Skip to main content
Post Migrated Away to math.stackexchange.com
added 26 characters in body
Source Link

startStart with a number equal to -1$-1$ modulo 2^n then$2^n$. Then, after one step, the number is (3*(-1)+1)/2=-1$\frac{3(-1)+1}{2}=-1$ modulo 2^n/2=2^(n-1)$\frac{2^n}{2}=2^{n-1}$, so inductively it will increase n$n$ steps.

start with a number equal to -1 modulo 2^n then after one step the number is (3*(-1)+1)/2=-1 modulo 2^n/2=2^(n-1), so inductively it will increase n steps

Start with a number equal to $-1$ modulo $2^n$. Then, after one step, the number is $\frac{3(-1)+1}{2}=-1$ modulo $\frac{2^n}{2}=2^{n-1}$, so inductively it will increase $n$ steps.

Source Link
user35593
  • 2.3k
  • 12
  • 19

start with a number equal to -1 modulo 2^n then after one step the number is (3*(-1)+1)/2=-1 modulo 2^n/2=2^(n-1), so inductively it will increase n steps