1) same kind arithmetic sequence of numbers
同类等差数列
1.
Based on the definition of basic prime number and same kind sequence of numbers, the guess of twinborn prime number is studied using the properties of same kind arithmetic sequence of numbers and the formula expressing the construction of twinborn prime number .
定义了基础素数与同类等差数列 ,用同类等差数列的性质和孪生素数组成公式探讨了孪生素数猜想 ,此外还讨论了三生素数的有关问
2) arithmetic progression
等差数列
1.
Study of quantitative diagnostic method of wind observation data ——utility of arithmetic progression;
测风记录定量诊断方法的研究——等差数列的利用
2.
A core property of the arithmetic progression is the same difference.
等差是等差数列最核心的本质特征。
3.
The article approaches the resolution of the sum of preceding N terms among a type of particular sequence,which consists of arithmetic product of arithmetic progression or arithmetic product reciprocal.
对由等差数列的乘积以及乘积的倒数所构成的一类特殊的数列的前n项和的求解作了探讨,得出两个定理及6个推论以及应用。
3) arithmetic series
等差数列
1.
The paper presents a further extension of a problem of sum limit from {n] to the arithmetic series {a_n].
将一和式极限问题中的特殊数列{n}推广到一般的等差数列{an},使得所得公式适用于求更广泛的和式的极限。
2.
If{a n} is arithmetic series with the first term a 1>0 ,and the common difference d>0 and s>0, writing as p n=mk=1(1+sa k), this paper gives a class of inequalities of the upper and lower bound of p n.
若 {an}是等差数列 ,首项a1>0 ,公差d >0 ,s >0 ,记Pn= nk =1 (1+ sak) 。
4) arithmetic sequence
等差数列
1.
A property of finite arithmetic sequences and its application;
关于有穷等差数列的一个性质
2.
The essay obtains a set of more universal inequality by studying several inequality of the power of positive term arithmetic sequence.
本文通过对正项等差数列方幂的若干不等式研究,得出了一组更具普遍性的不等式。
3.
Finally,a series of the combinatorial identity are abtained by applying two Vandermonde typed determinant and the property of the arithmetic sequence.
利用一元多项式的思想给出了Vandermonde行列式的一种计算法,接着利用此方法讨论了具有Vandermonde类型行列式的计算,最后使用两个Vandermonde类型行列式和等差数列的性质构造一系列组合恒等式。
6) arithmetic progression method
等差数列法
补充资料：等差数列
| 等差数列 arithmetic sequence 从第二项起每项与前一项之差等于一个常数的数列。又称算术数列 。 等差数列相邻两项的差称为公差。如果用d表示等差数列a1，a2，…，an，…的公差，那么d＝a2－a1＝a3－a2＝…＝an+1－an＝…，通项公式为an＝a1＋（n－1）d，前n项和公式为Sn＝  （a1＋an）或 。 |
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