How to prove all $c_{n},d_{n}$ to be integers if $(n+1)c_{n}=nc_{n-1}+2nd_{n-1}$ and $d_{n}=2c_{n-1}+d_{n-1}$? Let sequences $(c_n)$ and $(d_n)$ be given by
$$c_0=0,\:d_0=1$$
and recursively for $n\ge 1$ by
$$\begin{align}
c_n & =\frac{n}{n+1}c_{n-1}+\frac{2n}{n+1}d_{n-1} \\[2ex]
d_n & =2c_{n-1}+d_{n-1}
\end{align}$$
I'd like to show that all $c_{n},d_{n}$ are integers. (Creat by wang yong xi）
My try： Since 
$$\begin{align}(n+1)c_n & =nc_{n-1}+2nd_{n-1}\\[1ex]
d_n & =2c_{n-1}+d_{n-1}
\end{align}$$
we easily find $$c_{1}=1,\:d_{1}=1,\\
c_{2}=2,\:d_{2}=3,\\
c_{3}=6,\:d_{3}=7,$$
a.s.o.   How to prove that all the $c_{n},d_{n}$ are integers?
 A: COMMENT.- (Only as information for beginners) Yes, as suggested by lhf and Alex Ravsky, it is practically evident that the $(c_n, d_n)$ correspond respectively to the OEIS-sequences A0057 and A002426 (see comments above). The calculation of the first 10 pairs gives
$(0,1),(1,1),(2,3),(6,7),(16,19),(45,51),(126,141),(357,393),(1016,1107),(2907,3139),(8350,8953)$ 
what is in line with the aforementioned sequences whose integers respond to various interesting problems (geometric, numeric, combinatorial,etc).
In particular the numbers $d_n$ are solution of the recurrence of order two
$$nd_n=(2n-1)d_{n-1}+3(n-1)d_{n-2}$$ whose generating function is
$$f(x)=\dfrac{1}{\sqrt{(1+x)(1-3x)}}$$ so one has $$f(x)=1+x+3x^2+7x^3+19x^4+51x^5+141x^6+393x^7+1107x^8+3139x^9+8953x^{10}+\cdots$$
Who is not familiar with these methods, can see that the coefficients of the development of the generating function $f$ are exactly the successive values of $d_n$ which we have checked here up to $n = 10$ but this applies to any value of $n$.
A: Consider the polynomials $(1+x+x^2)^n$; the only coefficients that will be non zero are terms whose $x$ exponent is between $0$ and $2n$. Note two other things; firstly they will all be whole numbers and secondly they satisfy $a_{i}= a_{2n-i}$ (i.e they will be symmetrical around the central term).
Now we have define sequences $d_n$ to be the central coefficient and $c_n$ to $1$ off central. So we have
\begin{eqnarray*}
(1+x+x^2)^{n-1} =1+  \cdots +c_{n-1}x^{n-2}+ d_{n-1} x^{n-1} +c_{n-1} x^{n} + \cdots +x^{2n-2} \\
(1+x+x^2)^n =1+  \cdots +c_{n}x^{n-1}+ d_{n} x^{n} +c_{n} x^{n+1} + \cdots +x^{2n}. \\
\end{eqnarray*}
Multiply the first equation by $(1+x+x^2)$ and consider the coefficient of $x^{n}$ and we have
\begin{eqnarray*}
d_n=2c_{n-1}+d_{n-1}.
\end{eqnarray*}
Next differentiate both sides of the second equation 
\begin{eqnarray*}
n(1+2x)(1+x+x^2)^{n-1} =  \cdots +(n-1)c_{n}x^{n-2}+ nd_{n} x^{n-1} +(n+1)c_{n} x^{n} +\cdots +2nx^{2n-1} \\
\end{eqnarray*}
use the first equation and consider the coefficient $x^{n}$ and we have
\begin{eqnarray*}
(n+1)c_n=nc_{n-1}+2_nd_{n-1}.
\end{eqnarray*}
It suffices to show that the sequences $c$ and $d$ satisfy the initial conditions and remind ourselves of the observation made earlier that all of these values are  whole numbers.
