Consider an arithmetic function $g$ with codomain $\{a,b\}$ and a function $f$ which is analytic on some domain including $\{a,b\}$. We therefore have $$f(g(n))=\sum_{k=0}^\infty c_k (g(n)-a)^k$$ and noting that $(g(n)-a)^k=(b-a)^{k-1}(g(n)-a)$, we have $$f(g(n))=\sum_{k=0}^\infty c_k (b-a)^{k-1} (g(n)-a)=\frac{g(n)-a}{b-a}f(b).$$ This is clearly false; taking for example $f=\rm{Id}$ and $g(n)=(-1)^n$ gives $(-1)^n=\frac{(-1)^n+1}2$ and therefore $(-1)^n=1$. However, I can't find the error. Suggestions?
(I wasn't quite sure how to tag this; edits appreciated.)
The error is at $k = 0$, you can't write $c_0 (g(n)-a)^0 = c_0$ as $\dfrac{g(n)-a}{b-a}\cdot c_0$ unless $g(n) = b$ or $c_0 = 0$. We thus have
\begin{align} f(g(n)) &= c_0 + \frac{g(n)-a}{b-a}\sum_{k = 1}^\infty c_k (b-a)^k\\ &= c_0 + \frac{g(n)-a}{b-a}\bigl(f(b) - c_0\bigr)\\ &= f(a) + \frac{g(n)-a}{b-a}\bigl(f(b) - f(a)\bigr). \end{align}