Date: Sat, 24 May 2008 14:43:21 +0200
From: Richard Mathar <mathar(AT)strw.leidenuniv.nl>

%N A139768 Numbers n such that (10^(n+1) mod 9^(n+1))/(10^n mod 9^n)=10, or A139739(n+1)/A139739(n)=10.
%C A139768 Also, this is the set of numbers n such that 9*floor((10/9)^(n+1))==10*floor((10/9)^n). Here is a proof:

Proof:

9*floor[ 10^(n+1)/9^(n+1)] = 10 floor [ 10^n/9^n]
replace the floor[...] by a division with explicit remainder :

9*[ 10^(n+1)/9^(n+1) - {10^(n+1) mod 9^(n+1)}/ 9^(n+1)] 
   = 10 [ 10^n/9^n - {10^n mod 9^n}/9^n]

Divide both sides through 9

10^(n+1)/9^(n+1) - {10^(n+1) mod 9^(n+1)}/ 9^(n+1)
   = (10/9) [ 10^n/9^n - {10^n mod 9^n}/9^n]

Multiply both sides by 9^(n+1)

10^(n+1) - {10^(n+1) mod 9^(n+1)}
   = 10 [ 10^n - {10^n mod 9^n}]

expand right hand side

10^(n+1) - {10^(n+1) mod 9^(n+1)} = 10^(n+1) - 10 {10^n mod 9^n}]

eliminate common terms, switch sign

{10^(n+1) mod 9^(n+1)} = 10 {10^n mod 9^n}]

Divide through
{10^(n+1) mod 9^(n+1)}/{10^n mod 9^n} = 10
q.e.d, the original definition.