A048782 Smallest positive number of "triangular" shuffles of n(n+1)/2 cards needed to restore them to their original order.

1, 1, 2, 6, 10, 84, 70, 24, 36, 330, 210, 288, 1428, 252, 10080, 13680, 1260, 19320, 2160, 22440, 1692152, 50400, 140760, 3071880, 284088, 608753236, 617760, 15600, 35287560, 214138080, 30240, 1866600, 58333800, 8637552, 4034916600

Offset: 1

Author

John W. Layman, Jul 14 1999, Aug 22 2006

Keywords

• nonn

Comments

Lay out cards in a triangular array from left to right in rows with 1 card in row 1 (the top row), 2 cards in row 2, etc., then pick up by columns, in order from the top of the column to the bottom, first from column 1, then column 2, etc. See A122158 for analogous results for a different triangular shuffle.

Examples

			For n=3, successive shuffles give:
1.......1.......1
2.3.....2.4.....2.3
4.5.6...3.5.6...4.5.6,
returning the deck of 6 cards to its original order in 2 shuffles. Thus a(3)=2.
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 1..500

Crossrefs

Cf. A121052, A122158.

Programs

PARI

Perm(n)={concat(vector(n, i, vectorsmall(i, j, i+n*(j-1)-j*(j-1)/2)))}
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(v=Perm(n), q=0); for(i=1, #v, my(l=Follow(i, j->v[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 12 2017

Extensions

Edited by R. J. Mathar, Aug 02 2008