A275204 - OEIS

%I #34 Oct 22 2018 22:47:06

%S 1,1,1,1,3,1,1,4,5,1,1,6,8,7,1,1,7,15,12,9,1,1,9,22,25,16,11,1,1,10,

%T 31,43,35,20,13,1,1,12,41,68,65,45,24,15,1,1,13,54,99,113,87,55,28,17,

%U 1,1,15,66,143,178,159,109,65,32,19,1,1,16,82,193,273,267,205,131,75,36,21,1,1,18,98,258,398,430,357,251,153,85,40,23,1

%N Triangle read by rows: T(n,k) = number of domino stacks with n dominoes having a base of k dominoes.

%C The k-th column is the number of domino stacks having a base of k dominoes.

%C The n-th row is the number of domino stacks consisting of n dominoes.

%C A domino stack corresponds to a convex polyomino built with dominoes such that all columns intersect the base.

%H Alois P. Heinz, <a href="/A275204/b275204.txt">Rows n = 1..200, flattened</a>

%H T. M. Brown <a href="https://cs.uwaterloo.ca/journals/JIS/VOL20/Brown/brown2.html">Convex Domino Towers</a>, J. Integer Seq. 20 (2017).

%F T(n,k) = Sum_{i=1..k} (2*(k-i)+1)*T(n-k,i) where T(n,n)=1 and T(n,k)=0 if n or k is nonpositive or if n is less than k.

%F G.f.: x^k/(1-x^k) Sum_{S} (Product_{i=1..j} (2*(k_{i+1}-k_i)+1)*x^(k_i)/ (1-x^(k_i))) where the sum is over all subsets S of {1,..,k-1} such that S={k_1<k_2<..<k_{j-1}} and k_j=k.

%e Triangle begins:

%e   1;

%e   1,  1;

%e   1,  3,  1;

%e   1,  4,  5,  1;

%e   1,  6,  8,  7,  1;

%e   1,  7, 15, 12,  9,  1;

%e   ...

%e Dominoes are assumed to be horizontal, and each row must be a subset of the row below it. For n=4 and k=2, the bottom row has 2 dominoes. One possibility is to put both remaining dominoes in the next row up. Otherwise there will be one domino in the next row up, and it can be in three possible positions: right, center, or left. The last domino must be placed on top of it. So there are a total of four possible stacks, and therefore T(4,2) = 4. - _Michael B. Porter_, Jul 20 2016

%p T:= proc(n, k) option remember; `if`(k<1 or k>n, 0,

%p       `if`(n=k, 1, add((2*(k-i)+1)*T(n-k, i), i=1..k)))

%p     end:

%p seq(seq(T(n,k), k=1..n), n=1..15);  # _Alois P. Heinz_, Jul 19 2016

%t T[n_, n_] = 1; T[n_, k_] /; n<1 || k<1 || n<k = 0;

%t T[n_, k_] := T[n, k] = Sum[(2 (k-i) + 1) T[n-k, i], {i, 1, k}];

%t Table[T[n, k], {n, 1, 13}, {k, 1, n}] // Flatten (* _Jean-François Alcover_, Aug 17 2018 *)

%K nonn,tabl

%O 1,5

%A _Tricia Muldoon Brown_, Jul 19 2016

%E Data corrected by _Jean-François Alcover_, Aug 17 2018