关于Levine-O'Sullivan序列  

On Levine-O'Sullivan Sequence

作　　者：薛方刚 XUE Fanggang(School of Mathematics and Statistics,Nanjing University of Information Science and Technology,Nanjing,Jiangsu,210044,P.R.China)

机构地区：[1]南京信息工程大学数学与统计学院,南京江苏210044

出　　处：《数学进展》2024年第3期662-666,共5页Advances in Mathematics（China）

摘　　要：Levine-O’Sullivan序列Q={q1,q2,…}定义为q1=1,qn=max1≤k≤n-1{(+1)(n-qk)},n=2,3,…令Wn=(n+1)qn-nqn-1+1,s(n)=1/qn-1/(n(n+1))log(n(Wn+1))/(Wn-n).[Acta Math.Sinica(Chinese Ser.),2015,58(4):529-534]中证明了s(n)>1/n1.667对于所有的n>2500均成立,从而肯定了[Sci.China Math.,2013,56(5):951-966]中提出的猜想.本文证明了limn→∞log s(n)/log n=-√5+1/2.The Levine-O'Sullivan sequence Q={q_1,q_2,…}is defined as q_1=1 and q_n=max_(1≤k≤n-1){(k+1)(n-q_k)}for n=2,3,....Let Wn=(n+1)q_n-nq_(n-1)+1 and s(n)=1/q_n-1/n(n+1)log n((W_n+1))/(W_n-n).By confirming a conjecture posed in[Sci.China Math.,2013,56(5):951-966],it has been proved in[Acta Math.Sinica(Chinese Ser.),2015,58(4):529-534]that s(n)>1/n~(1.667)for all n>2500.In this paper,we prove that lim_(n→∞)log s(n)/log n=√5+1/2.

关 键 词：Levine-O’Sullivan序列 sum-free集合 FIBONACCI序列 

分 类 号：O157.5[理学—数学]