Hi guys, I stumbled into something interesting and I don't know if it's important but I'd like to share it with ya'll :-)

For example:

----- 2 -----: 2 Shuffles

----- 6 -----: 3 Shuffles

----- 14 -----: 4 Shuffles

----- 30 -----: 5 Shuffles

----- 62 -----: 6 Shuffles

In shuffles are when you split the card in half and interleave them with the top card becoming the second card from the top.

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14]

[8, 1, 9, 2, 10, 3, 11, 4, 12, 5, 13, 6, 14, 7]

[4, 8, 12, 1, 5, 9, 13, 2, 6, 10, 14, 3, 7, 11]

[2, 4, 6, 8, 10, 12, 14, 1, 3, 5, 7, 9, 11, 13]

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14]

4 Shuffles

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## Comments

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TopNewestIn general, if you in-shuffle \(N\) cards \(k\) times, and \(2^k \equiv 1\) mod \((N+1),\) you get the original deck back. Your case is \(N = 2^{n+1}-2\) and \(k = n+1.\)

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Thank you! I stumbled into this special case before knowing the formula. It all makes sense now

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This is very interesting. Do you have a proof of this? What happens when the number of cards is not of that form?

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From Patrick Corn, the number of times k you have to shuffle N cards is \(2^{k} = 1 (mod N + 1)\). What this means: what is the least k such that, when you exponentiate 2 ( \( 2^{1} , 2^{2}, 2^{3} \) )by k and you divide it by N + 1, will give you a remainder of one.

I might post another thing for a proof of this :-)

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This shuffle sends the nth card from the top to the "2n mod (N+1)"th position, so it's guaranteed to cycle once the number of shuffles k reaches the order of 2 in the cyclic group of N+1 elements. In general, fewer shuffles than this will be required, but you found a particular set of deck sizes where this k is the minimum.

Fun corollary: since 52+1 is prime, a regular card deck requires the worst case 52 shuffles before it first returns to its starting order. That is, unless you alter the order of interleaving so that the top card stays on top, in which case 8 shuffles suffice.

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