z) - v^2z^3/(1-z)}.$$
Observe that
$$[v^k] \frac{1}{1-vz^2/(1-z) - v^2z^3/(1-z)}
= \sum_{q=0}^k [v^k] \frac{v^q z^{2q}}{(1-z)^q} (1+vz)^q
\\ = \sum_{q=0}^k \frac{z^{2q}}{(1-z)^q} [v^{k-q}] (1+vz)^q
= \sum_{q=0}^k {q\choose k-q} \frac{z^{2q}}{(1-z)^q} z^{k-q}
\\ = \sum_{q=0}^k {q\choose k-q} \frac{z^{k+q}}{(1-z)^q}.$$
The coefficient extraction in $z$ now proceeds as follows:
$$[z^n] \frac{1}{z}\frac{1}{1-z}
\sum_{q=0}^k {q\choose k-q} \frac{z^{k+q}}{(1-z)^q}
= [z^{n+1}] \sum_{q=0}^k {q\choose k-q}
\frac{z^{k+q}}{(1-z)^{q+1}}
\\ = \sum_{q=0}^k {q\choose k-q}
[z^{n+1-k-q}] \frac{1}{(1-z)^{q+1}}.$$
We have the closed form which is
$$\bbox[5px,border:2px solid #00A000]{
\sum_{q=0}^k {q\choose k-q} {n+1-k\choose q}.}$$
Listing these values in a triangular format we find
$$1, 1, 1, 1, 2, 1, 1, 3, 3, 0, 1, 4, 6, 2, 0, 1, 5, 10, 7, 1,
\\ 0, 1, 6, 15, 16, 6, 0, 0, 1, 7, 21, 30, 19, 3, 0, 0, 1,
\\ 8, 28, 50, 45, 16, 1, 0, 0, \ldots$$
which points us to OEIS A078802 where the
above analysis is confirmed.
The above computation was supported by the following Maple code.
with(combinat);
ENUM :=
proc(n, k)
option remember;
local list, pos, res;
if k < 3 then return binomial(n,k) fi;
res := 0;
for list in choose(n, k) do
for pos to k-2 do
if list[pos] + 1 = list[pos + 1] and
list[pos + 1] + 1 = list[pos+2] then
break;
fi;
od;
if pos = k-1 then
res := res + 1;
fi;
od;
res;
end;
X1 := (n, k) ->
coeftayl(coeftayl(-1/z+1/z*1/(1-z-v*z^2-v^2*z^3),
v=0, k), z=0, n);
X2 := (n, k) ->
add(binomial(q,k-q)*binomial(n+1-k,q), q=0..k);
In particular for subsets of twenty elements of a set containing the
integer range from one to thirty the number of subsets without three
consecutive elements is
$$\bbox[5px,border:2px solid #00A000]{66.}$$
