On universal sums of polygonal numbers

For m = 3, 4,..., the polygonal numbers of order m are given by pm(n) =(m- 2) n2 + n(n= 0, 1, 2,...). For positive integers a, b, c and i, j, k 3 with max{i, j, k} 5, we call the triple(api, bpj, cpk)universal if for any n = 0, 1, 2,..., there are nonnegative integers x, y, z such that n = api(x) + bpj(y)+ cpk(z). We show that there are only 95 candidates for universal triples(two of which are(p4, p5, p6) and(p3, p4, p27)), and conjecture that they are indeed universal triples. For many triples(api, bpj, cpk)(including(p3, 4p4, p5),(p4, p5, p6) and(p4, p4, p5)), we prove that any nonnegative integer can be written in the form api(x) + bpj(y) + cpk(z) with x, y, z ∈ Z. We also show some related new results on ternary quadratic forms,one of which states that any nonnegative integer n ≡ 1(mod 6) can be written in the form x2+ 3y2+ 24z2 with x, y, z ∈ Z. In addition, we pose several related conjectures one of which states that for any m = 3, 4,...each natural number can be expressed as pm+1(x1) + pm+2(x2) + pm+3(x3) + r with x1, x2, x3 ∈ {0, 1, 2,...}and r ∈ {0,..., m- 3}.     

不动点集为L1(p)的对合

§ 1.　Introduction　　Let(M ,T)beasmoothinvolutiononasmoothclosedmanifold ,andFdenotesthefixedpointsetof (M ,T) .WhenF =RP( 2k) ,RP(m)∪RP(n) ,∪RP( 2l+ 1 ) (lfixed) ,∪pi =1 RP( 2li+ 1 ) ,∪ri =1 (S1 ) ki ,(Sn1 ×Sn2 ×… ×Snp)∪ {pt},etc .,theexistenceandtherepresentative (uptobordism)of(M ,T)havebeenstudiedin [2 ],[3],[5 ],[6],[7]and[8].Thepurposeofthepaperistodeterminetheexistenceandtherepretentativeuptobord ismofallinvolutionsfixingthelensspaceL1 ( p) ,whereL1 ( p)isa 3 dimens…     

走近函数f(x)={x}

题目 (2009年湖北文9)设x∈R,记不超过x的最大整数为[x],令{x}=x-[x],则{√5+1/2},[√5+1/2],√5+1/2 　　A.是等差数列但不是等比数列 　　B.是等比数列但不是等差数列 　　C.既是等差数列又是等比数列 　　D.既不是等差数列也不是等比数列……     

借用向量分解定理巧求f(x0)的范围

已知函数 f ( xi) ( i =1,2 ,3 ,… )的范围 ,求 f( x0 )的范围 .笔者在同行们研究的基础上 ,借用向量分解定理 ,使这类问题的解决更加简单、明了 ,可操作性强 ,便于实施 .例 1　已知一次函数 f( x) ,1≤ f ( 1)≤2 ,3≤ f ( 2 )≤ 4,试确定 f( 5 )的范围 .解　设一次函数为 f( x) =ax + b,则　 f( 1) =a+ b,f( 2 ) =2 a+ b,f( 5 ) =5 a+ b.记　 p1→ =a+ b,p2→ =2 a+ b,p=5 a+ b显然 p1→ ,p2→ 不共线 ,根据向量分解定理p=λ1 p1→ +λ2 p2→ 　 (λ1 ,λ2 为实数 ) ,即　 5 a+ b=λ1 ( a+ b) +λ2 ( 2 a+ b…     

一道征解题的推广

《中学数学有这样一道题:1034年第3;件门题与解答栏中设(1十x工:).二ao+a lx一{a:x名+…+a:。x“口,吐明a。=aZ一{一a,一:一a一i-a3十ae+…二al一!一a一+a,+,二…=3一1。现在我们将其推,’‘到一般清形:设(l一x一。x么又卜文).二‘。一卜a lx+a:x“+…+a:(、_,)工.(“),则a。+a‘·”二al+ak+1十a么k+1十,·‘二’·‘二ak_1十一,,a:七十a zk_l+a:、_:十…=k一‘.这.里n,幻寸自然数,且k》乳 证明:一戊们知达x“=l的k..根为eos(2敝/k)十葱5 in(2二兀厂k)(m=0,1,2,…,k一1入如呆记., 弓=‘o:(’二/k)一卜1 51”(见二/k), 则cos(几一,:二,k)一…     

我为高考设计题目

题93在数列{an}中,a1=1,且对任意的k∈N*,a2k-1,a2k,a2k+1成等比数列,其公比为qk.(1)若qk=2(k∈N*),求a1+a3+a5+…+a2k-1.(2)若对任意的k∈N*,a2k,a2k+1,a2k+2成等差数列,其公差为dk,设bk=1/qk-1.①求证:{bn}成等差数列,并指出其公差;②若d1=2,试求数列{dk}的前k项和Dk.     

integral frm n=0 to 1(1/x-[1/x])dx=1-C(尤拉常数)

数列。。二1一、一二十… 乙+生一In,.单调减且证明用l(x)的不连续点介1k(k二a。>0,因而{a,}是收敛数列①,即存在实数c使,im{‘l一己一十…十一1一、一,。:)二二。.,,,〔\或几/)实数。称为Euler常数.e=0·5772156649·…任何一个收敛数列都可定义一个实数.如数列{(‘+橇)”}单调增有上界,可定义实数“,即炊(l十一鑫)”一实数·应用极其广泛,但对尤拉常数。的应用知之甚少.下面就黎曼意义下的定积分.给出一个尤拉常数的应用.2,3,…)将仁0,l〕分为若一卜子区间,由题设了(x)的属性及定积分的几何意义,可将J;了(·)‘一示为无穷多个曲边三角形…     

ON MILIN-LEBDEV INEQUALITIES

Let \[\varphi (x) = \sum\limits_{k = 1}^\infty {{A_k}} {x^k},\Phi (x) = {e^{\varphi (x)}} = \sum\limits_{k = 1}^\infty {{D_k}} {x^k}\] \[\begin{gathered} \frac{1}{{{{(1 - x)}^\lambda }}} = \sum\limits_{k = 1}^\infty {{d_k}} (\lambda ){x^k} \hfill \ {\overline \Delta _n}(\lambda ) = {\lambda ^{2 - p}}\sum\limits_{k = 1}^\infty {{k^{p - 1}}} \mathop {|{A_k}|}\nolimits_{}^p - \sum\limits_{k = 1}^\infty {\frac{1}{k}} \hfill \\ \end{gathered} \] Milin-Lebedey proved that \[\sum\limits_{k = 0}^\infty {\frac{{|{D_k}{|^p}}}{{d_k^{p - 1}(\lambda )}}} \leqslant \exp \{ {\lambda ^{1 - p}}\sum\limits_{k = 1}^\infty {{k^{p - 1}}} |{A_k}{|^p}\} \] where p>l and \[\lambda \]>0. In this paper, we have proved the following theorems； Theorem 1. Let \[p \geqslant 1,\lambda > 0\] and \[F(x) = \sum\limits_{k = 0}^\infty {\frac{{|{D_k}{|^p}}}{{d_k^p(\lambda )}}} {x^p}\exp \{ - {\lambda ^{1 - p}}\sum\limits_{k = 1}^\infty {{k^{p - 1}}|{A_k}{|^p}{x^k}} \} (2)\] then F(x) is a decreasing function of x on [0, 1]. This theorem is stronger than the result (1). Theorem 2. Let \[p \geqslant 2,\lambda > 0\] and \[{{\bar Q}_n}(\lambda ) = \frac{1}{{n + 1}}\sum\limits_{k = 0}^n {\frac{{|{D_k}{|^p}}}{{d_k^p(\lambda )}}\exp } \{ - \frac{1}{{n + 1}}\sum\limits_{v = 1}^n {\overline {{\Delta _p}} } (\lambda )\} \] then \[{{\bar Q}_n}(\lambda )\] is a decreasing fimctLon of n(n=l, 2，...)In the case p=2 this is contained in the Miiin-Lebedev's result.     

用二进制解2003年(全国卷)理科压轴题

题　设 { an}是集合 { 2 t+2 s|0≤ s相似文献   

新题征展(93)

A 题组新编 　　1.(万尔遐)已知集合A = {x1,x2,…,xn},n∈N+且 n≥2. 　　(1)将集合A的第k个元素xk与数2k - 1相对应(1≤k≤n),则元素x5的对应数为__,元素__的对应数为128.……     

ON MILIN-LEBDEV INEQUALITIES

Let \[\varphi (x) = \sum\limits_{k = 1}^\infty {{A_k}} {x^k},\Phi (x) = {e^{\varphi (x)}} = \sum\limits_{k = 1}^\infty {{D_k}} {x^k}\] \[\begin{gathered} \frac{1}{{{{(1 - x)}^\lambda }}} = \sum\limits_{k = 1}^\infty {{d_k}} (\lambda ){x^k} \hfill \ {\overline \Delta _n}(\lambda ) = {\lambda ^{2 - p}}\sum\limits_{k = 1}^\infty {{k^{p - 1}}} \mathop {|{A_k}|}\nolimits_{}^p - \sum\limits_{k = 1}^\infty {\frac{1}{k}} \hfill \\ \end{gathered} \] Milin-Lebedey proved that \[\sum\limits_{k = 0}^\infty {\frac{{|{D_k}{|^p}}}{{d_k^{p - 1}(\lambda )}}} \leqslant \exp \{ {\lambda ^{1 - p}}\sum\limits_{k = 1}^\infty {{k^{p - 1}}} |{A_k}{|^p}\} \] where p>l and \[\lambda \]>0. In this paper, we have proved the following theorems； Theorem 1. Let \[p \geqslant 1,\lambda > 0\] and \[F(x) = \sum\limits_{k = 0}^\infty {\frac{{|{D_k}{|^p}}}{{d_k^p(\lambda )}}} {x^p}\exp \{ - {\lambda ^{1 - p}}\sum\limits_{k = 1}^\infty {{k^{p - 1}}|{A_k}{|^p}{x^k}} \} (2)\] then F(x) is a decreasing function of x on [0, 1]. This theorem is stronger than the result (1). Theorem 2. Let \[p \geqslant 2,\lambda > 0\] and \[{{\bar Q}_n}(\lambda ) = \frac{1}{{n + 1}}\sum\limits_{k = 0}^n {\frac{{|{D_k}{|^p}}}{{d_k^p(\lambda )}}\exp } \{ - \frac{1}{{n + 1}}\sum\limits_{v = 1}^n {\overline {{\Delta _p}} } (\lambda )\} \] then \[{{\bar Q}_n}(\lambda )\] is a decreasing fimctLon of n(n=l, 2，...)In the case p=2 this is contained in the Miiin-Lebedev's result.     

关于多项式系数的整除性

设 n是正整数 ,k1 ,k2 ,… ,ks 是适合 k1 +k2 +… +ks=n的非负整数 ,正整数 nk1 k2 … ks=n!k1 !k2 !… ks!称为多项式系数 .本文讨论了当n=a0 +a1 p+a2 p2 +… +arpr ,其中 p为素数且 p≤ n,0≤ ai相似文献   

R\'enyi公式的推广

在本文,作为著名的R\''enyi公式(其刻画了标号连通单圈图的计数显式)的自然推广,我们研究了标号匀称$(k+1)$秩$(p,~q)$超单圈的计数问题,给出了如下的计数显式:$$U_{p,~q}^{(k+1)}=\begin{cases} \frac{p!}{2[(k-1)!]^q}\cdot\sum_{t=2}^q \frac{q^{q-t-1}\cdot sgn(tk-2)}{(q-t)!}, & p=qk, \\ 0,& p\neq qk, \end{cases}$$其中$k,~p,~q$均为正整数.     

关于f[f(x)]的两个“不一定"

《中学生数学》2003年1月上期刊登的裴华明老师的“求f(x)表达式的几种方法”一文中有下面的例题及解答:“例4 已知f{f[f(x)]}=27x+13,求f(x).解因为复合函数f{f[f(x)]}不改变f(x)的次数,故可设F(x)=ax+b,…,故f(x)=3x+1.”     

2004年高考数学模拟训练试题

选择题　本大题共 12小题 ,每小题 5分 ,共 6 0分 .在每小题给出的四个选项中 ,只有一项符合题目要求 .1.设集合M ={x|x =4k± 1,k∈Z} ,N ={x|x =2k +1,k∈Z} ,其中Z表示整数集 ,则下列各项错误的是 (　　 )(A)M∪ (CZN) =Z .　 (B) (CZM )∩N = .(C)M =N . (D)M∪N =Z .2 .已知a ,b是两个单位向量 ,下列命题中错误的是 (　　 )(A) |a|=|b|.(B)a·b =1.(C)a与b方向相反时 ,a +b =0 .(D)a与b方向相同时 ,a =b .3.设命题 p :3≤ 4 ,q :5 6∈ [6 5 ,+∞ ) ,则三个复合命题 :“p且q” ,“p或q” ,“非 p”中 ,真命题的个数为 (　　 …     

一道习题的推广

《中学数学》1983年第4期问题征解中有这样一题,求证 1+2+3+…+1983|1~5+2~5+3~5+…+1983~5。事实上,我们有一般的结论:1°。1+2+3+…+n|1~5+2~5+3~5+…+n~5,甚至更一般的结论:2°。1+2+3+…+n|1~(2k+1)+2~(2k+1)+3~(2k+1)+…+n~(2k+1)。这里n、k为任意自然数,为了证明这一结论,我们要用到整数的两个性质。性质1。两个连续整数必互质。性质2。如果(p,q)=1,p|m,q|m则pq|m、((p,q)表示p与q的最大公约数)。此二性质都很容易用反证法证明,这里从略。我们来证明上述结论2°。证∵ 1+2+…+n=(1/2)n(n+1),记 S_(2k+1)(n)=1~(2k+1)+2~(2k+1)+…+n~(2k+1),     

表整数为一个素数与一个素数的k次方和

本文研究了小区间上的华罗庚定理。即令Ek(x) =# { {n≤x ;2 ｜n ,k是奇数 ,n ≠ p1+pk2 } ∪ {n≤x ;2 ｜n ,2｜k ,(p - 1 ) ｜k ,n 1 (modp) ,n≠ p1+pk2 } }。在GRH下 ,得到了对任意的k≥ 2 ,A >0 ,0 <ε<14,有Ek(x+H) -Ek(x) 相似文献   

On Some Cycles in Wenger Graphs

Let p be a prime,q be a power of p,and let Fq be the field of q elements.For any positive integer n,the Wenger graph Wn(q)is defined as follows:it is a bipartite graph with the vertex partitions being two copies of the(n+1)-dimensional vector space Fq^n+1,and two vertices p=(p(1),…,p(n+1))and l=[l(1),…,l(n+1)]being adjacent if p(i)+l(i)=p(1)l(1)i-1,for all i=2,3,…,n+1.In 2008,Shao,He and Shan showed that for n≥2,Wn(q)contains a cycle of length 2 k where 4≤k≤2 p and k≠5.In this paper we extend their results by showing that(i)for n≥2 and p≥3,Wn(q)contains cycles of length 2k,where 4≤k≤4 p+1 and k≠5;(ii)for q≥5,0相似文献   

关于多项式系数的整除性

设n是正整数,k1,k2,…+k1=n的非负整数,正整数[nk1k2…ks]=n！/k1!k2!…k5!称为多项式系数,本文讨论了当n=a0+a1p+a2p^2+…arp^r,其中p为素数且p≤n,0≤ai&;lt;p(0≤i≤r);ki=a0^(i)+a1^(i)p+…+ar^(i)p^r,其中ki≤0,∑^si=1,ki=n,0≤ak^(i)p(0≤i&;lt;s)时多项式系数的整除性问题,得出的结果推广了著名的Lucas定理^[1].     

基于0-1序列的正态自回归建模方法

<正> §1.引言 设{z_t}是一个零均值平稳正态AR(p)随机序列,即 z_t=φ_(p1)z_(t-1)+φ_(p2)z_(t-2)+…+φ_(pp)z_(t-p)+a_t, t=0,±1,…,(1.1)其中{a_t}是 N(0,σ_a~2)的正态白噪声,φ(B)= 1-sum from k=1 to p(φ_(pk)B~k的根都在单位圆外.众所周知,对(1.1)中的参数φ_(pk),k=1,…,p和σ_a~2,可以利用解Toeplitz方程