孫　蘋(píng), 胡　宏

(1.寧夏大學(xué) 數(shù)學(xué)與計(jì)算機(jī)學(xué)院, 寧夏 銀川　750021; 2.淮陰師范學(xué)院 數(shù)學(xué)科學(xué)學(xué)院, 江蘇 淮安　223300)

涉及二階線(xiàn)性遞歸序列的兩類(lèi)多項(xiàng)式的因式分解

孫蘋(píng)1,2, 胡宏2

(1.寧夏大學(xué) 數(shù)學(xué)與計(jì)算機(jī)學(xué)院, 寧夏 銀川750021; 2.淮陰師范學(xué)院 數(shù)學(xué)科學(xué)學(xué)院, 江蘇 淮安223300)

摘要:定義了與二階線(xiàn)性遞歸序列{wn}相關(guān)的序列{di,j}和{},及與序列{wn},{di,j}和{}相關(guān)的多項(xiàng)式 rn(x),ln(x),tn(x) 和,根據(jù){wn}的遞推關(guān)系和相關(guān)性質(zhì),研究了{(lán)di,j}和{}的相關(guān)性質(zhì),得到了一系列關(guān)于ln(x),tn(x)和的多項(xiàng)式的因式分解.

關(guān)鍵詞:二階線(xiàn)性遞歸序列; 多項(xiàng)式; 因式分解

0引言

二階線(xiàn)性遞歸序列{wn(a,b;P,Q)}是指滿(mǎn)足如下遞推關(guān)系的序列:

wn+1=Pwn-Qwn-1,w0=a,w1=b,n=1,2,3,…

(1)

一般地,記wn(a,b;P,Q)=wn,當(dāng)a=0,b=1時(shí),記wn(0,1;P,Q)=un(P,Q),簡(jiǎn)記為un.當(dāng)a=2,b=P時(shí),記wn(2,P;P,Q)=vn(P,Q),簡(jiǎn)記為vn.當(dāng)a=0,b=1,P=1,Q=-1時(shí),記wn(0,1;1,-1)=Fn,Fn就是著名的Fibonacci數(shù).當(dāng)a=2,b=1,P=1,Q=-1時(shí),記wn(2,1;1,-1)=Ln,Ln就是著名的Lucas數(shù)[1].當(dāng)a=0,b=1,P=1,Q=-2時(shí),記wn(0,1;1,-2)=Jn,Jn就是Jacobsthal數(shù).當(dāng)a=2,b=1,P=1,Q=-2時(shí),記wn(2,1;1,-2)=jn,jn是Jacobsthal-Lucas數(shù)[2].當(dāng)a=0,b=1,P=2,Q=-1時(shí),記wn(0,1;2,-1)=Pn,Pn就是Pell數(shù).當(dāng)a=2,b=2,P=2,Q=-1時(shí),wn(2,2;2,-1)=Qn,Qn是Pell-Lucas數(shù)[3].

文[4]中,令

(2)

(3)

vn(x)=m(n,1)xn+m(n,2)xn-1+…+m(n,n+1)

(4)

(5)

得到了多項(xiàng)式vn(x)和wn(x)的因式分解:

v2k+1(x)=(xF2k+1+F2k+2)(F2x2k+F4x2k-2+…+F2kx2+F2k+2)

(6)

v2k(x)-F2kF2k+2=x(xF2k+F2k+1)(F2x2k-2+F4x2k-4+…+F2k)

(7)

(8)

w2k(x)=(xF2k+1+F2k+2)(F2x2k-2+F4x2k-4+…+F2k)

(9)

(10)

(11)

rn(x)=w1xn+w2xn-1+…+wn+1,n=1,2,…

(12)

ln(x)=wn+1xn-Qwnxn-1+…+(-Q)nw1,n=1,2,…

(13)

tn(x)=dn,1xn+dn,2xn-1+…+dn,n+1,n=1,2,…

(14)

(15)

特別地, 當(dāng)i,h為正整數(shù)且h≥2時(shí),易得

di,i+h=Pdi,i+h-1-Qdi,i+h-2

(16)

(17)

1相關(guān)引理

由式(1)和(10)可得

引理1設(shè)n為任意的正整數(shù),則

dn,n=wnwn+1-(-Q)nab

(18)

引理2設(shè)n為任意的正整數(shù),則

(19)

引理3設(shè)i,j為任意的正整數(shù),i≤j時(shí),有

(20)

證由式(18)和(19)知j=i或i+1時(shí),式(20)成立.

假設(shè)當(dāng)i≤j

由歸納法知式(20)成立.由式(10)和(11) 可得

引理4設(shè)i,j為任意的正整數(shù),則

(21)

2主要結(jié)果

定理1設(shè)k為任意正整數(shù),則

(22)

(23)

證令n維列向量Xn-1=(xn-1,xn-2,…,x,1)T, 當(dāng)k≥1時(shí)有

t2k+1(x)=(d2k+1,1,d2k+1,2,…,d2k+1,2k+1,d2k+1,2k+2)X2k+1,

由式(10)和(20)可得

t2k+1(x)=(d1,2k+1,d2,2k+1,…,d2k+1,2k+1,d2k+1,2k+2)X2k+1=

(w2w2k+1+aQw2k+1,w2w2k+2-aQ2w2k,w4w2k+1+aQ3w2k-1,…,

w2k+2w2k+1+aQ2k+1w1,w2k+2w2k+2+aQ2k+1w2)X2k+1=

w2x2k(xw2k+1+w2k+2)+w4x2k-2(xw2k+1+w2k+2)+…+

w2k+2(xw2k+1+w2k+2)+aQx(w2k+1x2k-Qw2kx2k-1+…+Q2kw1)+aQ2k+1w2=

類(lèi)似地

t2k(x)=(d2k,1,d2k,2,…,d2k,2k,d2k,2k+1)X2k,

由式(10)和(20)可得

t2k(x)=(d1,2k,d2,2k,…,d2k,2k,d2k,2k+1)X2k=

(w2w2k+aQw2k,w2w2k+1-aQ2w2k-1,w4w2k+aQ3w2k-2,…,

w2kw2k+aQ2k-1w2,w2kw2k+1-aQ2kw1,w2kw2k+2-aQ2kw2)X2k=

w2x2k-1(xw2k+w2k+1)+w4x2k-3(xw2k+w2k+1)+…+w2kx(xw2k+w2k+1)

+w2kw2k+2-aQ2kw2+aQx(w2kx2k-1-Qw2k-1x2k-2+…-Q2k-1w1)=

綜上,定理1得證.

注1文[4]中定理4.1就是在定理1中取a=0,b=1,P=1,Q=-1時(shí)的特殊情況.

定理2設(shè)k為任意正整數(shù),則

(24)

(25)

證令n維列向量

由式(10),(20)和(21)可得

(w2w2k+2+aQw2k+2,w2w2k+3-aQ2w2k+1,w4w2k+2+aQ3w2k,…,

w2x2k(xw2k+2+w2k+3)+w4x2k-2(xw2k+2+w2k+3)+…+

aQ(w2k+2x2k+1-Qw2k+1x2k+…-Q2k-1w3x2+Q2kw2x-Q2k+1w1)+abQ2k+2=

類(lèi)似地

由式(10),(20)和(21)可得

w2x2k-1(xw2k+1+w2k+2)+w4x2k-3(xw2k+1+w2k+2)+…+

w2kx(xw2k+1+w2k+2)+aQ(w2k+1x2k-Qw2kx2k-1+…-Q2k-1w2x)=

aQ(w2k+1x2k-Qw2kx2k-1+…-Q2k-1w2x+Q2kw1)-abQ2k+1=

綜上定理2得證.

注2文[4]中定理4.2就是在定理2中取a=0,b=1,P=1,Q=-1時(shí)的特殊情況.

3相關(guān)推論

在式(22)～(25)中,當(dāng)a=0,b=1時(shí),可得

推論1設(shè)n為正整數(shù)時(shí),有

(u2n+1x+u2n+2)(u2x2n+u4x2n-2+…+u2nx2+u2n+2).

x(u2nx+u2n+1)(u2x2n-2+u4x2n-4+…+u2n-2x2+u2n).

推論2設(shè)n為正整數(shù)時(shí),有

在式(22)～(25)中,當(dāng)a=2,b=P時(shí),可得

推論3設(shè)n為正整數(shù)時(shí),有

(v2n+1x+v2n+2)(v2x2n+v4x2n-2+…+v2nx2+v2n+2).

2Q2nv2-v2nv2n+2=x(v2nx+v2n+1)(v2x2n-2+v4x2n-4+…+v2n-2x2+v2n).

推論4設(shè)n為正整數(shù)時(shí),有

x2(v2n+2x+v2n+3)(v2x2n-2+v4x2n-4+…+v2n-2x2+v2n).

x(v2n+1x+v2n+2)(v2x2n-2+v4x2n-4+…+v2n-2x2+v2n).

在式(22)～(25)中,當(dāng)a=0,b=1,P=1,Q=-2時(shí),可得

推論5設(shè)n為正整數(shù)時(shí),有

(J2n+1x+J2n+2)(J2x2n+J4x2n-2+…+J2nx2+J2n+2).

x(J2nx+J2n+1)(J2x2n-2+J4x2n-4+…+J2n-2x2+J2n).

推論6設(shè)n為正整數(shù)時(shí),有

在式(22)～(25)中,當(dāng)a=0,b=1,P=2,Q=-1時(shí),可得

推論7設(shè)n為正整數(shù)時(shí),有

(P2n+1x+P2n+2)(P2x2n+P4x2n-2+…+P2nx2+P2n+2).

x(P2nx+P2n+1)(P2x2n-2+P4x2n-4+…+P2n-2x2+P2n).

推論8設(shè)n為正整數(shù)時(shí),有

在式(22)～(25)中,當(dāng)a=2,b=1,P=1,Q=-1時(shí),可得

推論9設(shè)n為正整數(shù)時(shí),有

(L2n+1x+L2n+2)(L2x2n+L4x2n-2+…+L2nx2+L2n+2).

x(L2nx+L2n+1)(L2x2n-2+L4x2n-4+…+L2n-2x2+L2n).

推論10設(shè)n為正整數(shù)時(shí),有

x2(L2n+2x+L2n+3)(L2x2n-2+L4x2n-4+…+L2n-2x2+L2n).

x(L2n+1x+L2n+2)(L2x2n-2+L4x2n-4+…+L2n-2x2+L2n).

在式(22)～(25)中,當(dāng)a=2,b=1,P=1,Q=-2時(shí),可得

推論11設(shè)n為正整數(shù)時(shí),有

(j2n+1x+j2n+2)(j2x2n+j4x2n-2+…+j2nx2+j2n+2).

x(j2nx+j2n+1)(j2x2n-2+j4x2n-4+…+j2n-2x2+j2n).

推論12設(shè)n為正整數(shù)時(shí),有

x2(j2n+2x+j2n+3)(j2x2n-2+j4x2n-4+…+j2n-2x2+j2n).

x(j2n+1x+j2n+2)(j2x2n-2+j4x2n-4+…+j2n-2x2+j2n).

在式(22)～(25)中,當(dāng)a=2,b=2,P=2,Q=-1時(shí),可得

推論13設(shè)n為正整數(shù)時(shí),有

(Q2n+1x+Q2n+2)(Q2x2n+Q4x2n-2+…+Q2nx2+Q2n+2).

x(Q2nx+Q2n+1)(Q2x2n-2+Q4x2n-4+…+Q2n-2x2+Q2n).

推論14設(shè)n為正整數(shù)時(shí),有

x2(Q2n+2x+Q2n+3)(Q2x2n-2+Q4x2n-4+…+Q2n-2x2+Q2n)

x(Q2n+1x+Q2n+2)(Q2x2n-2+Q4x2n-4+…+Q2n-2x2+Q2n).

參考文獻(xiàn)：

[1]Zhao F Z,Wang T M.Some identities involving the Fibonacci and Lucas numbers[J].Ars Combinatoria,2004(72):311-318.

[2]張慧婷,王軍霞.有關(guān)Bell多項(xiàng)式與Jacobsthal數(shù)的恒等式[J].甘肅科學(xué)院學(xué)報(bào),2010,22(1):39-42.

[3]賈彥益,楊勝良.Pell-Lucas矩陣及其應(yīng)用[J].甘肅科學(xué)院學(xué)報(bào),2010,22(4):13-17.

[4]Clare Kimberling.Fusion,Fission and Factors[J].The Fibonacci Quarterly,2014,52(3):195-202.

[責(zé)任編輯：李春紅]

The Decomposition of two Kinds of Polynomials Involving 2-order Linear Recursive Sequence

SUN Ping1,2, HU Hong2

(1.School of Mathematics and Computer Science, Ningxia University, Yinchuan Ningxia 75000, China)(2.School of Mathematical Science, Huaiyin Normal University, Huaian Jiangsu 223300, China)

Abstract:In this paper,new sequences{di,j}and{i,j},and their related 2-order linear recursive sequence{wn}were defined,and some polynomials rn(x),ln(x),tn(x)andn(x)associated with {wn},{di,j}and{i,j}were defined.According to the recurrence relation and related properties of sequence{wn} and we studied the properties of{di,j}and{i,j},a series of ln(x),tn(x) andn(x) polynomial factorization were obtained.

Key words:2-order linear recursive sequences; polynomial; decomposition

收稿日期:2016-02-12

通訊作者:胡宏(1967-), 女, 江蘇金湖人, 教授, 研究方向?yàn)閿?shù)論與組合數(shù)學(xué). E-mail: hysyhh@163.com

中圖分類(lèi)號(hào):O157

文獻(xiàn)標(biāo)識(shí)碼:A

文章編號(hào):1671-6876(2016)02-0104-06