Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've been going through Concrete Mathematics and have a question on the inductive proof for the Josephus problem.

Recurrent relation:

$J(1) = 1$

$J(2n) = 2J(n) - 1$

$J(2n+1) = 2J(n) + 1$

Closed form hypothesis:

$J(2^m + l) = 2l + 1$, for $m \ge 0$ and $0\le l<2^m$

Inductive Proof (for the even case):


$J(2^m+l)= 2J(2^{m-1}+l/2)-1=2(2l/2+1)-1=2l+1$

How do the authors get from $2J(2^{m-1}+l/2)-1$ to $2(2l/2+1)-1$?

share|cite|improve this question
Because of the hypothesis – Yimin Feb 16 '13 at 21:10
@Yimin I understand that it comes from the hypothesis, but not how $2l/2+1$ can be plugged in for $J(2^{m-1} + l/2)$ – dbyrne Feb 16 '13 at 21:15
up vote 1 down vote accepted

$J(2^m + l) = 2l + 1$ is your hypothesis. You prove it for $n=0$ and $l=0$. Now comes the step. You assume the hypothesis for numbers smaller then m and l.

Then for $n=m-1$ and $k=\frac l2$, you have $J(2^n + k) = 2k + 1=\frac{2k} 2+1$ from the hypothesis. This is induction.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.