if such $(2n+4)x_{n+2}=b(2n+3)x_{n+1}+2a(n+1)x_{n}$ show $x_{n}\in Z$

Question

let $a,b$ be integer,and such $a\equiv 0,b\equiv 0\pmod 4$,and sequece $x_{n}$,such $x_{0}=1,x_{1}=\dfrac{b}{2}$and such $$(2n+4)x_{n+2}=b(2n+3)x_{n+1}+2\cdot a\cdot(n+1)x_{n}$$ show that $$x_{n}\in Z$$

I'm not sure I was wrong： I try let $$f(t)=\sum_{n=0}^{+\infty}x_{n}t^n$$,then $$f'(t)=\sum_{n=0}^{+\infty}nx_{n}t^{n-1}=\sum_{n=0}^{+\infty}(n+1)x_{n+1}t^n$$,so we have $$\sum_{n=0}^{+\infty}2(n+2)t^{n+2}=2(tf'(t)-\dfrac{b}{2}t)$$ other hand we have $$\sum_{n=0}^{+\infty}2(n+2)t^{n+2}=\sum_{n=0}^{+\infty}[b(2n+3)x_{n+1}+2a(n+1)x_{n}]t^{n+2}=2bt^2f'(t)+bt(f(t)-1)+2a(tf'(t)+tf(t))$$ so we have $$2tf'(t)-bt=2bt^2f'(t)+btf(t)-bt+2atf'(t)+2atf(t)$$ so we have $$(2-2bt-2a)f'(t)=(b+2a)f(t)$$ so we have $$(ln(f(t)))'=\dfrac{b+2a}{2-2bt-2a}$$ I fell I have wrong

Answer 1

Let $A=\frac{a}{4}$ and $B=\frac{b}{4}$, so that both $A$ and $B$ are integers. One can check by induction that

$$ x_n=\sum_{t=0}^{\lfloor \frac{n}{2}\rfloor}\frac{2^t}{t!}\bigg(\prod_{k=1}^{t} 2n-2t+2k-1\bigg)\binom{2(n-2t)}{n-2t} A^tB^{n-2t} \tag{1} $$

so it will suffice to show that $\frac{2^t}{t!}\prod_{k=1}^{t} 2n-2t+2k-1$ is an integer, or in other words

$$ \nu_p(t!) \leq \nu_p(2^t)+\nu_p(\prod_{k=1}^{t} 2n-2t+2k-1) \tag{2} $$

for every prime $p$ (recall that for any $m$, $\nu_p(m)$ denotes the exponent of $p$ in the prime factorization of $m$).

Suppose first that $p\neq 2$. Then $\nu_p(2^t)=0$, and among the $t$ successive odd numbers $2n-2t+2k-1$ (for $1\leq k\leq t)$, at least $\lfloor \frac{t}{p} \rfloor$ are divisible by $p$, at least $\lfloor \frac{t}{p^2} \rfloor$ are divisible by $p^2$ etc, so that $\nu_p(\prod_{k=1}^{t} 2n-2t+2k-1) \geq \sum_{i=1}^{\infty} \lfloor \frac{t}{p^i} \rfloor=\nu_p(t!)$ by Legendre's formula. So (2) follows in this case.

Finally, suppose that $p=2$. Then $\nu_p(2^t)=t$, and to show (2) it suffices to show that $\nu_2(t!) \leq t$ ; but by Legendre's formula again,

$$ \nu_2(t!)=\sum_{i\geq1} \lfloor \frac{t}{2^i} \rfloor \leq \sum_{i\geq1} \frac{t}{2^i} = t $$

This finishes the proof.