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A051732
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Number of rounds of shuffling required to restore a deck of n cards to its original order: shuffling is done by keeping first card, putting second at end of deck, keeping next, putting next at end and so on.
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7
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1, 1, 2, 2, 3, 5, 6, 6, 4, 9, 4, 28, 10, 9, 14, 12, 5, 70, 18, 24, 10, 7, 210, 126, 110, 60, 26, 120, 9, 29, 30, 60, 6, 33, 308, 42, 60, 990, 30, 374, 27, 41, 60, 2618, 840, 840, 420, 1386, 24, 15, 50, 644, 840, 53, 18, 1386, 14, 13300, 2520, 1260, 55, 6930, 50, 60, 7
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OFFSET
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1,3
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COMMENTS
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The shuffling process is the same as the 'deal one, skip one' method described in A289386 except that dealt cards are placed face up. With this variation the first card always remains the first card.
Equivalently, place the numbers 1..n-1 on a circle and cyclically mark the 2nd unmarked number until all numbers are marked. The sequence in which the numbers are marked defines a permutation. The order of this permutation is a(n). The numbers 1..n can also be used, but in that case the number 1 should be marked first.
(End)
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LINKS
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FORMULA
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a(2^n+1) = n+1. - Ripatti A. (ripatti(AT)inbox.ru), Feb 04 2010
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EXAMPLE
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a(6) = 5 because it takes 5 rounds of shuffling to return the cards to their original order as illustrated below:
1 2 3 4 5 6
1 3 5 2 6 4
1 5 6 3 4 2
1 6 4 5 2 3
1 4 2 6 3 5
1 2 3 4 5 6
(End)
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PROG
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(PARI)
P(n, i)={my(d=2*n+1-2*i); while(d<n, d*=2); 2*n-d}
Follow(s, f)={my(t=f(s), k=1); while(t>s, k++; t=f(t)); if(s==t, k, 0)}
CyclePoly(n, x)={my(q=0); for(i=1, n, my(l=Follow(i, j->P(n, j))); if(l, q+=x^l)); q}
a(n)={my(q=CyclePoly(n, x), m=1); for(i=1, poldegree(q), if(polcoeff(q, i), m=lcm(m, i))); m} \\ Andrew Howroyd, Nov 11 2017
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CROSSREFS
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KEYWORD
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easy,nonn,nice
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AUTHOR
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Marie-Christine Haton (Marie-Christine.Haton(AT)loria.fr)
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EXTENSIONS
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STATUS
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approved
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