A122158 Smallest positive number of "triangular" shuffles of n(n+1)/2 cards needed to restore them to their original order.

1, 2, 4, 20, 18, 12, 126, 33, 204, 1638, 1968, 2010, 504, 17043, 240, 222870, 4084080, 12462, 81396, 4200, 41790, 24254, 56560, 1377090, 30669496, 57420, 249480, 93840, 696010, 40680, 4627497204, 2122260, 4774320, 347190, 83800, 103458, 5017446420, 1686300

Offset

1,2

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 bottom of the column to the top, first from column 1, then column 2, etc. See A048782 for analogous results for a different triangular shuffle.

Links

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

Example

For n=3, successive shuffles give:
1.......4.......5.......3.......1
2.3.....2.1.....2.4.....2.5.....2.3
4.5.6...5.3.6...3.1.6...1.4.6...4.5.6
returning the deck of 6 cards to its original order in 4 shuffles. Thus a(3)=4.

Prog

(PARI)
Perm(n)={concat(Vecrev(vector(n, i, vector(i, j, n+1-i+(n-j+1)*(n-j)/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

Crossrefs

Cf. A121052, A048782.

Sequence in context: A132530 A316389 A133521 * A178223 A033180 A215815

Adjacent sequences: A122155 A122156 A122157 * A122159 A122160 A122161

Keyword

nonn

Author

John W. Layman, Aug 22 2006

Extensions

Terms a(17) and beyond from Andrew Howroyd, Nov 12 2017

Status

approved