[数列] 一个多项式序列

hbghlyj

多项式序列$\{f_k(n)\}$满足$\forall k\in\mathbf N_+,f_k(n)=nf_{k-1}(n)+f_k(n-1),f_k(1)=f_0(n)=1$,求证:
(1)$\forall k\in\mathbf N_+,\forall n\in\mathbf Z,f_k(n)\in\mathbf Z$
(2)k>1,若k为偶数，则$f_k(n)=\left(
\begin{array}{c}
n+k \\
k+1 \\
\end{array}
\right)\frac{g_k(n)}{g_k(1)}$;若k为奇数，则$f_k(n)=\left(
\begin{array}{c}
n+k \\
k+1 \\
\end{array}
\right)\left(
\begin{array}{c}
n+1 \\
2 \\
\end{array}
\right)\frac{g_k(n)}{g_k(1)}$,其中$g_k(n)$为$\mathbf Z$上的不可约多项式
(3)$g_{2k}$与$g_{2k+1}$的常数项相等
(4)$g_{2k}$的首项系数是$g_{2k+1}$的首项系数的2k+1倍
$f_1=\left(
\begin{array}{c}
n+1 \\
2 \\
\end{array}
\right)$
$g_2=3 n+{\color{red}{1}}$
$g_3={\color{red}{1}}$
$g_4=15 n^3+30 n^2+5 n-{\color{red}{2}}$
$g_5=3 n^2+7 n-{\color{red}{2}}$
$g_6=63 n^5+315 n^4+315 n^3-91 n^2-42 n+{\color{red}{16}}$
$g_7=9 n^4+54 n^3+51 n^2-58 n+{\color{red}{16}} $
$g_8=135 n^7+1260 n^6+3150 n^5+840 n^4-2345 n^3+540 n^2+404 n-{\color{red}{144}}$
$g_9=15 n^6+165 n^5+465 n^4-17 n^3-648 n^2+548 n-{\color{red}{144}}$
$g_{10}=99 n^9+1485 n^8+6930 n^7+8778 n^6-8085 n^5-8195 n^4+11792 n^3-2068 n^2-2288 n+{\color{red}{768}}$
$g_{11}=9 n^8+156 n^7+834 n^6+1080 n^5-1927 n^4-1252 n^3+4156 n^2-3056 n+{\color{red}{768}}$
$g_{12}=12285 n^{11}+270270 n^{10}+2027025 n^9+5495490 n^8+315315 n^7-12882870 n^6+5760755 n^5+14444430 n^4-15875860 n^3+2037672 n^2+3327584 n-{\color{red}{1061376}}$
$g_{13}=945 n^{10}+23625 n^9+201600 n^8+609210 n^7-113715 n^6-2207175 n^5+1817786 n^4+3161188 n^3-6544568 n^2+4388960 n-{\color{red}{1061376}}$
$g_{14}=405 n^{13}+12285 n^{12}+135135 n^{11}+621621 n^{10}+765765 n^9-1898325 n^8-2141139 n^7+6565559 n^6-990990 n^5-8790964 n^4+8132904 n^3-712672 n^2-1810176 n+{\color{red}{552960}}$
$g_{15}=27 n^{12}+918 n^{11}+11367 n^{10}+58794 n^9+76341 n^8-267246 n^7-298891 n^6+1285734 n^5-506956 n^4-2295480 n^3+3773216 n^2-2363136 n+{\color{red}{552960}}$
$g_{16}=6885 n^{15}+275400 n^{14}+4176900 n^{13}+28959840 n^{12}+78108030 n^{11}-47258640 n^{10}-413804820 n^9+379236000 n^8+1219182965 n^7-2258929400 n^6-253628440 n^5+3561649600 n^4-2900210608 n^3+149388928 n^2+679395072 n-{\color{red}{200005632}}$
$g_{17}=405 n^{14}+17955 n^{13}+303345 n^{12}+2352735 n^{11}+7068495 n^{10}-5795535 n^9-51527445 n^8+57167365 n^7+204007120 n^6-476029832 n^5+22886128 n^4+1043666128 n^3-1482037376 n^2+879400704 n-{\color{red}{200005632}}$
$g_{18}=161595 n^{17}+8241345 n^{16}+164826900 n^{15}+1600652340 n^{14}+7166307330 n^{13}+6261590790 n^{12}-46058007996 n^{11}-41937579684 n^{10}+286640389035 n^9-66589929679 n^8-997871657328 n^7+1358814133992 n^6+518597683632 n^5-2516039461488 n^4+1850554339200 n^3-34693350656 n^2-453757851648 n+{\color{red}{129369047040}}$
$g_{19}=8505 n^{16}+476280 n^{15}+10512180 n^{14}+113291136 n^{13}+566161974 n^{12}+547945776 n^{11}-4735037412 n^{10}-5204146752 n^9+38129744457 n^8-8170018024 n^7-180801912344 n^6+299422683008 n^5+89907082768 n^4-803164494208 n^3+1027247031040 n^2-583126898688 n+{\color{red}{129369047040}}$

hbghlyj

进展如下：第(1)问证完了，第(2)问证明了整除性，但不会证不可约性，第(3)问茫然不知从何处下手....请坛友指点迷津!

hbghlyj

(1)$f_k(n)=nf_{k-1}(n)+f_k(n-1),f_k(1)=1\Rightarrow f_k(n)=\sum\limits_{i=1}^nif_{k-1}(i)\Rightarrow f_k(n)\in\mathbf Z$
(3) 令$q_k(n)=\frac{g_k(n)}{g_k(1)}$,$f_{2k+1}(n)=nf_{2k}(n)+f_{2k+1}(n-1)\Rightarrow (n+2k+1)(n+1)q_{2k+1}(n)-(n-1)^2q_{2k+1}(n-1)=2(2k+2)q_{2k}(n)$,令n=0,$(2k+1)q_{2k+1}(0)-q_{2k+1}(-1)=2(2k+2)q_{2k}(0)$,为证明$q_{2k+1}(0)=(2k+2)q_{2k}(0)$,只需证明$(2k-1)q_{2k+1}(0)=q_{2k+1}(-1)$,然后怎么做？