Let \(P(x)=a_0x^d+a_1x^{d-1}+\dots +a_d\in \mathbb Q[x]\) with \(d\geq2\), \(a_0>0\) and such that \(a_0^{\frac 1{d-1}}\in\mathbb Q\). Let \(\{x_n\}_{n=0}^\infty\) be a sequence of positive integers such that \(x_{n+1}=P(x_n)\) for all \(n=0,1,\dots\) and \(\lim_{n\to\infty}x_n=\infty\). Set \(\alpha =\lim_{n\to\infty}x_n^{d^{-n}}\) and \(y_0=a_0^{\frac 1{d-1}}(x_0-\frac {a_1}{da_0})\). Then the author proves that the number \(\alpha >1\) is transcendental unless \(\frac 1{a_0}P(x-\frac {a_1}{da_0})+\frac {a_1}{da_0^2}=x^d\) (here \(\alpha =\mid y_0\mid \geq 2\) and \(\alpha\in\mathbb Z^+\)) or \(a_0^{\frac 1{d-1}}(P(a_0^{\frac {-1}{d-1}}-\frac {a_1}{da_0})+\frac {a_1}{da_0})=2T_d(\frac x2)\) (here \(\alpha =\frac {\mid y_0\mid}2+\sqrt {\frac {y_0^2}4-1})\), \(\mid y_0\mid \geq 3\), \(\mid y_0\mid\in\mathbb Z^+\)), where \(T_d(x)\) is the Chebyshev polynomial of the first kind of degree \(d\).