%I #33 Jan 31 2018 12:34:29
%S 1,3,3,8,17,8,21,130,130,21,55,931,2604,931,55,144,6871,54732,54732,
%T 6871,144,377,50778
%N Square array T(n,k) n >= 1, k >= 1 read by antidiagonals: T(n, k) is the number of distinct Bojagi boards with dimensions n X k that have a unique solution.
%C Bojagi is a puzzle game created by David Radcliffe.
%C A Bojagi board is a rectangular board with some cells empty and some cells containing positive integers. A solution for a Bojagi board partitions the board into rectangles such that each rectangle contains exactly one integer, and that integer is the area of the rectangle.
%H Taotao Liu, Thomas Ledbetter <a href="/A296106/a296106.cs.txt">C# Program</a>
%H David Radcliffe, <a href="https://naturalmath.com/2014/10/bojagi-cute-multiplication-puzzles-by-and-for-families/">Rules of puzzle game Bojagi</a>
%F T(n,1) = A088305(n), the even-indexed Fibonacci numbers.
%F T(n,1) = Sum_{i=1..n} i*T(n-i,1) if we take T(0,1) = 1.
%e Array begins:
%e ======================================
%e n\k| 1 2 3 4 5 6
%e ---+----------------------------------
%e 1 | 1 3 8 21 55 144 ...
%e 2 | 3 17 130 931 6871 ...
%e 3 | 8 130 2604 54732 ...
%e 4 | 21 931 54732 ...
%e 5 | 55 6871 ...
%e 6 | 144 ...
%e ...
%e As a triangle:
%e 1;
%e 3, 3;
%e 8, 17, 8;
%e 21, 130, 130, 21;
%e 55, 931, 2604, 931, 55;
%e 144, 6871, 54732, 54732, 6871, 144;
%e ...
%e If n=1 or k=1, any valid board (a board whose numbers add up to the area of the board) has a unique solution.
%e For n=2 and k=2, there are 17 boards that have a unique solution. There is 1 board in which each of the four cells has a 1.
%e There are 4 boards which contain two 2's. The 2's must be adjacent (not diagonally opposite) in order for the board to have a unique solution.
%e There are 8 boards which contain one 2 and two 1's. The 1's must be adjacent in order for the board to have a solution. The 2 can be placed in either of the remaining two cells.
%e There are 4 boards which contain one 4. It can be placed anywhere.
%Y Cf. A088305.
%K hard,nonn,tabl,more
%O 1,2
%A _Taotao Liu_, Dec 04 2017