1) Sequence gradation
序列递变
2) recursive sequence
递归序列
1.
The relation between the iteration of projective function and the linear recursive sequences of order 2 is given.
先给出射影函数的迭代与 2阶线性递归序列的关系 ,进而得到此递归序列与Bernoulli数的一个恒等
3) recursive sequence
递推序列
1.
This paper proves that the Diophantine Equation has only positive integral solution with the methods of recursive sequence,congruence and quadratic remainder.
利用一种初等的证明方法,即递推序列、同余式和平方剩余的方法,对不定方程x2-11y4=38的正整数解进行了研究,证明了不定方程x2-11y4=38仅有正整数解(x,y)=(7,1)。
2.
In this paper the author has proved that the Diophantine equation x2-3y4=22 has only positive integral solutions(x,y) =(5,1),(85,7) with the methods of recursive sequence,congruence and quadratic remainder.
利用一种初等的证明方法,即递推序列,同余式和平方剩余的方法,对一个不定方程x2-3y4=22的正整数解进行了研究,证明了不定方程x2-3y4=22仅有正整数解(x,y)=(5,1),(85,7)。
4) recurrent sequence
递归序列
1.
This paper proves that the Diophantine equation has only integer solution with the help of the Pell method taking an integer>1 as module to make inconsistency,the natures of recurrent sequences and equivalent Pell equation.
采用对方程取某个正整数M>1为模来制造矛盾的同余法和利用递归序列的性质,以及Pell方程的性质,证明不定方程x3-1=13y2仅有整数解(x,y)=(1,0)。
2.
In this paper,the author has proved, with two method of contradictor recurrent sequences and congruence when modules of some positive integer M>1, that the Diophantine equation x~3+1=19y~2 has only integer solution(x,y)=(1,0).
利用两种初等的方法,即对方程取某个正整数M>1为模来制造矛盾的同余法和递归序列法,证明了不定方程x3 -1=19y2 仅有整数解(x,y)=(1,0),从而进一步的证明了方程x2 -19y2 =-13无整数解;方程x2 -3r2 =-3仅有整数解(1。
3.
With the method of recurrent sequence and congruences,proved that the Diophantine equation x3+1 =37y2has only integer solution(x,y)=(-1,0),(11,±6).
利用递归序列,同余式证明了丢番图方程x 3+1=37y2,仅有整数解(x,y)=(-1,0),(11,±6)。
5) recurrence sequences
递归序列
1.
Formulas for simple and direct computations for Euler--Bernoulli polynomials of n variablesare presented,some identities containing recurrence sequences and Euler--Bernooulli polynomials of n variables have been established.
给出简捷计算n元Euler-Bernoulli多项式的公式，建立一些包含递归序列和上述多项式的恒等式。
6) Recurrence sequence
递推序列
1.
For the special recurrence sequence ｛a_n｝, the characteristic polynomial of which is C(x)=(x-1)r ,the article give a general solution of that and an abundant and essential condition of judging that:existing integer r:r≥1,for whichever n≥0, Δra_n=0 is certainly be obtained.
对于特征多项式为C(x) =(x- 1) r的特殊递推序列an,给出了它的通解求法以及判定它的充要条件 :存在整数r≥ 1,对任意的n≥ 0 ,Δran =0成立。
补充资料：递变
1.交替变化;演变。
说明：补充资料仅用于学习参考，请勿用于其它任何用途。
参考词条