1) Fractional Recurrent Sequence
分式递归数列
1.
The Period of a Class of Fractional Recurrent Sequence;
一类分式递归数列的周期性
2) fractional linear recursive sequence of number
分式线性递归数列
3) recursive sequence
递归数列
1.
In this paper,the author has proved that the Diophantine equation x3+64=21y2 has only an integer solution(x,y)=(-4,0),(5,±3) and then gives all integer solution of x3+64=21y2 by using the elementary methods such as recursive sequence,congruent fomula and quadratic residue.
利用递归数列、同余式和平方剩余几种初等方法,证明了不定方程x3+64=21y2仅有整数解(x,y)=(-4,0),(5,±3);给出了x3+64=21y2的全部整数解。
2.
In this paper,the author has proved that the Diophantine equation x3+27=7y2 has only an integer solution(x,y)=(-3,0),(1,±2) and then gives all integer solution of x3+27=7y2 by using the elementary methods such as recursive sequence,congruent fomula and quadratic residu
利用递归数列、同余式和平方剩余几种初等方法,证明了不定方程x3+27=7y2仅有整数解(x,y)=(-3,0),(1,±2);给出了x3+27=7y2的全部整数解。
3.
By applying the properties of determinant, the general formula of a usual type of linear recursive sequence is studied.
应用行列式的有关性质，研究了一般形式的线性递归数列的通项公式。
4) recurrent sequence
递归数列
1.
In this paper,the authors use recurrent sequence to prove that the diophantine equation x3-1=26y2 has only integer solutions(x,y)=(1,0),(3,±1),(313,±1086).
用递归数列方法证明了方程x3-1=26y2全部整数解是(1,0),(3,±1),(313,±1086)。
2.
By using congruence and recurrent sequence,the author has proved that the Diophantine equation x3+1=86y2 has only the integer solution(x,y)=(-1,0),(7,±2).
利用递归数列、同余式证明了丢番图方程x3+1=86y2仅有整数解(x,y)=(-1,0),(7,±2)。
3.
By using the method of recurrent sequence,the Diophantine equation x3-8=13y2 has been proven to have the only integer solution(x,y)=(5,±3) with gcd(x,y)=1.
利用同余式和递归数列的方法,证明了不定方程x3-8=13y2仅有适合(x,y)=1的整数解(x,y)=(5,±3)。
5) fractional recursive progression
分式递推数列
1.
With matrix in higher mathematics,this article attempts to solve problems concerning fractional recursive progression and to find out the bonding point of matrix with fractional recursive progression.
以高等代数中矩阵为工具 ,解决分式递推数列问题 ,从而寻求到矩阵与分式递推数列的结合
6) polynomial recursion sequenee
多项式递归序列
1.
A polynomial recursion sequenee is constructed to obtain a calculational method of ni=1 m which takes the calculation of ni=1i m as its special conditio
构造一个多项式递归序列 ,得到 ni =1[a +(i -1)d] m 的一种求法 ,使 ni=1im 的计算成特殊情
补充资料：递归数列
| 递归数列 recursive sequence 一种用归纳方法给定的数列。例如，等比数列可以用归纳方法来定义，先定义第一项a1的值(a1≠0)，对于以后的项，用递推公式an+1＝qan（q≠0，n＝1，2，…）给出定义。一般地，递归数列的前k项a1，a2，…，ak为已知数，从第k＋1项起，由某一递推公式an+k＝f（an，an+1，…，an+k-1）( n＝1，2，…）所确定。k称为递归数列的阶数。例如 ，已知 a1＝1，a2＝1，其余各项由公式an+1＝an＋an-1（n＝2，3，…）给定的数列是二阶递归数列。这是斐波那契数列，各项依次为 1，1，2，3，5，8，13，21，…，同样，由递归式an+1－an ＝an－an-1( a1，a2为已知，n＝2，3，… ) 给定的数列，也是二阶递归数列，这是等差数列。 |
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参考词条