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%I #10 Nov 13 2017 03:03:37
%S 1,1,2,6,10,84,70,24,36,330,210,288,1428,252,10080,13680,1260,19320,
%T 2160,22440,1692152,50400,140760,3071880,284088,608753236,617760,
%U 15600,35287560,214138080,30240,1866600,58333800,8637552,4034916600
%N Smallest positive number of "triangular" shuffles of n(n+1)/2 cards needed to restore them to their original order.
%C 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.
%H Andrew Howroyd, <a href="/A048782/b048782.txt">Table of n, a(n) for n = 1..500</a>
%e For n=3, successive shuffles give:
%e 1.......1.......1
%e 2.3.....2.4.....2.3
%e 4.5.6...3.5.6...4.5.6,
%e returning the deck of 6 cards to its original order in 2 shuffles. Thus a(3)=2.
%o (PARI)
%o Perm(n)={concat(vector(n, i, vectorsmall(i, j, i+n*(j-1)-j*(j-1)/2)))}
%o Follow(s, f)={my(t=f(s), k=1); while(t>s, k++; t=f(t)); if(s==t, k, 0)}
%o 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}
%o 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
%Y Cf. A121052, A122158.
%K nonn
%O 1,3
%A _John W. Layman_, Jul 14 1999, Aug 22 2006
%E Edited by _R. J. Mathar_, Aug 02 2008
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