compostion of p-adic power series This question arises from $p$-adic Numbers: an Introduction written by Gouvea. It is problem 148 in the book.
Let $f(x)=1+x+x^2/2!+\cdots+ x^n/n!+\cdots$ be the exponential series as a formal power series and $g(x)=2x^2-2x$. Let $h(x)=f(g(x))=\sum_{n\geq 0} a_nx^n$ be their formal compostion (not necessarily the compostion as functions), which has rational coefficients.
How to prove that ${\rm ord}_2(a_n)\ge 1+ n/4$ when $n\ge 2$? (Here ${\rm ord}_2$ is the 2-adic valuation on $\mathbb{Q}_2$.)
This is an important part of the construction of a  counterexample that says that the value of a formal compostion series $f(g(x))$ where it converges sometimes may not agree with the compostion of the individual series $g(x)$ and $f(x)$ where they converge.
The first few terms of $h(x)$ look like this:
$h(x)=1-2x+4x^2-\frac{16}{3}x^3+\frac{20}{3}x^4-\frac{104}{15}x^5+\frac{304}{45}x^6\dots$
The author thinks this question is rather hard and he recommends  Cyclotomic Field I and II by Lang (Chapter 14, Section 2) to look for a hint about why this particular power series is interesting.
I have already written down the formula $a_n$.
$$a_n=(-1)^n\sum_{k=0}^{\lfloor n/2\rfloor}\frac{2^{n-k}}{k!(n-2k)!}$$
 A: Gouvea says the problem is hard because it is hard. It is the type of problem where the solution uses something far beyond what it takes to understand the statement of the problem alone. It is not reasonable to expect a full solution to be written here for you.
Here is something that at first does not look relevant. For two $p$-adic formal power series $A(x) = \sum a_nx^n$ and $B(x) = \sum b_nx^n$, set $C(x) = A(x)B(x) = \sum c_nx^n$ as a product of formal power series, so
$c_n = \sum_{j=0}^n a_jb_{n-j}$ for all $n \geq 0$. Letting $v = {\rm ord}_p$, show that if $v(a_n) \geq Kn+L$ and $v(b_n) \geq Kn + L$ for all $n$, where $K > 0$ and $L \geq 0$ then $v(c_n) \geq Kn+L$ for all $n$ too. Thus the space of $p$-adic power series $\sum a_nx^n$ whose coefficients tend to $0$ in absolute value at least as quickly as a “common geometric series” (meaning $|a_n|_p \leq (1/p)^{Kn+L} = (1/p)^L(1/p^K)^n$) is closed under multiplication.
The application of this idea to your example needs something not in the book, which is partly why Gouvea calls it a hard problem: the $2$-adic Artin-Hasse series
$$
{\rm AH}_2(x) = e^{x + x^2/2+x^4/4+x^8/8+\cdots},
$$
which despite the way it looks actually has coefficients in $\mathbf Z_2$, so the $2$-adic valuations of its coefficients are all nonnegative, unlike the coefficients of $e^x$ or even $e^{2x}$. (I am not saying the coefficients of this series tend to $0$ in the $2$-adics: they don’t, as infinitely many coefficients will be $2$-adic units, which is not obvious and also irrelevant.)
The series you call $h(x)$ is $e^{2x^2-2x}$. Let $E(x) = e^{x+x^2/2}$, so $h(x) = E(-2x)$. Note the roots of $x+x^2/2$ are $0$ and $-2$, so you would naively think $E(-2)$ is $e^0 = 1$, but it is $-1$, and likewise $h(1)$ is not $1$ but instead $-1$. To prove $E(x)$ converges at $-2$ or $h(x)$ converges at $1$, write
$$
E(x) = {\rm AH}_2(x)\prod_{k\geq 2} e^{-x^{2^k}/2^k},
$$
so replacing $x$ with $-2x$ turns this into
$$
h(x) = {\rm AH}_2(-2x)\prod_{k\geq 2}e^{-2^{2^k-k}x^{2^k}}.
$$
and use the result I mentioned at the start on all the series that are factors on the right side of this equation. To get a sense of why the right side is very good for you, the first series ${\rm AH}_2(-2x)$ converges at $1$ since its coefficients tend to $0$ in $\mathbf Q_2$ at least like $2^n$, and the factor at $k=2$ in the product is $e^{-4x^4}$, which also converges at $1$ since the coefficients of $e^{4x}$ tend to $0$ in $\mathbf Q_2$. The rate of convergence at $1$ only gets better with the factors at larger $k$.
That approach to the problem is hard, as Gouvea writes about the problem overall, but it does lead to success, in contrast to writing
$$
h(x) = e^{-2x+2x^2} = e^{-2x}e^{2x^2}
$$
since neither of the two factors on the right converges at $1$ in $\mathbf Q_2$.
Follow Gouvea’s advice to look at Lang’s book. Or you could look in Section $2$ of Chapter $7$ of Robert’s A Course in $p$-adic Analysis (especially the theorem on p. $396$) or here (especially Corollary $3.9$).
