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%I #10 Apr 28 2020 15:41:13
%S 1,2,3,3,5,5,6,7,15,20,11,24,24,14,6,28,17,120,55,180,21,18,60,42,90,
%T 153,140,429,56,152,60,70,483,3640,180,272,72,1260,180,252,174,1260,
%U 36,442,1404,660,47,496,240,481,48,98,570,572
%N a(n) is the order (or period) of the "Yummie" permutation applied to a set of n objects.
%C 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").
%H Andrew Howroyd, <a href="/A161172/b161172.txt">Table of n, a(n) for n = 1..2048</a>
%H Colm Mulcahy, <a href="http://www.maa.org/community/maa-columns/past-columns-card-colm/the-yummie-deal-and-variations">The Yummie Deal and Variations </a>, Card Colm, MAA Online, April 2009
%e 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.
%o (PARI)
%o P(n,i)={if(i%2, n-(i\2), P(n\2, (n-i)\2+1))}
%o Follow(s, f)={my(t=f(s), k=1); while(t>s, k++; t=f(t)); if(s==t, k, 0)}
%o 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))}
%o a(n)={lcm(Cycles(n))} \\ _Andrew Howroyd_, Apr 28 2020
%Y Cf. A051732, A161173, A289386.
%K nonn
%O 1,2
%A _Colm Mulcahy_, Jun 04 2009
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