J. Korean Math. Soc. 39 (2002), No. 4, pp. 621–651 (f, 2)–ROTATIONAL EXTENDED STEINER TRI

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(延世大韩国语教程4)39

延世大学韩国语教程4 第39课单词语法课文讲解(1)【제39과신문보고알았어요】1. 생활과신문生活与报纸【단어】1. 붐비다: 拥挤，杂乱，忙乱。

한산하다. 冷清，僻静。

감기가유행일때는사람이붐비는곳에가지마라.流行感冒的时候不要去人群拥挤的地方。

서울역에사람들이붐벼서아이를찾기가힘들었다. 首尔车站人群拥挤，不好找孩子。

2. 충격적: 令人震惊的。

충격요법. 电击疗法。

충격사. 休克致死。

충격을주다(받다). （受到）震动。

대통령이총에맞아살해당한것은국민들에게대단히충격적인사건이었다.总统中枪被杀是令国民非常震惊的事件。

충격적인소식을듣고그녀는정신을잃었다.听到了令人震惊的消息后，她精神失常了。

3. 흥미: 兴趣。

흥미있다. 有兴趣。

흥미를끌다. 引起兴趣。

흥미롭다. 有兴趣。

우리아이는공부에흥미가없나봐요. 我们家孩子好像对学习没兴趣。

요즘인기있는주말연속극이어떻게결론지어질지무척흥미롭다.最近受人欢迎的周末连续剧怎么结尾非常令人感兴趣。

4. 단: 只，只是，单。

단신. 单身。

단독. 单独。

단짝. 配角，搭档。

담배를단하루만이라도끊어보라는날이바로금연의날이다.只说戒烟一天的那天正是戒烟日。

이순신장군은거북선단한척으로적의배들을물리쳤다.李舜臣将军只用一艘龟船就击退了敌船。

【문법】1. -ㅁ/-음单纯名词形词尾“-ㅁ/-음”的意义①“-ㅁ/-음”使谓词或者以谓词为中心的词组、主谓结构的子句所表达的意思事物化、概念化，作句子的某一个成分。

국가의법률을지킴은매개공민의의무다. 遵守国家法律是每个公民的义务。

②“-ㅁ/-음”用于动词词干后，使动词具有名词的功能，大体上表现为动作或状态已经完结的具体的事实。

“분명하다,마땅하다, 당연하다, 확실하다, 이상하다, 옳다, 그르다”等形容词和“밝혀지다, 알려지다, 알다, 모르다, 기억하다, 짐작하다, 배우다, 후회하다, 깨닫다”等动词以及“사실이다, 잘못이다, 불만이다,유감이다”等体词谓词形作谓语时，采用“-ㅁ/-음”名词形。

AANA随机变量序列的强收敛性

AANA随机变量序列的强收敛性杨英钟;王宽程【摘要】利用截尾方法,研究在矩条件下,AANA随机变量序列的强收敛性,由于AANA随机变量序列比NA随机变量序列要弱,故本文所得的结论对NA随机变量序列仍然成立.【期刊名称】《贵阳学院学报（自然科学版）》【年(卷),期】2018(013)004【总页数】3页(P14-16)【关键词】AANA序列;强收敛性;三级数定理【作者】杨英钟;王宽程【作者单位】闽南理工学院,福建泉州362700;闽南理工学院,福建泉州362700【正文语种】中文【中图分类】O211.41 引言定义1[1] 称Xn;n∈N为渐近几乎负相依(简称AANA)随机变量序列，如果存在非负序列q(n)→0n→∞，对任意的n，k≥1都有：CovfXn，gXn+1，…，Xn+k≤qn.其中，f和g是任何两个使上述方差存在且对每个变元均为非降的连续函数.称q(n);n∈N为该序列的混合系数。

AANA序列是一类广泛的负相依序列，文献[1]介绍了AANA序列的定义并指出NA序列[2]是AANA序列，其混合系数满足(q(n)=0，n≥1)，反之不真，文中还举出了一些例子.因此研究AANA序列的收敛性质是十分有意义的.有关更多AANA序列的成果，可以参见文献[3-10] .本文利用随机变量的截尾方法，研究矩条件下AANA序列的强收敛性.由于AANA序列比NA序列更弱，因此本文所得的结论对NA序列仍然成立.本文一律以C记与n无关的正实数，且C在不同的地方可表示不一样的值，即使在同一个式子里也是如此，记Sn，“≤”表示通常的大“O”。

2 所需引理引理1[1] 设Xn;n∈N为AANA序列，并且混合系数是q(n);n∈N，若fn;n∈N 皆是单调非降(或者单调非增)连续函数，那么f nxn;n∈N仍然是AANA序列，其混合系数仍然是q(n);n∈N.引理2[6]设Xn;n∈N为零均值的AANA序列，当1<p≤2时，混合系数q(n);n∈N，那么存在仅依赖于p的正数Cp，使得≤.引理3[11] 令Xn，n≥1为AANA随机变量序列，混合系数为qn，n≥1满足∞.令c为某个正数，记Yn=XnIXn≤c，n≥1.如果满足-∞∞，∞，∞，则收敛， a.s.3 定理及其证明定理1 设Xn，n≥1为零均值的AANA随机变量序列，混合系数为qn，n≥1满足∞，如果下列二条件之一成立：∞，0<β≤1，(1)∞，1≤β≤2.(2)其中M>0，an是常数列且满足0≤an↑∞，则.证明：当β>0，Xn≥Man>0时，有≥1，从而PXn≥Man=EIXn≥Man≤.由(1)式可得∞.当β≥1，Xn≥Man>0时，有≥1，由(2)式同样可得∞.从而当(1)式或(2)式成立时，有∞.(3)记.由引理1知，仍是AANA随机变量序列，从而是AANA的. 对0<β<1，有≤ManEIXn>Man+EXnIXn≤Man≤≤≤.对β≥1，由EXn=0，有≤EManIXn>Man+EXnIXn≤Man ≤≤≤.所以当(1)式或(2)式成立时，均有∞.(4)最后，如果0<β≤2，由Cr不等式有≤因而如果(1)式成立，则有≤∞.如果(2)式成立，则有≤∞.总之都有∞ .(5)由(3)—(5)式及引理3可知.s收敛.再由Kronecker引理有.推论1令Xn，n≥1为零均值的AANA随机变量序列，混合系数为qn，n≥1满足∞，设an，n≥1是一常数列，且满足0<an↑∞，若存在某个β∈1，2，使得∞，则.说明推论1给出了与独立情形相类似的经典强大数定律[12] .推论2 令Xn，n≥1为零均值的AANA随机变量序列，混合系数为qn，n≥1满足∞，设an是一常数列，满足0<an↑∞，bn是正实数序列且∞，如果存在r≥2，使得∞，则.证明：当r=2时，由推论1 结论成立，现设r>2，记.于是有.从而有hn=hnIhn≤bn+hnIhn>bn≤≤.由假设可得∞，又因为≤故由推论1知结论成立，说明如果我们令qn=0，n≥1，则AANA可变为NA列，因此以上定理及推论对于NA列仍然成立。

Banach空间混合型泛函方程的稳定性问题

Banach空间混合型泛函方程的稳定性问题成立花【摘要】The equivalence of two mixed type functional equations in Banach space is given.By using the fixed point method,the stability of the mixed functional equation in Banach space is built up step bystep.Furthermore,the better upper bound is gotten.%首先整合资源,给出Banach空间一类混合型泛函方程的等价性证明,其次利用不动点的择一性逐步分类地研究了Banach空间中混合型泛函方程的稳定性问题,得到较为精确的上界.【期刊名称】《中山大学学报（自然科学版）》【年(卷),期】2017(056)006【总页数】4页(P68-71)【关键词】混合型泛函方程;广义Hyers-Ulam-Rassias稳定性;不动点的择一性【作者】成立花【作者单位】西安工程大学理学院,陕西西安710048【正文语种】中文【中图分类】O175.25泛函方程的稳定性起源于Ulam：在什么条件下，一个从群到度量群的近似加法映射附近存在唯一的加法映射？即：设G1是一个群，且设G2是一个具有度量d(·,·)的度量群，对于任意的ε>0,是否存在δ>0,使得对于任意的x,y∈G1,当映射f:G1→G2满足如下条件d(f(xy),f(x)f(y))≤δ时，是否存在一个满足d(f(x),g(x))≤ε的群同态g:G1→G2？紧接着，Hyers关于可加函数在Banach空间的稳定性问题给出肯定性的答案。

随之,一大批数学家开始在各种特殊空间提出特定的泛函方程的存在性问题[1-8]。

也有学者专注研究特定空间中单次或混合型泛函方程的稳定性问题或广义稳定性(Hyers-Ulam-Rassias) (简称HUR)问题[3-5]。

Gram矩阵在不等式中的应用

Gram矩阵在不等式中的应用张宾【摘要】using the positive semi-definite of the Gram matrix t;（x1 ,x2… ,xn） , tins paper first studied the applicationof the matrix G（x1 ,x2… ,xn） on the absolute value of the maximum inner product spaces and integral average inner product space, and then studied the Gram determinant inequalities between F（x1 ,x2… ,xn.） and F（xi）. Finally ,through changing the conditions of Ostrowski inequality ,the author obtained some inequalities of two vectors in the inner product space.%利用了Gram矩阵G（x1，x2，…，xn）的半正定性，首先研究了Gram 矩阵在绝对值最大值内积空间和积分平均内积空间中的应用，然后研究了Gram 行列式r（x1，x2，…xn）与r（xi）的不等式关系．最后通过改变Ostrowski不等式的条件，得到了空间中两个向量的内积所满足的不等式．【期刊名称】《湖北民族学院学报（自然科学版）》【年(卷),期】2012(000)002【总页数】5页(P191-195)【关键词】Gram矩阵;半正定性;内积空间;Ostrowski不等式【作者】张宾【作者单位】湖北民族学院预科教育学院,湖北恩施445000【正文语种】中文【中图分类】O151定义1[1] 设x1,x2,…,xn是内积空间中的n个向量，矩阵:称为由x1,x2,…,xn生成的Gram矩阵.通常用G(x1,x2,…,xn)来记上述Gram矩阵.其行列式称为Gram行列式，通常用Γ(x1,x2,…,xn)表示.引理1[5] 设x1,x2,…,xn是内积空间中的n个向量，则G(x1,x2,…,xn)为半正定矩阵，当且仅当x1,x2,…,xn线性无关时G(x1,x2,…,xn)为正定的.引理2 设x,y,z是实内积空间中的三个非零向量，则:注：由引理1和引理2可得式(1)当且仅当x1,x2,…,xn线性相关时等号成立.引理3[6] 设f是区间[0,1]上的正实值函数，f是严格对数凸函数且在区间(0,1)上严格单调递减，x,y,z,ω∈[0,1],ω≤xyz,则有:其中引理4(Ostrowski不等式)[2] 设a=(a1,a2,…,an),b=(b1,b2,…,bn)是两个线性无关的n维向量，若x=(x1,x2,…,xn)满足(x,a)=0,(x,b)=1,则有等号成立当且仅当定理1 设x,y是实内积空间中的向量，若x=(ξ1,ξ2,…,ξn),y=(η1,η2,…,ηn),则有:其中1≤i≤n.证明定义实内积空间的内积为设zi=(0,…,0,1,0,…,0),其中第i个分量为1，其余分量为0，则在此内积空间中(x,zi)=|ξi|,(y,zi)=|ηi|,‖zi‖=1,由式(1)：(1)代入可得：其中1≤i≤n.注：类似的可进一步得到在无穷维实内积空间中:其中i≥1.定理2 设f,g是区间(a,b)上的Lebesgue可积函数，满足:m≤f(x)≤M,n≤g(x)≤N,∀x∈(a,b),其中m+M≠0,n+N≠0,则有:证明在L2(a,b)上定义内积由此内积定义可得:由式(1)得:整理得：|‖g‖2(f,1)-(f,g)(g,1)|≤‖f‖dist(g,Span{f})dist(g,Span{1})≤‖g‖dist(f,Span{g})dist(g,Span{1}).代入得：‖f‖dist(g,Span{f})dist(g,Span{1})≤(b-a)2‖g‖dist(f,Span{g})dist(g,Span{1})由题设m≤f(x)≤M,n≤g(x)≤N,∀x∈(a,b)得：因此进而可得:‖f-sg‖≤‖‖‖‖‖‖≤dist(f,Span{g})=‖g-sf‖≤‖g-‖+‖-sf‖≤+‖f-‖≤+.记由内积定义可得: ‖f‖‖g‖由引理2可知‖f‖dist(g,Span{f})=‖g‖dist(f,Span{g}).从而有:≜综上可得结论：定理3 设x1,x2,…,xn为内积空间中的向量，‖xi‖≤1,则有:其中证明因为Γ(x1,x2,…,xn)≤Γ(x1)Γ(x2)…Γ(xn).由Gram矩阵的性质知，这些行列式都非负且小于等于1，所以:Γα(x1,x2,…,xn)≤Γα(x1)Γα(x2)…Γα(xn),其中α≥0,结合引理3得：其中注：同理可得下面结论.定理4 设x1,x2,…,xn为内积空间中的向量，‖xi‖≤1,1≤k≤n,则有:其中根据引理4(Ostrowski不等式)[2]和引理2可得以下结论：定理5 设a,b,x是实内积空间的三个非零向量，满足(a,x)=0,‖x‖=1，则有:当且仅当时等号成立.证明在式(1)中用a,b,x代替式(1)中的x,y,z得:因为(a,x)=0,‖x‖=1,故(a,b)2≤‖a‖2[‖b‖2-(b,x)2]，从而有等号成立当且仅当a,b,x线性相关.设x=λa+μb，由假设条件(a,x)=0,‖x‖=1可得:‖a‖2λ+(a,b)μ=0,‖a‖2λ2+2(a,b)λμ+‖b‖2μ2=1，解得λ=∓故当且仅当时等号成立.定理6 设设a,b,x是实内积空间的三个非零向量，满足(a,x)=0,‖x‖=1,p≥2,则有:当且仅当时等号成立.证明由G(a,b,x)得半正定性得Γ(a,b,x)≥0,即:展开得：‖a‖2‖b‖2‖x‖2-‖a‖2(b,x)2-‖b‖2(a,x)2-‖x‖2(a,b)2+2(a,b)(a,x)(b,x)≥0.代入题设条件(a,x)=0,‖x‖=1得:‖a‖2‖b‖2≥‖a‖2(b,x)2+(a,b)2,由函数性质可得：‖a‖p‖b‖p≥‖a‖p(b,x)p+(a,b)p,p≥2,将上式整理得：等号成立当且仅当定理7 设x,a,b,c是实内积空间中的向量，满足(x,a)=0,(x,b)=1,(x,c)=0，则有:其中ε=2‖a‖‖b‖|(b,c)||(a,c)|-2(a,b)(b,c)(a,c)>0证明将Γ(x,a,b,c)展开整理后可得[4]:‖c‖2Γ(x,a,b)+Q(x,a,b,c)≥ [‖x‖‖a‖|(b,c)|-‖x‖‖b‖|(a,c)|]2+[‖x‖‖a‖|(〗b,c)|-‖a‖‖b‖|(x,c)|]2+[‖x‖‖b‖|(a,c)|-‖a‖‖b‖|(x,c)|]2.其中:Q(x,a,b,c)= ‖x‖2‖y‖2|(z,ω)|2+‖x‖2‖z‖2|(y,ω)|2+‖y‖2‖z‖2|(x,ω)|2-2(x,ω)(x,y)(z,ω)(y,z)-2(z,ω)(x,z)(x,y)(y,ω)-2(y,ω)(x,z)x(z,ω)(y,z)+(z,ω)2(x,y)2+(y,ω)2(x,z)2+(x,ω)2(y,z)2代入题设条件(x,a)=0,(x,b)=1,(x,c)=0得:‖x‖2[‖a‖2‖b‖2‖c‖2-‖a‖2(b,c)2-‖b‖2(a,c)2-‖c‖2(a,b)2+2‖a‖‖b‖(b,c)(a,c)]≥Γ(a,c)所以可得结论其中:ε=2‖a‖‖b‖|(b,c)||(a,c)|-2(a,b)(b,c)(a,c)>0.定理8 设设x,a,b,c是实内积空间中的向量，满足‖x‖=1,(x,a)=0,(x,b)=0，则有:证明将Γ(x,a,b,c)展开整理后可得:‖x‖2Γ(a,b,c)+Q(x,a,b,c)≥[‖a‖‖b‖|(c,x)|-‖a‖‖c‖|(b,x)|]2+[‖a‖‖b‖|(c,x)|-‖b‖‖c‖|(a,x)|]2+[‖a‖‖c‖|(b,x)|-‖b‖‖c‖|(a,x)|]2.代入题设条件‖x‖=1,(x,a)=0,(x,b)=0可得:Γ(a,b,c)+[‖a‖2‖b‖2+(a,b)2](c,x)2≥2‖a‖2‖b‖2(c,x)2,上式整理即可得到定理9 设x1,x2,…,xn,y是实内积空间中的向量，满足x1,x2,…,xn线性无关，且(x1,y)=1,(xi,y)=0,i=2,3,…,n，则有证明将Γ(y,x1,x2,…,xn)展开得[3]:‖y‖2[‖x1‖2-btG-1(x2,…,xn)b]Γ(x2,…,xn)-Γ(x2,…,xn)=‖y‖2Γ(x1,x2,…,xn)-Γ(x2,…,xn).由Γ(y,x1,x2,…,xn)≥0可得：‖y‖2Γ(x1,x2,…,xn)-Γ(x2,…,xn)≥0,故定理10 设设x1,x2,…,xn,y是实内积空间中的向量，满足x1,x2,…,xn线性无关，且‖y‖=1,(xi,y)=0,i=1,2,…,n-1，则有证明将Γ(y,x1,x2,…,xn)展开得:Γ(y,x1,x2,…,xn)= Γ(x1,x2,…,xn-1)[‖xn‖2-(xn,y)2-btG-1(x1,x2,…,xn-1)b]= Γ(x1,x2,…,xn-1)[‖xn‖2-btG-1(x1,x2,…,xn-1)b]-Γ(x1,x2,…,xn-1)(xn,y)2=Γ(x1,x2,…,xn)-Γ(x1,x2,…,xn-1)(xn,y)2[8].这里b=((x1,xn),(x2,xn),…(xn-1,xn))t.由Γ(y,x1,x2,…,xn)≥0可得：Γ(x1,x2,…,xn)-Γ(x1,x2,…,xn-1)(xn,y)2≥0，从而可得结论由Gram矩阵的性质可知当且仅当x1,x2,…,xn,y线性相关时等号成立.参考文献:[1] MA J G.An Identity in real Inner ProductSpaces[M].Zhizhou:JIPAM,China,2007.[2] DRAGOMIR S S . A Ge nerralization of Gr¨uss’inequality in inner Product Spaces and Application[J].J Math Anal Appl,1999,237:74-82. [3] ALI′C M, J. PEˇCARI′C. On some inequalities of ˇZ. Madevski and A. M. Ostrowski [J].RadHAZU,Matematiˇcke znanosti,1997,472:77-82.[4] BEESACK P R.On Bessel’s inequality and Ostrowski’s [J].Univ Beograd Publ Elektrotehn Fak,Ser Mat Fiz,1975,498/541:69-71.[5] CHO Y J, MATI′C M, PEˇCARI′C J. On Gram’s determinant in 2-inner product spaces[J].J Korean Math Soc,2001,38(6):1125-1156.[6] Vlad Ciobotariu-Boer.An Intergral Inequality for 3-convexFunction[M].Cluj-Napoca Romania,2000.[7] Ostrowski A Vorlesungen.Über Differential andIntegralrechnung[J].Basel,1951,2:289-294.。

秩-1坡矩阵的周长及保持算子

秩-1坡矩阵的周长及保持算子张廷海【摘要】坡代数是一类积不超过其因子的加法幂等半环，它概括了布尔代数、模糊代数和有界分配格。

本文定义了整坡代数上的秩-1矩阵的周长是其左、右因子的非零元素之和，并给出了其上的秩-1矩阵和的周长所满足的不等式，获得了保持其周长的算子，这概括和发展了布尔矩阵、模糊矩阵上的相关结论。

%Inclines are the additively idempotent semirings in which products are less than or equal to fac-tors.They generalize Boolean algebra,fuzzy algebra and distributive lattice.This paper defines that the perimeter of rank-1 matrices over inclines is the sum of nonzero elements of its left and right factors, and proves the inequality about the perimeter of the sum of rank-1 matrices over inclines and gives the preserving operators of rank-1 matrices over inclines.The results in the present paper generalize and de-velop correlated results over Boolean matrices and fuzzy matrices.【期刊名称】《广西师范大学学报（自然科学版）》【年(卷),期】2014(000)003【总页数】4页(P57-60)【关键词】坡矩阵;秩-1;周长;控制;算子【作者】张廷海【作者单位】江西师范大学数学与信息科学学院，江西南昌 330022【正文语种】中文【中图分类】O151.21自从20世纪80年代我国学者曹志强提出了坡代数和坡矩阵以来[1],有许多学者研究了有关坡代数和坡矩阵的各类性质及其应用。

滁州学院2010年度发表期刊论文一览表

单位球上Bloch型空间上的Toeplitz算子遗传聚类算法改进及其仿真一种混合均值聚类算法的实现与编者语专家的对话数学教育现状呼唤数学理解性教学弦图的 L(3,2,1) 标号概率加法公式新定理 MKdV方程和SG方程的可积曲面的构造决策矩阵排序的投影法及其相关定理改进的优劣系数法及其区间数推广谈数学史在数学分析教学中的积极作用面向应用能力培养的网页类课程教学改革与实践
中文系师范专业学生专业思想的调查
23 数学系 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48
1.李正红(T) 1.许志才(T) 1.许志才(T) 1.许志才(T) 1.翟明清(T) 1.翟明清(T) 1.丁楠(T) 1.丁楠(T) 2.李庆宏(T) 1.黄日朋(T) 1.黄日朋(T) 1.黄述亮(T) 1.黄述亮(T) 1.黄述亮(T) 1.黄述亮(T) 1.李庆宏(T) 1.庞玲玲(T) 1.谭玉明(T) 1.谭玉明(T) 1.王兵(T) 1.王大星(T) 1.王龙芹(T) 1.王圣祥(T) 1.王雄亮(T) 1.王雄亮(T) 1.王雄亮(T)
Journal of Mathematical Analysis and 计算机应用与软件计算机工程与应用中小学数学中小学数学运筹学学报西南民族大学学报(自然科学版) 浙江大学学报.理学版大学数学数学的实践与认识滁州学院学报滁州学院学报
一类四类二类四类四类二类三类二类四类三类三类三类三类三类二类三类三类二类三类三类三类四类二类二类二类三类
49 50 51 52 53 54 55 56 57 58 59 计算机系 60 61 62 63 64 65 66 67 68 69 电子系 70 71 72 73 74

Bull. London Math. Soc. 39 (2007) 482–486 C 2007 London Mathematical Society doi10.1112bl

ordinates in the interval (0, t]. Then, for t 2,
N (t) =
t
log
t
−
t
7 + + S(t) + O
1
,
2π 2π 2π 8
t
(1.1)
where, if t is not the ordinate of a zero, S(t) denotes the value of (1/π) arg ζ(1/2 + it) obtained by continuous variation along the straight line segments joining 2, 2 + it, and 1/2 + it, starting with the value 0 (see [10]). If t is the ordinate of a zero, we set
There has been some earlier work on Theorem 2. In place of the constant 1/2, Ramachandra and Sankaranarayanan [8] obtained 1.119 . . . in 1993, and Fujii [2] obtained .67 in 2004. For a recent survey of this area, see [4]. It is interesting that Brumer [1] obtained the same constant 1/2 for a similar bound for the rank of an elliptic curve in terms of the conductor.

韩国小学教科书“数与计算”的编写特点及中韩比较——以一年级上册为例

韩国小学教科书“数与计算”的编写特点及中韩比较——以
一年级上册为例
孔凡哲;高春月
【期刊名称】《小学教学：数学版》
【年(卷),期】2010(000)006
【摘要】韩国作为我国的邻邦，深厚的历史渊源使中韩两国彼此的小学数学课程
在很多方面可以相互借鉴。

中韩小学数学教科书的比较研究已多有涉及，而已有的研究大多表现为宏观比较分析，而对具体内容的深入细致的比较，尚不多见。

因而，本文将针对小学数学一年级上册的知识点，对韩国小学数学教科书（韩国天才教育出版社）一年级上册的“数与计算”领域，以我国现行的人教版、北师大版小学数学教科书为比照对象，进行比较分析。

韩国教科书的“数与计算”领域，不论是在学习范围上，还是具体知识点的学习上都与我国有一定差异。

【总页数】4页(P50-53)
【作者】孔凡哲;高春月
【作者单位】东北师范大学教师教育研究院;国家基础教育实验中心
【正文语种】中文
【中图分类】G623.2
【相关文献】
1.中韩户外明星真人秀节目创意比较研究——以《两天一夜》的中国版和韩国版为例 [J], 袁源;
2.中韩综艺节目的比较及启示——以湖南卫视和韩国MBC电视台为例 [J], 赵亮宇;
3.中韩信用证法律比较研究——以韩国新湖商社案与韩国国民银行案为例 [J], 曹伟峰
4.韩国小学教科书"数与计算"的编写特点及中韩比较——以一年级上册为例 [J], 孔凡哲; 高春月
5.中韩高校幼儿教师职前艺术教育课程设置及教学模式比较研究
——以遵义师范学院和韩国庆南大学为例 [J], 侯菲菲;郑国凤
因版权原因，仅展示原文概要，查看原文内容请购买。

奥苏贝尔认知同化学习理论在化学教学导入课中的运用

2011年8月第23期科技视界Science &Technology Vision※本文系湖南省教育厅教改课题《面向中学新课改在高师成人教育中开展科学探究教学的研究与实践》。

作者简介:于江雁(1986—),女,云南保山人,湖南师范大学化学化工学院研究生,研究方向为化学教学论研究。

熊士荣(1953—),男,湖南长沙人,湖南师范大学化学化工学院副教授,硕士生导师,主要从事化学课程与学习理论研究。

奥苏贝尔认知同化学习理论在化学教学导入课中的运用于江雁熊士荣(湖南师范大学化学化工学院湖南长沙410081)【摘要】本文以奥苏贝尔认知同化学习理论中的意义学习为核心,将先行组织者策略、学习中的动机因素、发现学习三个方面运用于化学教学的导入课当中,将学习理论和教学实际结合起来,以期取得良好的教学效果。

【关键词】导入课;先行组织者策略;动机因素;发现学习1问题的提出教学是一门科学,也是一门艺术。

课的导入是化学教学的重要和必要环节,是衡量教师教学艺术水平高低的标志之一。

合适地导入新课,能给听课者以耳目一新的美感和艺术上的享受。

可以激发学生的求知欲,把学生带到教师的思路上来,而且可以使学生明确要求,集中精力,引发思维,以积极的心理状态投入到课堂学习中去,从而达到渲染良好的课堂气氛、提高课堂教学效果的目的。

这些都与奥苏贝尔所倡导的意义学习是一脉相承的,意义学习有两个条件:(1)学生表现出一种意义学习的心向,即表现出一种在新学的内容与自己已有的知识之间建立联系的倾向;(2)学习内容对学生具有潜在意义,即能够与学生已有的知识结构联系起来[1]。

本文主要以意义学习为核心,将先行组织者策略、学习中的动机因素、发现学习三个方面运用于化学教学的新课导入当中,以期取得良好的教学效果。

2奥苏贝尔认知同化学习理论在化学教学导入课中的运用2.1先行组织者策略在导课中的运用[2]奥苏贝尔就如何贯彻“逐渐分化”和“整合协调”的原则,提出了具体应用的策略即先行组织者策略。

2023年韩国奥林匹克数学竞赛第二轮试题

2023年韩国奥林匹克数学竞赛第二轮试题韩国奥林匹克数学竞赛第二轮试题（2023年）一、选择题1. 下列哪个数是素数？A. 15B. 22C. 31D. 402. 若 a + b = 10，且 a^2 + b^2 = 58，求 a 和 b 的值。

3. 三个相邻正整数的乘积是 2100，求这三个数。

4. 若一个数字的个位数是其十位数的平方，且这个数字是 3 位数，求该数字。

5. 对于任意正整数 n，证明：n^3 + (n+1)^3 + (n+2)^3 恰好可以被 9 整除。

二、填空题1. 写出满足以下条件的两个质数的乘积：两个质数之和是 30。

__________, __________2. 若 a + b = 15，且 a^3 + b^3 = 2915，则 a 和 b 的值分别为__________, __________。

3. 一个 4 位数，各位数字之和为 22，千位和百位数字相等，求该数。

4. 证明：1 + 2 + 3 + ... + n = n(n+1)/2（提示：使用数学归纳法）。

三、解答题1. 若点 A(x1, y1) 到点 B(4, -4) 的距离为 5，且点 A 在直线 y = 3x - 1 上，求点 A 的坐标。

2. 设数列 {an} 满足 a1 = 1，an+1 = 2an + 3。

求数列的通项公式。

3. 已知三角形 ABC，∠CAB = 45°，∠ABC = 75°。

设 O 是 AC 上的一点，且∠OBC = 15°。

求证：∠AOB = 120°。

4. 已知正整数 a，b，c 满足 a^2 + b^2 = c^2。

若 a，b，c 均为素数，求 a，b，c 的值。

四、解答题（含证明）1. 已知 a 和 b 是整数，且满足 a^2 - 3ab + 2b^2 = 0。

证明 a = b = 0。

2. 设函数 f(x) = (x + 1)(2x - 3) - (x - 2)(3x + 4)，求函数 f(x) 的最小值。

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J.Korean Math.Soc.39(2002),No.4,pp.621–651(f,2)–ROTATIONAL EXTENDED STEINERTRIPLE SYSTEMS WITH f=2AND f=3Chung Je ChoAbstract.An extended Steiner triple system of order v,denotedby EST S(v),is said to be(f,k)–rotational if it admits an automor-phism consisting of exactly fﬁxed elements and k cycles of lengthv−f k .In this paper,we obtain a necessary and suﬃcient conditionfor the existence of(f,2)–rotational extended Steiner triple systemswith f=2and f=3.1.IntroductionAn extended Steiner triple system of order v,denoted EST S(v),is an ordered pair(V,B)where V is a set of v elements and B is a set of triples of(not necessarily distinct)elements of V,called blocks,such that every unordered pair of(not necessarily distinct)elements of V occurs in exactly one block of B.In an EST S(v),there are three types of blocks of the forms{a,a,a},{a,a,b},{a,b,c},where a,b and c are distinct elements.If there is a block of the form {a,a,a},then such an element a is called an idempotent,but the element b which forms a block{b,b,c}with b=c is a nonidempotent.We will denote by EST S(v,ρ)an extended Steiner triple system of order v withρidempotents.In1972,Johnson and Mendelsohn[6]obtained a necessary condition for the existence of an EST S(v,ρ),and,in1978,Bennett and Mendelsohn[1]showed the necessary condition was also suﬃcient.Received October15,2001.Revised March11,2002.2000Mathematics Subject Classiﬁcation:05B05,05B07,05B10.Key words and phrases:extended Steiner triple system,automorphism of an ex-tended Steiner triple system.This research was supported by the KISTEP in2000-2001.622Chung Je ChoTheorem1.1[1,6].Let0≤ρ≤v.Then there exists an EST S(v,ρ) if and only if(i)v≡0(mod3)andρ≡0(mod3),or(ii)v≡1,2(mod3)andρ≡1(mod3),but(iii)when v is even,ρ≤v2,and(iv)whenρ=v−1,v=2.An automorphism of an EST S(v,ρ)(V,B)is a permutation of V which preserves the blocks of B.An EST S(v,ρ)is said to be(f,k)–rotational if it admits an automorphism consisting of exactly fﬁxedelements and k cycles of length v−fk .Solutions to the existence problemof an(f,k)–rotational EST S(v,ρ)wereﬁrst given when f=0,k=1, as the existence of cyclic extended triple systems[2],when f=0,k≥2, as that of k–regular extended triple systems[3]which were completely determined for k=2,3and partially for k=4,when f=1,k≥1,as that of k–rotational extended triple systems[2]which were completely determined for k=1,2and partially for k=3.If(V,B)is an EST S(v,ρ)with an automorphismα,then the cyclic group generated byα,<α>,acts on the set V.So the whole blocks B are partitioned into disjoint orbits of blocks under the cyclic group <α>.We say that a set of blocks which are taken from each of the orbits exactly one is called a set of starter blocks for the system under the automorphismα.An(A,k)-system is a set of ordered pairs{(a r,b r)|r=1,2,...,k} that partition the set{1,2,...,2k}with the property that b r−a r= r for r=1,2,...,k;and there exists an(A,k)-system if and only if k≡0or1(mod4)[8,9].A(B,k)-system is a set of ordered pairs {(a r,b r)|r=1,2,...,k}that partition the set{1,2,...,2k−1,2k+1} with the property that b r−a r=r for r=1,2,...,k;and there exists a (B,k)-system if and only if k≡2or3(mod4)[7,8].A(C,k)-system is a set of ordered pairs{(a r,b r)|r=1,2,...,k}that partition the set {1,2,...,k,k+2,...,2k+1}with the property that b r−a r=r for r= 1,2,...,k;and there exists a(C,k)-system if and only if k≡0or3(mod 4)[8].A(D,k)-system is a set of ordered pairs{(a r,b r)|r=1,2,...,k} that partition the set{1,2,...,k,k+2,...,2k,2k+2}with the property that b r−a r=r for r=1,2,...,k;and there exists a(D,k)-system if and only if k≡1or2(mod4)and k=1[8].In this paper,we obtain a necessary and suﬃcient condition for the existence of(f,2)–rotational extended Steiner triple systems with f=2 and f=3.(f,2)–rotational extended Steiner triple systems6232.(2,2)–rotational extended Steiner triple systemsSuppose that there exists a(2,1)–rotational EST S(v,ρ)whose ele ment-set is V={∞1,∞2}∪Z v−2and with an automorphismα=(∞1)(∞2)(01···v−3).Then either{∞1,∞1,∞1}and{∞1,∞2,∞2},or{∞2,∞2,∞2}and {∞2,∞1,∞1}must be blocks.So there must exist starter blocks of the forms{∞1,0,a}and{∞2,0,b}for some a=b in Z v−2and then at least one of the pairs{∞1,0}and {∞2,0}occur twice.Thus there is no(2,1)–rotational extended Steiner triple systems.In a(2,k)–rotational EST S(v,ρ),if{a,a,a}is a block then the length of its orbit is either1or v−2k;so we have the following lemma.Lemma2.1.If there exists a(2,k)–rotational EST S(v,ρ),thenρ=1or1+n·v−2 kfor n=0,1,...,k.Remark2.2.If there exists a(2,2)–rotational EST S(v,ρ),thenρ=1,v2,or v−1.But ifρ=v−1,v must be2(by Theorem1.1);so ifρ=v−1then v=2andρ=1.Lemma 2.3.A necessary condition for the existence of a(2,2)–rotational EST S(v,ρ)is(i)ρ=1and v≡2,4(mod6),or(ii)ρ=v2and v≡0,2(mod6).Proof.(i)Ifρ=1,then,by Theorem1.1,v≡1or2(mod3).Butv−2 2must be an integer.Thus v satisﬁes both v≡1,2(mod3)andv≡0(mod2)and hence v≡2or4(mod6).(ii)Ifρ=v2then,by Theorem1.1,v2≡0or1(mod3)and hencev≡0or2(mod6).624Chung Je ChoNow,we will construct(2,2)–rotational EST S(v,ρ)s.We assume that our(2,2)–rotational EST S(v,ρ)s have the element-set,V={∞1,∞2}∪Z v−22×{1,2},and the corresponding automorphism isα=(∞1)(∞2)0111···v−22−110212···v−22−12,where we write for brevity x i instead of(x,i).Lemma2.4.There exists a(2,2)–rotational EST S(v,1)for v=8 and20.Proof.If v=8,then the following triples{∞1,∞1,∞1},{∞1,∞2∞2},{∞1,01,02},{∞2,01,12},{01,01,22},{01,11,21},{02,02,12}form a set of starter blocks for a(2,2)–rotational EST S(8,1).If v=20,the following triples{∞1,∞1,∞1},{∞1,∞2,∞2},{01,11,31},{01,01,41},{02,02,01},{02,12,21},{02,22,71},{02,42,81}{∞1,02,61},{∞2,02,31},{02,32,62}form a set of starter blocks of a(2,2)–rotational EST S(20,1).Lemma2.5.If v≡8(mod12),then there exists a(2,2)–rotational EST S(v,1).Proof.Let v=12t+8=2(6t+3)+2and let t≥0be an arbitrary integer.The cases t=0and1have been treated in Lemma2.4.If t≥2,then the following triples{∞1,∞1,∞1},{∞1,∞2,∞2},{01,r1,(b r+t)1},r=1,2,...,t,(f,2)–rotational extended Steiner triple systems625 where{(a r,b r)|r=1,2,...,t}is a(C,t)-system if t≡0,3(mod4),or a(D,t)-system if t≡1,2(mod4),and{02,(2t+1)2,(4t+2)2},{02,02,(3t+1)2},{01,(2t+1)1,(−c2t+1)2},{∞1,02,(c3t+1)1},{∞2,02,(d3t+1)1},{02,r2,(d r)1},r=1,2,...,2t,2t+2,...,3t,where{(c r,d r)|r=1,2,...,3t+1}is an(A,3t+1)-system if t≡0,1 (mod4),or a(B,3t+1)-system if t≡2,3(mod4),and6t+3should be read as0,and{01,01,02}if t≡0or1(mod4),or{(6t+2)1,(6t+2)1,02}if t≡2or3(mod4),form a set of starter blocks for a(2,2)–rotational EST S(v,1).Lemma2.6.If v≡4(mod12),then there exists a(2,2)–rotational EST S(v,1).Proof.Let v=12t+4=2(6t+1)+2and let t≥0be an arbitrary integer.If t=0,the following triples{∞1,∞1,∞1},{∞1,∞2,∞2},{∞1,01,02},{∞2,01,01},{∞2,02,02} form a set of starter blocks for a(2,2)–rotational EST S(4,1).If t≥1,then the following triples{∞1,∞1,∞1},{∞1,∞2,∞2},{01,r1,(b r+t)1},r=1,2,...,t,where{(a r,b r)|r=1,2,...,t}is an(A,t)-system if t≡0,1(mod4),or a(B,t)-system if t≡2,3(mod4),and{02,02,(3t)2},{∞1,02,(c3t)1},{∞2,02,(d3t)1},{02,r2,(d r)1},r=1,2,...,3t−1,626Chung Je Chowhere{(c r,d r)|r=1,2,...,3t}is an(A,3t)-system if t≡0,3(mod4), or a(B,3t)-system if t≡1,2(mod4),and6t+1should be read as0, and{01,01,02}if t≡0,3(mod4),or{(6t)1,(6t)1,02}if t≡1,2(mod4),form a set of starter blocks for a(2,2)–rotational EST S(v,1).Lemma2.7.If v≡2or26(mod48),then there exists a(2,2)–rotational EST S(v,1).Proof.Let v=12t+2and let t≡0(mod2).If t=0,then the triples{∞1,∞1,∞1},{∞1,∞2,∞2}are starter blocks for a(2,2)–rotational EST S(2,1).If t≥2is an even integer,then the following triples{∞1,∞1,∞1},{∞1,∞2,∞2},{∞2,01,02},{∞1,01,(3t)1},{∞1,02,(3t)2},{01,(2t)1,(4t)1},and01,01,5t21,01,(2r−1)1,t2+t−1+r1,r=1,2,...,t2,{01,(2r)1,(3t−r)1},r=1,2,...,t2−1,{02,r2,(b r)1},r=1,2,...,3t−1,where{(a r,b r)|r=1,2,...,3t−1}is an(A,3t−1)-system if t≡2(mod 4),or a(B,3t−1)-system if t≡0(mod4),and{02,02,(6t−1)1}if t≡2(mod4),or{02,02,(6t−2)1}if t≡0(mod4),form a set of starter blocks for a(2,2)–rotational EST S(v,1).(f,2)–rotational extended Steiner triple systems627 Lemma2.8.If v≡14or38(mod48),then there exists a(2,2)–rotational EST S(v,1).Proof.Let v=12t+2and let t≡1(mod2).Then the following triples{∞1,∞1,∞1},{∞1,∞2,∞2},{∞2,01,02},{∞1,01,(3t)1},{∞1,02,(3t)2},{01,(2t)1,(4t)1},and01,01,3t−121,01,(2r)1,t+12+t−1+r1,r=1,2,...,t−12,{01,(2r−1)1,(3t−r)1},r=1,2,...,t−1 2,{02,r2,(b r)1},r=1,2,...,3t−1,where{(a r,b r)|r=1,2,...,3t−1}is an(A,3t−1)-system if t≡3(mod 4),or a(B,3t−1)-system if t≡1(mod4),and{02,02,(6t−1)1}if t≡3(mod4),or{02,02,(6t−2)1}if t≡1(mod4),form a set of starter blocks for a(2,2)–rotational EST S(v,1).Lemma2.9.There exists a(2,2)–rotational EST S(10,1).Proof.The following triples{∞1,∞1,∞1},{∞1,∞2,∞2},{∞1,01,21},{∞1,02,22},{∞2,01,02},{01,01,11},{02,12,21},{02,02,31}form a set of starter blocks of a(2,2)–rotational EST S(10,1).Lemma2.10.If v≡10(mod12),then there exists a(2,2)–rotational EST S(v,1).628Chung Je ChoProof.Let v=12t+10and let t be any nonnegative integer.The case t=0has been treated in Lemma2.9.If t≥1,then the following triples{∞1,∞1,∞1},{∞1,∞2,∞2},{∞2,01,02},{∞1,01,(3t+2)1},{∞1,02,(3t+2)2},and{01,01,(3t+1)1}if t≡0,1(mod4),or{01,01,(3t)1}if t≡2,3(mod4),and{01,r1,(b r+t)1},r=1,2,...,t,where{(a r,b r)|r=1,2,...,t}is an(A,t)-system if t≡0,1(mod4),or a(B,t)-system if t≡2,3(mod4),and{02,r2,(b r)1},r=1,2,...,3t+1,where{(a r,b r)|r=1,2,...,3t+1}is an(A,3t+1)-system if t≡0,1 (mod4),or a(B,3t+1)-system if t≡2,3(mod4),and{02,02,(6t+3)1}if t≡0,1(mod4),or{02,02,(6t+2)1}if t≡2,3(mod4),form a set of starter blocks for a(2,2)–rotational EST S(v,1).Lemmas2.3,2.5,2.6,2.7,2.8and2.10together yield the following theorem.Theorem2.11.There exists a(2,2)–rotational EST S(v,1)if and only if v≡2or4(mod6).Lemma2.12.There exists a(2,2)–rotational EST S(8,4)(f,2)–rotational extended Steiner triple systems 629Proof.The following triples{∞1,∞1,∞1},{∞1,∞2,∞2},{01,01,01},{01,22,22},{01,11,21},{∞1,01,02},{∞2,01,12},{02,12,22}form a set of starter blocks of a (2,2)–rotational EST S (8,4).Lemma 2.13.There exists no (2,2)–rotational EST S (20,10)Proof.If there exists a (2,2)–rotational EST S (20,10),by counting of the number of orbits,there must exist two orbits of length 3.Con-sequently,there exists a cyclic Steiner triple system of order 9,which is non-existing system.Lemma 2.14.If v ≡8(mod 12)and v =20,then there exists a (2,2)–rotational EST S v,v 2 .Proof.Let v =12t +8=2(6t +3)+2.The case t =0has been treated in Lemma 2.12.By Lemma 2.13,t =1.Let t ≥2be an integer.Then the following triples{∞1,∞1,∞1},{∞1,∞2,∞2},{01,01,01}{01,r 1,(b r +t )1},r =1,2,...,t,where {(a r ,b r )|r =1,2,...,t }is a (C,t )-system if t ≡0,3(mod 4),or a (D,t )-system if t ≡1,2(mod 4),and{01,(2t +1)1,(4t +2)1},{∞1,02,(c 2t +1)1},{∞2,02,(d 2t +1)1},{02,(2t +1)2,(4t +2)2},{{02,r 2,(d r )1},r =1,2,...,2t,2t +2,...,3t +1},where {(c r ,d r )|r =1,2,...,3t +1}is an (A,3t +1)-system if t ≡0or 1(mod 4),or a (B,3t +1)-system if t ≡2or 3(mod 4),and 6t +3should be read as 0,and{01,02,02}if t ≡0,1(mod 4),or {(6t +2)1,02,02}if t ≡2,3(mod 4),form a set of starter blocks for a (2,2)–rotational EST S v,v 2 .630Chung Je ChoLemma 2.15.There exists no (2,2)–rotational EST S (12,6).Proof.Suppose that there exists a (2,2)–rotational EST S (12,6).Then we may say that it has starter blocks of the forms{∞1,∞1,∞1},{∞1,∞2,∞2},{a 1,a 1,a 1},{∞1,a 1,b 2},{∞2,x 1,y 2},{u 1,v 1,w 2},{a 1,b 1,c 2},{x 2,y 2,z 1},{r 2,s 2,t 1},{a 2,a 2,x 1}.We see that we need 11orbits of the form {a 1,b 2}.But there are only 5such orbits.Lemma 2.16.If v ≡0(mod 12)and v =12,then there is a (2,2)–rotational EST S v,v 2 .Proof.Let v =12t =2(6t −1)+2.By Lemma 2.15,t =1.Let t ≥2be an integer.Then the following triples{∞1,∞1,∞1},{∞1,∞2,∞2},{01,01,01}and{02,12,(3t −1)2}if t ≡1,2(mod 4),or{02,22,(3t −1)2}if t ≡0,3(mod 4)and{01,r 1,(b r +t −1)1},r =1,2,...,t −1,where {(a r ,b r )|r =1,2,...,t −1}is an (A,t −1)-system if t ≡1,2(mod 4),or a (B,t −1)-system if t ≡0,3(mod 4),and let {(c r ,d r )|r =1,2,...,3t −1}be an (A,3t −1)-system if t ≡2,3(mod 4),or a (B,3t −1)-system if t ≡0,1(mod 4)(if d r =6t −1then d r should be read as 0),then{01,(3t −1)1,(d 3t −1)2}and{01,(3t −2)1,(d 3t −2)2}if t ≡1,2(mod 4),or {01,(3t −3)1,(d 3t −3)2}if t ≡0,3(mod 4)(f,2)–rotational extended Steiner triple systems 631and{02,r 2,(−c r )1},r =2,3,...,3t −3if t ≡2,3(mod 4),orr =1,3,...,3t −4,3t −2if t ≡0,1(mod 4)and{∞2,01,02,},{01,(c 1)2,(c 1)2},{∞1,01,(d 1)2}if t ≡0,1(mod 4)or{∞2,01,(6t −2)2,},{01,(c 2)2,(c 2)2},{∞1,01,(d 2)2}if t ≡2,3(mod 4),form a set of starter blocks for a (2,2)–rotational EST S v,v 2.Lemma 2.17.There exists a (2,2)–rotational EST S (6,3).Proof.The following triples{∞1,∞1,∞1},{∞1,∞2,∞2},{01,01,01},{01,12,12},{∞1,01,11},{∞1,02,12},{∞2,01,02}form a set of starter blocks of a (2,2)–rotational EST S (6,3).Lemma 2.18.If v ≡6or 18(mod 48),then there exists a (2,2)–rotational EST S v,v2 .Proof.Let v =12t +6=2(6t +2)+2and let t ≡0or 1(mod 4).The case t =0has been treated in Lemma 2.16.Let t ≥2.Then the following triples{∞1,∞1,∞1},{∞1,∞2,∞2},{01,01,01},{∞1,01,(3t +1)1},{∞1,02,(3t +1)2},{∞2,01,02},and{01,r 1,(b r +t )1},r =1,2,...,t,632Chung Je Chowhere{(a r,b r)|r=1,2,...,t}is an(A,t)-system,and{02,r2,(b r)1},r=1,2,...,3t,where{(a r,b r)|r=1,2,...,3t}is an(A,3t)-system if t≡0(mod4),or a(B,3t)-system if t≡1(mod4),and{02,02,(6t+1)1}if t≡0(mod4),or{02,02,(6t)1}if t≡1(mod4),form a set of starter blocks for a(2,2)–rotational EST Sv,v2.Definition2.19.A(R1,3k)-system is a set{(a r,b r)|r=1,2,...,3k} such that(i){a r,b r|r=1,2,...,3k}=Z6k+2\{0,9k+42},and(ii)b r−a r=r+1for r=1,2,...,3k−1and b3k−a3k=3k−2. Lemma2.20.If k≡2(mod4),then there exists a(R1,3k)-system. Proof.If k≡2(mod4),form ordered pairs(r,3k+2−r),r=1,2, (3)2,(3k+1+2r,6k−2r),r=1,2,...,3k−64,(3k+2+2r,6k+3−2r),r=1,2,...,3k−24,3k+22,9k2,(3k+2,6k).Definition2.21.A(R2,3k)-system is a set{(a r,b r)|r=1,2,...,3k} such that(i){a r,b r|r=1,2,...,3k}=Z6k+2\{0,9k−32},and(ii)b r−a r=r+1for r=1,2,...,3k−1and b3k−a3k=3k−3.Lemma 2.22.If k≡3(mod4)and k=3,then there exists a (R2,3k)-system.Proof.Case1.k≡7(mod8):(r,3k+1−r),r=1,2,...,3k−12 (3k+8+4r,6k−4r),r=0,1,...,3k−138,3k+12,9k+12,(6k+1,3k+4),9k−12,9k+52.(3k+r,6k−r),r=1,2,3,5,6,7,9,10,11,13,...,3k−72.Case2.k≡3(mod8)and k=3:(f,2)–rotational extended Steiner triple systems 633(r,3k +1−r ),r =1,2,...,3k −12,(3k +8+4r,6k −4r ),r =0,1,...,3k −178, 3k +12,9k +12,(6k +1,3k +4), 9k +32,9k +92,(3k +r,6k −r ),r =1,2,3,5,6,7,9,10,11,13,...,3k −52. Lemma 2.23.There exists a (2,2)-rotational EST S (42,21).Proof.The following triples{∞1,∞1,∞1},{∞1,∞2,∞2},{01,01,01},{∞1,01,71},{∞1,02,72},{∞2,01,02},{01,11,41},{01,21,71},{01,61,82},{01,81,92},{01,91,142},{02,12,92},{01,162,162},{02,22,51},{02,32,171},{02,42,131},{02,52,71},{02,62,161},{02,72,81},form a set of starter blocks of a (2,2)-rotational EST S (42,21).Lemma 2.24.If v ≡30or 42(mod 48),then there exists a (2,2)–rotational EST S v,v 2 .Proof.Let v =12k +6=2(6k +2)+2and let k ≡2or 3(mod 4).The case k =3has been treated in Lemma 2.23.Let k =3.Then the following triples{∞1,∞1,∞1},{∞1,∞2,∞2},{01,01,01},{∞1,01,(3k +1)1},{∞1,02,(3k +1)2},{∞2,01,02},and{01,r 1,(b r +k −1)1},r =1,2,...,k −1,where {(a r ,b r )|r =1,2,...,k −1}is an (A,k −1)-system if k ≡2(mod 4),or a (B,k −1)-system if k ≡3(mod 4),and{01,(3k −2)1,(b 3k )2}if k ≡2(mod 4),or {01,(3k −3)1,(b 3k )2}if k ≡3(mod 4)634Chung Je Choand{02,12,(3k)2},{01,(3k−1)1,(b3k−2)2},{01,(3k)1,(b3k−1)2},{02,(r+1)2,(−a r)1},r=1,2,...,3k−3,where{(a r,b r)|r=1,2,...,3k}is an(R1,3k)-system if k≡2(mod4), or a(R2,3k)-system if k≡3(mod4),and01,9k+422,9k+422if k≡2(mod4),or01,9k−322,9k−322if k≡3(mod4),form a set of starter blocks for a(2,2)–rotational EST Sv,v2.Definition2.25.A(S1,k−1)-system is a set{(a r,b r)|r=1,2,..., k−1}such that(i){a r,b r|r=1,2,...,k−1}={1,2,...,k,k+2,...,2k−1},and(ii)b r−a r=r for r=1,2,...,k−1.Lemma2.26.If k≡2or3(mod4),then there exists a(S1,k−1)-system.Proof.Case1.k=4t+2.t=0:(1,2).t=1:(10,11),(2,4),(6,9),(1,5),(3,8).t≥2:(r,4t+2−r),r=1,2,...,2t,(4t+3+r,8t+4−r),r=1,2,...,t−1,(5t+2+r,7t+3−r),r=1,2,...,t−1,(2t+1,6t+2),(4t+2,6t+3),(7t+3,7t+4).Case2.k=4t+3.t=0:(1,2),(3,5).t=1:(11,12),(3,5),(10,13),(2,6),(4,9),(1,7).t≥2:(r,4t+4−r),r=1,2,...,2t+1,(f,2)–rotational extended Steiner triple systems 635(4t +4+r,8t +5−r ),r =1,2,...,t −1,(5t +3+r,7t +4−r ),r =1,2,...,t −1,(2t +2,6t +3),(6t +4,8t +5),(7t +4,7t +5).Definition 2.27.A (S 2,k −1)-system is a set {(a r ,b r )|r =1,2,...,k −1}such that(i){a r ,b r |r =1,2,...,k −1}={1,2,...,k,k +2,...,2k −2,2k },and(ii)b r −a r =r for r =1,2,...,k −1.Lemma 2.28.If k ≡0or 1(mod 4),then there exists a (S 2,k −1)-system.Proof.Case 1.k =4t .t =1:(2,3),(6,8),(1,4).t =2:(1,2),(14,16),(3,6),(8,12),(5,10),(7,13),(4,11).t ≥3:(r,4t +1−r ),r =1,2,...,2t −1,(4t +1+r,8t −3−r ),r =1,2,...,t −2,(5t −1+r,7t −3−r ),r =1,2,...,t −3,(2t,6t −2),(2t +1,6t −3),(6t −1,8t −3),(8t −2,8t ),(7t −3,7t −2).Case 2.k =4t +1.t =1:(2,3),(8,10),(4,7),(1,5).t ≥2:(r,4t +2−r ),r =1,2,...,2t ,(4t +4+r,8t +1−r ),r =1,2,...,2t −2,(2t +1,4t +4),(4t +3,8t +2). Definition 2.29.A (S 3,3k )-system is a set {(a r ,b r )|r =1,2,...,2k −1,2k +1,...,3k }such that(i){a r ,b r |r =1,2,...,2k −1,2k +1,...,3k }=1,2,...3k 2,3k +42...,6k −1 ,and (ii)b r −a r =r for r =1,2,...,2k −1,2k +1,...,3k −1,andb 3k −a 3k =3k −1.636Chung Je ChoLemma 2.30.If k ≡2(mod 4),then there exists a (S 3,3k )-system.Proof.If k ≡2(mod 4),form ordered pairs(r,3k +1−r ),r =1,2,...,3k −22,(3k +r,6k −r ),r =1,2,...,k −22(k >2), 9k −22−r,9k −22+r ,r =1,2,...,k −1, 3k 2,9k −22 , 11k −22,11k 2.Definition 2.31.A (S 4,3k )-system is a set {(a r ,b r )|r =1,2,...,2k −1,2k +1,...,3k }such that(i){a r ,b r |r =1,2,...,2k −1,2k +1,...,3k }= 1,2,...3k −12,3k +32...,6k −1,and(ii)b r −a r =r for r =1,2,...,2k −1,2k +1,...,3k −1,andb 3k −a 3k =3k −1.Lemma 2.32.If k ≡3(mod 4),then there exists a (S 4,3k )-system.Proof.If k ≡3(mod 4),form ordered pairs(r,3k −r ),r =1,2,...,3k −32,(3k +r,6k −1−r ),r =1,2,...,k −32(k >3), 9k −32−r,9k −32+r ,r =1,2,...,k −1,3k −12,9k −32 ,(3k,6k −1), 11k −32,11k −12 .Definition 2.33.A (S 5,3k )-system is a set {(a r ,b r )|r =1,2,...,2k −1,2k +1,...,3k }such that(i){a r ,b r |r =1,2,...,2k −1,2k +1,...,3k }= 1,2,...,3k −22,3k +22...,6k −1,and(ii)b r −a r =r for r =1,2,...,2k −1,2k +1,...,3k −1,andb 3k −a 3k =3k −2.Lemma 2.34.If k ≡0(mod 4),then there exists a (S 5,3k )-system.(f,2)–rotational extended Steiner triple systems 637Proof.If k ≡0(mod 4),form ordered pairs(r,3k +1−r ),r =1,2,...,3k −22,(3k +r,6k −1−r ),r =1,2,...,k −42(k >4), 9k −22−r,9k −22+r ,r =1,2,...,k −1, 3k +22,9k −22,(3k +1,6k −1), 11k −22,11k 2 .Definition 2.35.A (S 6,3k )-system is a set {(a r ,b r )|r =1,2,...,2k −1,2k +1,...,3k }such that(i){a r ,b r |r =1,2,...,2k −1,2k +1,...,3k }= 1,2,...3k −32,3k +12,...,6k −1,and(ii)b r −a r =r for r =1,2,...,2k −1,2k +1,...,3k −1,andb 3k −a 3k =3k −2.Lemma 2.36.If k ≡1(mod 4),then there exists a (S 6,3k )-system.Proof.If k ≡1(mod 4),form ordered pairs(r,3k −r ),r =1,2,...,3k −32(k >1),(3k −1+r,6k −1−r ),r =1,2,...,k −12(k >1), 9k −32−r,9k −32+r ,r =1,2,...,k −1(k >1), 3k +12,9k −32 , 11k −32,11k −12.Lemma 2.37.If v ≡2(mod 12),then there exists a (2,2)–rotationalEST S (v,v 2).Proof.Let v =12k +2and let k be a nonnegative integer.The case k =0is trivial.If k ≥1,then the following triples{∞1,∞1,∞1},{∞1,∞2,∞2},{01,01,01},{∞2,01,02},{∞1,01,(3k )1},{∞1,02,(3k )2},{01,(2k )1,(4k )1},{02,(2k )2,(4k )2},{01,r 1,(b r +k −1)1},r =1,2,...,k −1,638Chung Je Chowhere {(a r ,b r )|r =1,2,...,k −1}is either a (S 1,k −1)-system if k ≡2or 3(mod 4),or a (S 2,k −1)-system if k ≡0or 1(mod 4),and02,02,3k2 1if k ≡0(mod 4),or 02,02,3k −121if k ≡1(mod 4),or 02,02,3k +221if k ≡2(mod 4),or 02,02,3k +12 1if k ≡3(mod 4)and{02,(a 3k )1,(b 3k )1},{02,r 2,(b r )1},r =1,2,...,2k −1,2k +1,...,3k −1,where {(a r ,b r )|r =1,2,...,2k −1,2k +1,...,3k }is a (S t ,3k )-systemsuch thatt =3if k ≡2(mod 4),or t =4if k ≡3(mod 4),or t =5if k ≡0(mod 4),or t =6if k ≡1(mod 4),form a set of starter blocks of a (2,2)–rotational EST Sv,v 2 .Lemmas 2.3,2.12through 2.37together yield the following theorem.Theorem 2.38.There exists a (2,2)–rotational EST S v,v 2if and only if v ≡0or 2(mod 6)and v =12,20.Now,we can conclude the following theorem.Theorem 2.39.There exists a (2,2)–rotational EST S (v,ρ)if and only if(i)ρ=1and v ≡2,4(mod 6),or (ii)ρ=v 2and v ≡0,2(mod 6)and v =12,20.(f,2)–rotational extended Steiner triple systems 6393.(3,2)-rotational extended Steiner triple systemsAs before (2,2)–rotational EST S (v,ρ)s,we assume that our (3,2)–rotational EST S (v,ρ)s have the element-setV ={∞1,∞2,∞3}∪(Z v −32×{1,2})and the corresponding automorphism isα=(∞1)(∞2)(∞3) 0111··· v −32−1 1 0212··· v −32−1 2,where we also write for brevity x i instead of (x,i ).By an elementary argument,we have the following necessary condi-tion for the existence of a (3,k )–rotational EST S (v,ρ).Lemma 3.1.If there exists a (3,k )–rotational EST S (v,ρ),thenρ=0or 3+n ·v −3kfor each n =0,1,···,k .Remark 3.2.If there exists a (3,2)–rotational EST S (v,ρ),then ρis0,3,v +32or v.Lemma 3.3.A necessary condition for the existence of a (3,2)–rotational EST S (v,ρ)is (i)ρ=v and v ≡1,3(mod 6),v =13,21,or(ii)ρ=v +32and v ≡3,5(mod 6),v =5,or (iii)ρ=0,3and v ≡3(mod 6).Proof.(i)This follows from the existence of a (3,2)–rotational Steiner triple systems [4,5].(ii)If ρ=v +32,thenv +32≡0or 1(mod 3)since the existence of EST S (v,ρ)implies ρ=0or 1(mod 3).Thus v ≡3or 5(mod 6).If v =5,then ρ=5+32=4=v −1;so v must be 2.(iii)If ρ=0or 3,then v ≡0(mod 3).But v −32is an integer;so v ≡3(mod 6). The following theorem is a consequence of the existence of a (3,2)–rotational Steiner triple system of order v [4,5].640Chung Je ChoTheorem 3.4.There exists a(3,2)–rotational EST S(v,v)if and only if v≡1or3(mod6),v=13,21.Lemma3.5.There exists a(3,2)–rotational EST S(9,3).Proof.The following triples{∞1,∞2,∞3},{01,01,11},{02,02.12},{∞1,∞1,∞1},{∞2,∞2,∞2},{∞3,∞3,∞3},{∞1,01,02},{∞2,01,12},{∞3,01,22}form a set of starter blocks of a(3,2)–rotational EST S(9,3).Lemma3.6.If v≡9(mod12),then there exists a(3,2)–rotational EST S(v,3).Proof.Let v=12t+9=2(6t+3)+3and let t be a nonnegative integer.The case t=0has been treated in Lemma3.5.Let t≥1.Then the following triples{∞1,∞2,∞3},{∞i,∞i,∞i},i=1,2,3,and if t≡0,1(mod4),then{∞1,01,02},{0i,0i,(3t+1)i},i=1,2,or if t≡2,3(mod4),then{∞1,(6t+2)1,02},{01,01,(3t)1},{02,02,(3t+1)2},and{01,r1,(b r+t)1},r=1,2,...,t,where{(a r,b r)|r=1,2,...,t}is an(A,t)-system if t≡0,1(mod4)or a(B,t)-system if t≡2,3(mod4),and{02,r2,(d r)1},r=1,2, (3){∞2,02,(d3t+1)1},{∞3,02,(c3t+1)1},where{(c r,d r)|r=1,2,...,3t+1}is an(A,3t+1)-system if t≡0, 1(mod4)or a(B,3t+1)-system if t≡2,3(mod4),and here6t+3 should be regarded as0,form a set of starter blocks for a(3,2)–rotational EST S(v,3).(f,2)–rotational extended Steiner triple systems641Definition3.7.Let t be a positive integer.A(Z6t\{0,3t,9t−12,6t−1},3t−2)-system is a set of ordered pairs of integers{(a r,b r)|r= 1,2,...,3t−2}such that(i){a r,b r|r=1,2,...,3t−2}=Z6t\{0,3t,9t−12,6t−1},and(ii)b r−a r=r,for r=1,2,...,3t−2.Lemma 3.8.If t≡1or3(mod4),then there exists a(Z6t\{0,3t,9t−12,6t−1},3t−2)-system.Proof.Let t≡3(mod4).Then the following ordered pairs form a(Z6t\{0,3t,9t−12,6t−1},3t−2)-system:(r,3t−r),r=1,2,...,3t−12,(3t+r,6t−1−r),r=1,2,...,3t−32.Definition3.9.Let t be a positive integer.A(Z6t\{0,3t,6t−2,6t−1},3t−2)-system is a set of ordered pairs of integers{(a r,b r)|r= 1,2,...,3t−2}such that(i){a r,b r|r=1,2,...,3t−2}=Z6t\{0,3t,6t−2,6t−1},and(ii)b r−a r=r,for r=1,2,...,3t−2.Lemma3.10.If t≡0(mod4)and t≥4,then there exists a(Z6t\ {0,3t,6t−2,6t−1},3t−2)-system.Proof.Let t≡0(mod4)and if t=4,then the following ordered pairs form a(Z24\{0,12,22,23},3t−2)-system:(16,17),(5,7),(18,21),(4,8),(14,19),(3,9),(13,20),(2,10),(6,15),(1,11).If≥8,then the following ordered pairs form a(Z6t\{0,3t,6t−2,6t−1},3t−2)–system:(r,3t−r),r=1,2,...,3t−12,(3t+r,6t−3−r),r=1,2,...,t−2,(4t−1+r,5t−2−r),r=1,2,...,t−44,9t−62−r,9t−42+r,r=1,2,...,t−84,3t 2,9t−62,4t−1,9t−42,(5t−2,6t−3),19t−124,19t−84.642Chung Je ChoDefinition3.11.Let t be a positive integer.A(Z6t\{0,3t2,3t,6t−1},3t−2)-system is a set of ordered pairs of integers{(a r,b r)|r= 1,2,...,3t−2}such that(i){a r,b r|r=1,2,...,3t−2}=Z6t\{0,3t2,3t,6t−1},and(ii)b r−a r=r,for r=1,2,...,3t−2.Lemma3.12.If t≡2(mod4),then there exists a(Z6t\{0,3t2,3t,6t−1},3t−2)-system.Proof.Let t≡2(mod4).Then the following ordered pairs form a (Z6t\{0,3t2,3t,6t−1},3t−2)-system:(r,3t−r),r=1,2,...,3t−22,(3t+r,6t−1−r),r=1,2,...,3t−22.Lemma3.13.There exists a(3,2)–rotational EST S(15,3).Proof.The following triples{∞1,∞2,∞3},{02,12,21},{02,02,22},{∞1,∞1,∞1},{∞2,∞2,∞2},{∞3,∞3,∞3},{02,01,41},{01,01,11},{∞1,01,31},{∞1,02,32},{∞2,02,31},{∞3,02,51}form a set of starter blocks of a(3,2)–rotational EST S(15,3).Lemma3.14.If v≡3(mod12),then there exists a(3,2)–rotational EST S(v,3).Proof.Let v=2·6t+3and let t be a nonnegative integer.The case t=0is trivial and the case v=15has been treated in Lemma3.13. Let t≥2.Then the following triples{∞1,∞2,∞3},{02,(3t)1,(6t−1)1},{∞i,∞i,∞i},i=1,2,3,{∞1,0i,(3t)i},i=1,2,{∞2,02,01},{02,02,(3t−1)2}(f,2)–rotational extended Steiner triple systems643 and∞3,02,9t−121if t≡1,3(mod4),or∞3,02,3t21if t≡2(mod4),or{∞3,02,(6t−2)1}if t≡0(mod4),and{01,01,(3t−2)1}if t≡1,2(mod4),or{01,01,(3t−3)1}if t≡0,3(mod4)and{01,r1,(b r+t−1)1},r=1,2,...,t−1,where{(a r,b r)|r=1,2,...,t−1}is a(A,t−1)-system if t≡1,2(mod 4)or a(B,t−1)-system if t≡0,3(mod4)and{02,r2,(d r)1},r=1,2,...,3t−2,where{(c r,d r)|r=1,2,...,3t−2}is a(Z6t\{0,3t,9t−12,6t−1},3t−2)-system if t≡1,3(mod4),or a(Z6t\{0,3t2,3t,6t−1},3t−2)-system ift≡2(mod4),or a(Z6t\{0,3t,6t−2,6t−1},3t−2)-system if t≡0(mod 4),form a set of starter blocks for a(3,2)–rotational EST S(v,3).Now we can conclude the following theorem.Theorem3.15.There exists a(3,2)–rotational EST S(v,3)if and only if v≡3(mod6).In a(3,2)–rotational EST S(v,3),if the blocks{∞i,∞i,∞i},i= 1,2,3,and{∞1,∞2,∞3}are replaced by the blocks{∞1,∞1,∞2}, {∞2,∞2,∞3}and{∞3,∞3,∞1},we get a(3,2)–rotational EST S(v,0). Thus we have the following theorem.Theorem3.16.There exists a(3,2)–rotational EST S(v,0)if and only if v≡3(mod6).。