%I #33 Dec 12 2018 05:54:19
%S 1,2,4,6,24,56,120,720,1712,5040,9408,40320,92800,362880,3628800,
%T 4948992,7918592,39916800,479001600,984237056,6085088256,6227020800,
%U 63352393728,87178291200,168662855680,1307674368000,14782316470272,20922789888000,38238313152512
%N Chocolate numbers.
%C Given an m X n chocolate bar, let A(m,n) be the number of ways to break it into m*n unit pieces where each break occurs along a gridline. Order matters, and the pieces are distinguishable. Then this sequence lists the values A(m,n) in increasing order as m and n range over the positive integers.
%C The sequence of factorials, A000142, is a subsequence as A(1,n) = A(n,1) = (n-1)!.
%C For m,n>1, A(m,n) is divisible by 2^(m+n-2).
%H Caleb Ji, Tanya Khovanova, Robin Park, Angela Song, <a href="http://arxiv.org/abs/1509.06093">Chocolate Numbers</a>, arXiv:1509.06093 [math.CO], 2015.
%H Caleb Ji, Tanya Khovanova, Robin Park, Angela Song, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL19/Khovanova/khova9.html">Chocolate Numbers</a>, Journal of Integer Sequences, Vol. 19 (2016), #16.1.7.
%F A(m,n)=1 for max(m,n)<2 and A(m,n) = Sum_{i=1..m-1} C(m*n-2,i*n-1) *A(i,n) *A(m-i,n) + Sum_{i=1..n-1} C(m*n-2,i*m-1) *A(m,i) *A(m,n-i) else.
%e For n = m = 2, there are two ways for the first break: breaking it horizontally or vertically. After that we need two more breaks that can be done in any order. Thus A(2,2) = 4, and 4 belongs to the sequence.
%t terms = 29;
%t A[m_, n_] := A[m, n] = If[Max[m, n] < 2, 1, Sum[A[i, n] Binomial[m n - 2, i n - 1] A[m - i, n], {i, 1, m - 1}]] + Sum[A[m, i] Binomial[m n - 2, i m - 1] A[m, n - i], {i, 1, n - 1}];
%t Table[A[m, n], {m, 1, terms}, {n, 1, terms}] // Flatten // Union //
%t Take[#, terms]&
%Y Cf. A257281, A261747, A000142, A261964.
%K nonn
%O 1,2
%A _Caleb Ji_, _Tanya Khovanova_, _Robin Park_, _Angela Song_, Aug 30 2015
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