<s>[INST] Prove the series has positive integer coefficients How can I show that the Maclaurin series for
$$
\mu(x) = (x^4+12x^3+14x^2-12x+1)^{-1/4}
\\
= 1+3\,x+19\,{x}^{2}+147\,{x}^{3}+1251\,{x}^{4}+11193\,{x}^{5}+103279\,
{x}^{6}+973167\,{x}^{7}+9311071\,{x}^{8}+\cdots
$$
has positive integer coefficients?  (I have others to do, too, but this one will be a start.)
possibilities
(a) The coefficients $Q(n)$ satisfy the recurrence
$$
(n+1)Q(n)+(12n+21)Q(n+1)+(14n+35)Q(n+2)+(-12n-39)Q(n+3)+(n+4)Q(n+4) = 0
$$
(b) $\mu(x)$ satisfies the differential equation
$$
(x^3+9x^2+7x-3)\mu(x)+(x^4+12x^3+14x^2-12x+1)\mu'(x)=0
$$
(c) factorization of $x^4+12x^3+14x^2-12x+1$ is
$$
\left( x-\sqrt {5}+3+\sqrt {15-6\,\sqrt {5}} \right)  \left( x-\sqrt 
{5}+3-\sqrt {15-6\,\sqrt {5}} \right)  \left( x+\sqrt {5}+3-\sqrt {15+
6\,\sqrt {5}} \right)  \left( x+\sqrt {5}+3+\sqrt {15+6\,\sqrt {5}}
 \right) 
$$
(d) $(1-8X)^{-1/4}$ has positive integer coefficients, But if $X$ is defined by $1-8X=x^4+12x^3+14x^2-12x+1$, then $X$ does not have integer coefficients.
(e) Can we compute the series for $\log\mu(x)$
$$3\,x+{\frac {29}{2}}{x}^{2}+99\,{x}^{3}+{\frac {3121}{4}}{x}^{4}+{
\frac {32943}{5}}{x}^{5}+{\frac {348029}{6}}{x}^{6}+\dots
$$
and then recognize that its exponential has integer coefficients?
 [/INST] Here is a proof for positivitiy.  Use the recurrence
$$
(n+1)Q(n)+(12n+21)Q(n+1)+(14n+35)Q(n+2)+(-12n-39)Q(n+3)+(n+4)Q(n+4) = 0
$$
$Q(0)=1,Q(1)=3,Q(2)=19,Q(3)=147$.  Prove by induction that
$Q(n) \ge 3 Q(n-1)$ for $n \ge 1$.  This is true for the first few terms.
Assume true up to $Q(n+3)$, then prove it for $Q(n+4)$ as follows:
$$
(n+4)Q(n+4) = (12n+39)Q(n+3)-(14n+35)Q(n+2)-(12n+21)Q(n+1)-(n+1)Q(n)
\\
\ge \left[(12n+39)-\frac{14n+35}{3}-\frac{12n+21}{9}-\frac{n+1}{27}\right]Q(n+3)
\\
=\left[\frac{161}{27} n + \frac{692}{27}\right] Q(n+3)
\gt (3n+12)Q(n+3)=(n+4)3Q(n+3)
$$
and therefore $Q(n+4) > 3 Q(n+3)$.
</s>