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


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)
Title sequences b2n-1 and b2n-1+1 are divisible by b+1
Canonical name SequencesB2n1AndB2n11AreDivisibleByB1
Date of creation 2013-03-22 16:14:19
Last modified on 2013-03-22 16:14:19
Owner perucho (2192)
Last modified by perucho (2192)
Numerical id 6
Author perucho (2192)
Entry type Derivation
Classification msc 11A63