请输入您要查询的字词:

 

单词 SequencesB2n1AndB2n11AreDivisibleByB1
释义

sequences b2n-1 and b2n-1+1 are divisible by b+1


Consider the alternating geometric finite series

Sm+1(μ)=i=0m(-1)i+μbi,(1)

where μ=1,2 and b2 an integer. Multiplying (1) by -b and subtracting from it

(b+1)Sm+1(μ)=i=0m(-1)i+μbi-i=0m(-1)i+1+μbi+1

and by elemental manipulations, we obtain

Sm+1(μ)=(-1)μ[1-(-1)m+1bm+1]b+1=i=0m(-1)i+μbi.(2)

Let μ=1, m=2n-1. Then

b2n-1b+1=-i=02n-1(-1)ibi.(3)

Likewise, for μ=2, m=2n-2

b2n-1+1b+1=i=02n-2(-1)ibi,(4)

as desired.

Palindromic numbers of even length

As an application of above sequences, let us consider an even palindromic numberMathworldPlanetmath (EPN) of arbitrary length 2n which can be expressed in any base b as

(EPN)n=k=0n-1bk(b2n-1-k+bk)=k=0n-1bkbk[(b2n-1+1)+(b2k-1)],(5)

where 0bkb-1.It is clear, from (3) and (4), that (EPN)n is divisible by b+1. Indeed this one can be given by

(EPN)n=(b+1)k=0n-1j=02(n-1-k)bk(-1)jbk+j.(6)
随便看

 

数学辞典收录了18232条数学词条,基本涵盖了常用数学知识及数学英语单词词组的翻译及用法,是数学学习的有利工具。

 

Copyright © 2000-2023 Newdu.com.com All Rights Reserved
更新时间:2025/5/4 20:01:45