A275204 Triangle read by rows: T(n,k) = number of domino stacks with n dominoes having a base of k dominoes.
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, 31, 43, 35, 20, 13, 1, 1, 12, 41, 68, 65, 45, 24, 15, 1, 1, 13, 54, 99, 113, 87, 55, 28, 17, 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
Offset: 1
Examples
Triangle begins: 1; 1, 1; 1, 3, 1; 1, 4, 5, 1; 1, 6, 8, 7, 1; 1, 7, 15, 12, 9, 1; ... 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
Links
- Alois P. Heinz, Rows n = 1..200, flattened
- T. M. Brown Convex Domino Towers, J. Integer Seq. 20 (2017).
Programs
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Maple
T:= proc(n, k) option remember; `if`(k<1 or k>n, 0, `if`(n=k, 1, add((2*(k-i)+1)*T(n-k, i), i=1..k))) end: seq(seq(T(n,k), k=1..n), n=1..15); # Alois P. Heinz, Jul 19 2016 -
Mathematica
T[n_, n_] = 1; T[n_, k_] /; n<1 || k<1 || n
Jean-François Alcover, Aug 17 2018 *)
Formula
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.
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
Extensions
Data corrected by Jean-François Alcover, Aug 17 2018
Comments