# Proof for OEIS A002326 multiplicative order of 2 (mod 2n - 1)

I am trying to prove the amount of out-shuffles needed to return a for example set of cards to its original state. The sequence I am talking about is the A002326 sequence.

If you number for example 12 cards like this: 0,1,2,3,4,5,6,7,8,9,10,11 and shuffle 10 times the cards returned to their original state, this is because every number except for 0 and 11 are connected to each other in this 'circuit', thus needing (2n-2) shuffles. But if you would take 10 cards not all the numbers are connected to each other. Click here for the pattern of 10 cards. (To get to the next number simply multiply by 2 and if necessary substract (n2-1) where n is the amount of chips per stack)

For numbers like these where there are multiple circuits it gets interesting. I found that:

1. The biggest circuit length is always equal to the length of the circuit of chip number 1.
2. The length of all other circuits divide into the length of chip number 1.

Therefore it can be stated that the amount of shuffles needed to return to the original state is equal to the length of circuit 1. But I cannot prove the 2 rules above, I just know it is true at least for the first 6000 cards. Can someone please help?

If we know that $2^k \equiv 1 \bmod 2n{-}1$, which is the cycle length of $1$ for the smallest $k>0$, then we also know that $a\cdot2^k\equiv a \bmod 2n{-}1$. So even if $a$ has a shorter cycle, it must divide $k$ in order to fulfill this equivalence. So this effectively gives you your two rules.