关于Smarandache-Pascal数列的几个猜想

On Several Conjectures Related to the Smarandache-Pascal Sequences

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.