OFFSET
1,2
COMMENTS
The Yummie permutation is done as follows. Start with a packet of n cards (numbered 1 to n from top to bottom), and deal them into two piles, first to a spectator (pile A), and then to yourself (pile B), saying "You, me," silently to yourself over and over. Then, pick up pile B and deal again, first to the spectator, thereby adding to the existing pile A, and then to yourself, forming a new pile B. Repeat, picking up the diminished pile B, and dealing "You, me" as before. Eventually, just one card remains in pile B; place it on top of pile A. The sequence of cards in pile A determines the Yummie permutation ("You, me" said fast sounds like "Yummie").
LINKS
Andrew Howroyd, Table of n, a(n) for n = 1..2048
Colm Mulcahy, The Yummie Deal and Variations , Card Colm, MAA Online, April 2009
EXAMPLE
a(9) = 15, because when the Yummie permutation is applied to {1,2,3,4,5,6,7,8,9} we get {6,2,4,8,9,7,5,3,1}, which corresponds to the product of a disjoint five cycle and a three cycle, and hence has order 15.
PROG
(PARI)
P(n, i)={if(i%2, n-(i\2), P(n\2, (n-i)\2+1))}
Follow(s, f)={my(t=f(s), k=1); while(t>s, k++; t=f(t)); if(s==t, k, 0)}
Cycles(n)={my(L=List()); for(i=1, n, my(k=Follow(i, j->P(n, j))); if(k, listput(L, k))); vecsort(Vec(L))}
a(n)={lcm(Cycles(n))} \\ Andrew Howroyd, Apr 28 2020
CROSSREFS
KEYWORD
nonn
AUTHOR
Colm Mulcahy, Jun 04 2009
STATUS
approved