std::iota is very limited

how about std::generate ?

int n = -2;
std::generate(x.begin(), x.end(), [&n]{ return n+=2; }); 
int n = 10;
std::generate(x.begin(), x.end(), [&n]{ return n--;})

But how would one do:

x = range(0,20,2)

Alternatively to std::generate() (see other answer), you can provide your own unary function to std::iota(), it just have to be called operator++():

#include <iostream>
#include <functional>
#include <numeric>
#include <vector>

template<class T>
struct IotaWrapper
{
    typedef T type;
    typedef std::function<type(const type&)> IncrFunction;

    type value;
    IncrFunction incrFunction;

    IotaWrapper() = delete;
    IotaWrapper(const type& n, const IncrFunction& incrFunction) : value(n), incrFunction(incrFunction) {};

    operator type() { return value; }
    IotaWrapper& operator++() { value = incrFunction(value); return *this; }
};

int main()
{
    IotaWrapper<int> n(0, [](const int& n){ return n+2; });
    std::vector<int> v(10);
    std::iota(v.begin(), v.end(), n);

    for (auto i : v)
        std::cout << i << ' ';
    std::cout << std::endl;
}

Output: 0 2 4 6 8 10 12 14 16 18

Demo

---

Here is an idea of how one could implement Range():

struct Range
{
    template<class Value, class Incr>
    std::vector<Value> operator()(const Value& first, const Value& last, const Incr& increment)
    {
        IotaWrapper<Value> iota(first, [=](const int& n){ return n+increment; });
        std::vector<Value> result((last - first) / increment);
        std::iota(result.begin(), result.end(), iota);
        return result;
    }
};

Demo