<s>[INST] monotonicity from 4 term-recursion. In determining the monotonicity of coefficients in a series expansion (which appeared in one of my study), I come across the following problem. 
Let $p\ge 2$ be an integer, and $$6p^3(i+3)d_{i+3}=6p^2(i+2+p)d_{i+2}+3p(p-1)(i+1+2p)d_{i+1}+(p-1)(2p-1)(i+3p)d_i,~~i\ge0$$ with $d_0=d_1=d_2=1$. How to show $d_i>d_{i+1}$ for all $i\ge3$? 
By easy calculation, $d_3=1, d_4=\frac{18p^3+11p^2-6p+1}{24p^3}$. For a recursion of 2 or 3 terms, it is easy to proceed with induction, but what about 4 terms in this case?
 [/INST] I am a little bit upset that my previous answer to the question
was downvoted by somebody. As far as I understand the idea of MO is
not necessarily to produce final answers/solutions/responses but
mostly to provide the ideas of approaching a given problem.
It's a question of time to solve this particular problem (I still
wonder about where it comes from). Denote
$$
F(x)=\sum_{i=0}^\infty d_ix^i=1+x+x^2+x^3+\dots
$$
the generating function of the sequence. The given recursion can
be then written as the differential equation of order 1,
$$
\frac{F'}{F}=\frac{3p(2p^2+2p(p-1)x+(p-1)(2p-1)x^2)}{6p^3-6p^2x-3p(p-1)x^2-(p-1)(2p-1)x^3}
$$
which can be explicitly solved:
$$
F(x)=\biggl(1-\frac xp-\frac{(p-1)x^2}{2p^2}-\frac{(p-1)(2p-1)x^3}{6p^3}\biggr)^{-p}
$$
(one checks that the expansion indeed starts $1+x+x^2+x^3+\dots$).
The desired property $d_i>d_{i+1}$ for $i\ge3$ is equivalent to showing that
the expansion
$$
1-(1-x)F(x)=c_0x^4+c_1x^5+c_2x^6+\dots
$$
has all coefficients $c_0,c_1,c_2,\dots$ positive. In turn, the latter is
equivalent to the positivity of all coefficients in the expansion
$$
\frac{d}{dx}\bigl(1-(1-x)F(x)\bigr)=4c_0x^3+5c_1x^4+6c_2x^5+\dots.
$$
We have
$$
\frac{d}{dx}\bigl(1-(1-x)F(x)\bigr)
=\biggl(1-\frac xp-\frac{(p-1)x^2}{2p^2}-\frac{(p-1)(2p-1)x^3}{6p^3}\biggr)^{-p}
$$
$$
-p(1-x)\biggl(\frac1p+\frac{(p-1)x}p+\frac{(p-1)(2p-1)x^2}{2p^3}\biggr)
\cdot\biggl(1-\frac xp-\frac{(p-1)x^2}{2p^2}-\frac{(p-1)(2p-1)x^3}{6p^3}\biggr)^{-p-1}
$$
$$
=\frac{(p-1)(2p-1)(3p-1)}{6p^3}x^3
\cdot\biggl(1-\frac xp-\frac{(p-1)x^2}{2p^2}-\frac{(p-1)(2p-1)x^3}{6p^3}\biggr)^{-p-1}
$$
$$
=\frac{(p-1)(2p-1)(3p-1)}{6p^3}x^3
\cdot\Biggl(1+\sum_{n=1}^\infty\frac{(p+1)(p+2)\dots(p+n)}{n!}
\biggl(\frac xp+\frac{(p-1)x^2}{2p^2}+\frac{(p-1)(2p-1)x^3}{6p^3}\biggr)^n\Biggr),
$$
and the desired positivity follows.
</s>