摘要
设Fn表示数列Fibonacci数列的第n项,an表示{an=an-1+an-3+an-4}的第n项.得到如下结果:Fi+s)2,a6=(∑m+2Fi+s)2,a4=(∑m+1Fi+s)2且an=an-1+an-3+an-4,则(i)a2n=(∑m+n-1Fi+s)2,设a1=1,a2=(∑mi=3i=ni=1i=2Fi+s);(ii)a2n+1=(∑m+n-1Fi+s)(∑m+n-1Fi+s)+(-1)n+1X(m,s).其中X(m,Fi+s)(∑m+na2n-1+a2n-2+a2n-3=2(∑m+n-2i=ni=n+1i=n-1i=ns)=(Fm+s+1-Fs+1)(Fm+s+2-Fs+2)-1.从而肯定回答了徐道提出的一个猜测.
We denote respectively by F_n and by a_n the n'th term of Fibonacci Numbers and {a_n=a_(n-1)+a_(n-3)+a_(n-4)}.In this paper,we obtain the following results:Suppose that a_1=1,a_2=(∑mi=1F_(i+s))~~2,a_4=(∑m+1i=2F_(i+s))~2,a_6=(∑m+2i=3F_(i+s))~2 and a_n=a_(n-1)+a_(n-3)+a_(n-4).Then we have:(i)a_(2n)=(∑m+n-1i=nF_(i+s))~2,a_(2n-1)+a_(2n-2)+a_(2n-3)=2(∑m+n-2i=n-1F_(i+s))(∑m+n-1i=nF_(i+s));(ii)a_(2n+1)=(∑m+n-1i=nF_(i+s))(∑m+ni=n+1F_(i+s))+(-1)^(n+1)X(m,s),where X(m,s)=(F_(m+s+1)-F_(s+1))(F_(m+s+2)-F_(s+2))-1.Moreover a conjecture posed by Xu is affirmed.
出处
《商丘师范学院学报》
CAS
2004年第2期73-74,共2页
Journal of Shangqiu Normal University
