摘要
对任意数列{bn},它的Smarandache-Pascal数列是通过{bn}定义的一个新的数列{Tn},其中T1=b1,T2=b1+b2,T3=b1+2b2+b3.一般地,当n≥2时,Tn+1=∑k=0nCnk·bk+1,其中Ckn=n!/k!(n-k)!为组合数.利用初等方法以及组合数和Fibonacci数的性质研究并解决猜想:设{Tn}是由{bn}={F8n+1}={F1,F9,F17,…}生成的Smarandache-Pascal数列,则有恒等式Tn+1≡49(Tn-Tn-1),其中n≥2.
For any fixed sequence {bn }, its Smarandache-Pascal sequences are a new sequence defined by {bn } , in which T1 =b1 , T2 =b1 +b2 and T3 =b1 +2b2 +b3. Generally,Tn+1=∑k=0^nCn^n·bk+1 for all n≥2, where Cn^k=n!/k!(n-k)! is the combination number. In this paper, we use the elementary method and the properties of the combination number and Fibonacci number to prove the conjecture: For { bn } = (F8n+1 } , we have the identity Tn+1≡49(Tn-Tn-1) for all n≥2.
出处
《西南大学学报(自然科学版)》
CAS
CSCD
北大核心
2013年第10期67-70,共4页
Journal of Southwest University(Natural Science Edition)
基金
国家自然科学基金(11071194)
陕西省教育厅科学研究计划项目(2013JK0566)