A225232 The number of FO3C2 moves required to restore a packet of n playing cards to its original state (order and orientation).

2, 4, 4, 12, 6, 24, 8, 40, 10, 60, 12, 84, 14, 112, 16, 144, 18, 180, 20, 220, 22, 264, 24, 312, 26, 364, 28, 420, 30, 480, 32, 544, 34, 612, 36, 684, 38, 760, 40, 840, 42, 924, 44, 1012, 46, 1104, 48, 1200, 50, 1300, 52, 1404, 54, 1512, 56, 1624, 58, 1740, 60, 1860, 62, 1984

Offset: 3

Author

Colm Mulcahy, May 03 2013

Keywords

• nonn
• easy

Comments

Each FO3C2 move Flips Over the top 3 cards as a unit and then Cuts 2 cards from the top to bottom. - Mulcahy

References

• Colm Mulcahy, Mathematical Card Magic: Fifty-Two New Effects, A K Peters, 2013, chapter 9.

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 3..1000
• Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).

Crossrefs

The even numbered terms are A046092.

Cf. A106232.

Programs

PARI

a(n)={
    if(n<6,return(if(n>3,4,2)));
    n--;
    my(deck=vector(n,i,i),original=deck,steps);
    while(1,
        steps+=2;
        deck=concat(deck[5..n],-[deck[2],deck[1],deck[4],deck[3]]);
        if(deck==original,return(steps))
    )
}; \\ Charles R Greathouse IV, May 03 2013

PARI

a(n)=if(n%2,n-1,n*(n-2)/2) \\ Charles R Greathouse IV, May 06 2013

PARI

Vec(2*x^3*(x^2-2*x-1)/((x-1)^3*(x+1)^3) + O(x^100)) \\ Colin Barker, Jun 04 2014

Formula

Assume n >=3. For odd n we have a(n) = n-1 and for even n we have a(n) = (n-2)n/2. Equivalently, a(2k+1) = 2k and a(2k) = 2k(k-1).

a(n) = 3*a(n-2)-3*a(n-4)+a(n-6). - Colin Barker, Jun 04 2014

G.f.: 2*x^3*(x^2-2*x-1) / ((x-1)^3*(x+1)^3). - Colin Barker, Jun 04 2014

Extensions

a(10), a(12)-a(64) from Charles R Greathouse IV, May 03 2013