A221564 The number of FO4C3 moves required to restore a packet of n playing cards to its original state (order and orientation), where each move Flips Over the top four (4) as a unit and then Cuts three (3) cards from the top to bottom.

2, 4, 4, 4, 12, 12, 6, 24, 24, 8, 40, 40, 10, 60, 60, 12, 84, 84, 14, 112, 112, 16, 144, 144, 18, 180, 180, 20, 220, 220, 22, 264, 264, 24, 312, 312, 26, 364, 364, 28, 420, 420, 30, 480, 480, 32, 544, 544, 34, 612, 612, 36, 684, 684, 38, 760, 760, 40, 840, 840

Offset: 4

Author

Colm Mulcahy, May 04 2013

Keywords

• nonn
• easy

Comments

Conjecture: a(3k+1) = 2k.

The top card remains on top but is flipped over with each move. The remaining cards split into three cycles either of length 2*floor((n-1)/3) or 2*ceiling((n-1)/3). - Andrew Howroyd, Apr 27 2020

Links

• Andrew Howroyd, Table of n, a(n) for n = 4..1000
• Index entries for linear recurrences with constant coefficients, signature (0,0,3,0,0,-3,0,0,1).

Crossrefs

Cf. A225232.

Programs

PARI

a(n)={2*((n-1)\3)*if(n%3==1, 1, (n-1)\3+1)} \\ Andrew Howroyd, Apr 27 2020

PARI

Vec(2*x^4*(1 + 2*x + 2*x^2 - x^3) / ((1 - x)^3*(1 + x + x^2)^3) + O(x^40)) \\ Colin Barker, Apr 29 2020

Formula

a(3*n+1) = 2*n; a(3*n) = a(3*n-1) = 2*n*(n-1). - Andrew Howroyd, Apr 27 2020

From Colin Barker, Apr 29 2020: (Start)

G.f.: 2*x^4*(1 + 2*x + 2*x^2 - x^3) / ((1 - x)^3*(1 + x + x^2)^3).

a(n) = 3*a(n-3) - 3*a(n-6) + a(n-9) for n>12.

(End)

Extensions

a(16) corrected and terms a(17) and beyond from Andrew Howroyd, Apr 27 2020