Let the continued fraction be
\[
a_0=3,\quad a_n=n^4+(n+1)^4+2\bigl(n^2+(n+1)^2\bigr),\quad b_n=-n^8.
\]
Recurrence: \(P_{-1}=1,P_0=3\), \(P_n=a_nP_{n-1}+b_nP_{n-2}\).
Compute the first few terms:
\[
P_1=80,\; P_2=9072,\; P_3=2985984,\; P_4=2280960000,\; P_5=3493601280000,\dots
\]
Let \(P_n=(n!)^4 R_n\). Then \(R_0=3, R_1=80, R_2=567, R_3=2304, R_4=6875, R_5=16848, R_6=36015,\dots\)
The fifth differences are constant at 240, so \(R_n\) is a quintic polynomial. Solving for coefficients gives
\[
R_n=2n^5+11n^4+24n^3+26n^2+14n+3,
\]
which factorizes as
\[
R_n=(n+1)^4(2n+3).
\]
Direct verification shows this \(R_n\) satisfies the recurrence, hence
\[
\boxed{P_n=(n!)^4(n+1)^4(2n+3)}.
\]
Let \(y_n=Q_n/P_n\). From the continued fraction identity
\[
P_nQ_{n-1}-P_{n-1}Q_n=(-1)^{n-1}b_1\cdots b_n=-(n!)^8,
\]
dividing by \(P_nP_{n-1}\) yields
\[
y_n-y_{n-1}=\frac{(n!)^8}{P_nP_{n-1}}.
\]
Substituting \(P_n\) gives
\[
z_n:=y_n-y_{n-1}= \frac{1}{(n+1)^4(2n+1)(2n+3)}.
\]
Initial value \(y_0=1/3\). Let \(m=n+1\); then for \(n\ge1\) we have \(m\ge2\),
\[
y_n=\frac13+\sum_{m=2}^{n+1}\frac{1}{m^4(2m-1)(2m+1)}.
\]
Let \(n\to\infty\); we need the series
\[
S=\sum_{m=2}^\infty\frac{1}{m^4(2m-1)(2m+1)}.
\]
Partial fraction decomposition:
\[
\frac{1}{m^4(4m^2-1)}=\frac{16}{4m^2-1}-\frac{4}{m^2}-\frac{1}{m^4}.
\]
Thus
\[
S=16\sum_{m=2}^\infty\frac{1}{4m^2-1}-4\sum_{m=2}^\infty\frac{1}{m^2}-\sum_{m=2}^\infty\frac{1}{m^4}.
\]
We know
\[
\sum_{m=2}^\infty\frac{1}{4m^2-1}=\frac16,\qquad
\sum_{m=2}^\infty\frac{1}{m^2}=\zeta(2)-1,\qquad
\sum_{m=2}^\infty\frac{1}{m^4}=\zeta(4)-1.
\]
Hence
\[
S=\frac{8}{3}-4\zeta(2)+4-\zeta(4)+1=\frac{23}{3}-4\zeta(2)-\zeta(4).
\]
Then
\[
y=\lim_{n\to\infty}y_n=\frac13+S=8-4\zeta(2)-\zeta(4).
\]
\[
C=\frac1y=\frac{1}{8-4\zeta(2)-\zeta(4)}=-\frac{1}{\zeta(4)+4\zeta(2)-8}.
\]
The right-hand side is the desired continued fraction. QED.
