Here’s a shuffle I’ve not seen before:
- Take an ordinary deck of 52 cards and place them, face up, in the following pattern:
Going from the top of the deck to the bottom, placing cards down left-to-right, put 13 cards in the top row:
11 cards in the next row:
then 9 cards in the next row:
then 7 cards in the next row:
then 5 cards in the next row:
then 3 cards in the next row:
and finally, the remaining 4 cards in the last row: - Now, take the left-most card in each row and move it to the right of the others (effectively, this is a cyclic shift of that row of cards to the left).
- Finally, reassemble the deck by reversing the order of the placement.
In memory of the great German mathematician Edmund Landau (1877-1938, see also this bio), I call this the Landau shuffle. As with any card shuffle, this shuffle permutes the original ordering of the cards. To restore the deck to it’s original ordering you must perform this shuffle exactly 180180 times. (By the way, this is called the order of the shuffle.) Yes, one hundred eighty thousand, one hundred and eighty times. Moreover, no other shuffle than this Landau shuffle will require more repetitions to restore the deck. So, in some sense, the Landau shuffle is the shuffle that most effectively rearranges the cards.
Now suppose we have a deck of 
Question: What is the largest possible order of a shuffle of this deck (and how do you construct it)?
This requires a tiny bit of group theory. You only need to know that any permutation of 
The Landau function 
Example: If 
Oddly, my favorite mathematical software program SageMath does not have an implementation of the Landau function, so we end with some SageMath code.
def landau_function(n):
L = Partitions(n).list()
lcms = [lcm(P) for P in L]
return max(lcms)
Here is an example (the time is included to show this took about 2 seconds on my rather old mac laptop):
sage: time landau_function(52)
CPU times: user 1.91 s, sys: 56.1 ms, total: 1.97 s
Wall time: 1.97 s
180180
first, the 9 of Spades 
next (so it is below the 
. 
, 10 with 10, Jack with 11, Queen with 12. Therefore, the ranks may be identified with the additive group of integers (mod 13), namely: 
.
.
then you know that 13 cards later (and If you reach the end of the deck, continue counting with the top card) is the 
, 13 cards after that is the 
, and 13 cards after that is the 
.
to each element of the deck. Some people call this a “false shuffle” of “false cut.”
and the card drawn is 
then
. What is the card the player picked?
(which is Clubs 
.
and 
” Mathematics of Computation Vol. 29, No. 129 (Jan., 1975), pp. 243-269
squares of integers 
whose rows all add to a constant C, whose columns all add to C, whose diagonals both add to C, and whose “broken diagonals” all add to C. An example was given by the German artist Albrecht Durer in the 1514 engraving called 
, let 
be two independent, uniform random variables taken from 
, and let 
. The last banana game concerns the case 
. Here, I’m interested in investigating the question: What is 
? 
.
, we have
.
is defined as 
, but there’s a ![E(Z)=\sum_{k=0}^{n-1} P(Z>k)=\sum_{k=0}^{n-1}[1-P(Z\leq k)]](https://s0.wp.com/latex.php?latex=E%28Z%29%3D%5Csum_%7Bk%3D0%7D%5E%7Bn-1%7D+P%28Z%3Ek%29%3D%5Csum_%7Bk%3D0%7D%5E%7Bn-1%7D%5B1-P%28Z%5Cleq+k%29%5D&bg=ffffff&fg=323232&s=0&c=20201002)
.
of degree m+1 in the variable n that gives the value of the sum of the m-th powers of the first n consecutive positive integers:
.
but they do and Faulhaber’s results establish this existence.) These polynomials were discovered by (German) mathematician 
to 
. It was about 100 years later, in the early 1700s, that (Swiss) mathematician 
,
,
lower order terms.
be m independent, uniform random variables taken from 
. Again we ask: What is 
case generalizes to:
to make 2018.
where p, q, r, s are primes.
where p, q, r, s are primes.
using the fewest number of distinct digits.
for integers 
Yet Another Mathblog 
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