A051732 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.
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
Offset: 1
Examples
From _Andrew Howroyd_, Nov 11 2017: (Start) 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)
Links
- Andrew Howroyd, Table of n, a(n) for n = 1..2000
- Gary Antonick, The Polka Dot Puzzle, NY Times Wordplay blog.
Programs
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PARI
P(n,i)={my(d=2*n+1-2*i); while(d
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
Formula
a(2^n+1) = n+1. - Ripatti A. (ripatti(AT)inbox.ru), Feb 04 2010
Extensions
Name clarified by Andrew Howroyd, Nov 11 2017
Comments