Algebraic independence of the generating functions of the stern polynomials and their twisted analogues (Q494603)

Let \(\{ a_n(x)\}_{n=0}^\infty\) and \(\{ b_n(x)\}_{n=0}^\infty\) be two recurrent sequences of polynomials such that \(a_0(x)=b_0(x)=0\), \(a_1(x)=b_1(x)=1\) and \(a_{2n}(x)=a_n(x^2)\), \(b_{2n}(x)=-b_n(x^2)\), \(a_{2n+1}(x)=xa_n(x^2)+ a_{n+1}(x^2)\), \(b_{2n+1}(x)=-(xb_n(x^2)+ b_{n+1}(x^2))\) for all \(n\in\mathbb Z^+\). Set \[ A(x,z)=\sum_{n=0}^\infty a_{n+1}(x)z^n, \] \[ B(x,z)=\sum_{n=0}^\infty b_{n+1}(x)z^n\, \text{ and}\, P(x,z)=1+z+xz^2. \] Let \(\mathbb D\) is an open unit disk. Then for \(\alpha,\beta\in \overline {\mathbb Q}^\times\cap \mathbb D\) we have that \(P(\alpha^{2^k},\beta^{2^k})\not= 0\) holds for any \(k\in\mathbb N_0\) if and only if the numbers \(A(\alpha,\beta)\) and \(B(\alpha,\beta)\) are algebraically independent over \(\mathbb Q\). The proof is in the spirit of Mahler.