I have reproduced with a simple code below something I am missing.
Class B owns class A as member and uses same typenames .
I want to avoid replicating these templates so that in the main() I can call something like B b(a, 3.0) which would reuse template from A. Is that possible ?
#include <iostream>
#include <vector>
template<int N, typename T=double>
struct A
{
    A(T val) : vecA(N, val) {}
    void print() { for (auto i : vecA) std::cout << i << ";"; }
    std::vector<T> vecA;
};
template<int N, typename T>
struct B
{
    B(const A<N,T> & in, T scal) : a(in), scalB(scal) {}
    void print() { a.print(); std::cout << " | " << scalB << std::endl; }
    A<N,T> a;
    T scalB;
};
int main()
{
    A<5,float> a(2.0);
    B<5,float> b(a, 3.0);   // This is redundancy. Can I do something like B b(a,3.0) ?
    b.print();
}
                Template parameters can be deduced from the types of a function template's arguments, but not of a class template's constructor arguments. So you could do what the standard library does (for example with make_pair) and write a function to deduce them for you:
template <int N, typename T>
B<N,T> make_b(const A<N,T> & in, T scal) {
    return B<N,T>(in, scal);
}
auto b = make_b(a, 3.0f);
Note that this requires a float argument 3.0f, to match the type of a. You could parametrise that type separately, if you want to allow conversions on the second argument:
template <int N, typename T, typename T2>
B<N,T> make_b(const A<N,T> & in, T2 scal) {
    return B<N,T>(in, scal);
}
auto b = make_b(a, 3.0);
                        No, since you create object of type B<int, T>.
With C++11 you can use auto and write simple creation function
template<int N, typename T>
B<N, T> create(const A<N, T>& in, T scal)
{
   return B<N, T>(in, scal);
}
// in main
auto b = create(a, 3.0f);
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