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a(n) is the number of symmetric domino towers with n bricks.
3

%I #24 Aug 12 2022 09:23:04

%S 1,1,3,3,7,9,19,25,53,71,149,203,423,583,1209,1681,3473,4863,10017,

%T 14107,28987,41019,84113,119513,244645,348829,712987,1019731,2081547,

%U 2985097,6086375,8749185,17820657,25671983,52241825,75402907,153316715,221673707,450393329,652234089

%N a(n) is the number of symmetric domino towers with n bricks.

%C A domino tower is a stack of bricks, where (1) each row is offset from the preceding row by half of a brick, (2) the bottom row is contiguous, and (3) each brick is supported from below by at least half of a brick.

%C The number of (not necessarily symmetric) domino towers with n blocks is given by 3^(n-1).

%C a(n) is odd for all n.

%C The not necessarily symmetric case is described in the Miklos Bona reference. Similar considerations lead to a decomposition of symmetric towers into half pyramids which are enumerated by the Motzkin numbers. - _Andrew Howroyd_, Mar 12 2021

%D Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, pages 25-27.

%H Andrew Howroyd, <a href="/A320314/b320314.txt">Table of n, a(n) for n = 1..1000</a>

%H Peter Kagey, <a href="https://math.stackexchange.com/q/2949131/121988">Symmetric Brick Stacking</a>, Mathematics Stack Exchange.

%F G.f.: (x + 2*x^3*M(x^2) + x^2*M(x^2))/((1-x^3*M(x^2))*(1-x^2*M(x^2))) where M(x) is the g.f. of A001006. - _Andrew Howroyd_, Mar 12 2021

%e For n = 4, the a(4) = 3 symmetric stacks are

%e +-------+

%e | |

%e +---+---+---+---+

%e | | |

%e +---+---+---+---+,

%e | |

%e +-------+

%e +-------+ +-------+

%e | | | |

%e +---+---+---+---+---+---+, and

%e | | |

%e +-------+-------+

%e +-------+-------+-------+-------+

%e | | | | |

%e +-------+-------+-------+-------+.

%o (PARI) seq(n)={my(h=(1 - x^2 - sqrt(1-2*x^2-3*x^4 + O(x^3*x^n)))/(2*x^2)); Vec((x + 2*x*h + h)/((1-x*h)*(1-h)))} \\ _Andrew Howroyd_, Mar 12 2021

%Y Cf. A000244, A001006, A168368, A264746.

%K nonn

%O 1,3

%A _Peter Kagey_, Oct 10 2018

%E a(20)-a(40) from _Andrew Howroyd_, Oct 25 2018