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%I #24 Sep 14 2021 05:54:56
%S 1,1,1,1,1,1,1,2,2,1,1,2,4,2,1,1,3,6,6,3,1,1,2,9,4,9,2,1,1,4,11,20,20,
%T 11,4,1,1,2,21,7,49,7,21,2,1,1,4,24,54,115,115,54,24,4,1,1,3,43,12,
%U 343,4,343,12,43,3,1,1,4,62,190,850,1225,1225,850,190,62,4,1
%N Number A(n,k) of tilings of a k X n rectangle using integer-sided rectangular tiles of equal area; square array A(n,k), n>=0, k>=0, read by antidiagonals.
%H Alois P. Heinz, <a href="/A220777/b220777.txt">Antidiagonals n = 0..26, flattened</a>
%e A(3,5) = 7, because there are 7 tilings of a 5 X 3 rectangle using integer-sided rectangular tiles of equal area:
%e ._____. ._____. ._____. ._____. ._____. ._____. ._____.
%e | | | | | | |_____| |_____| |_____| | | | | |_|_|_|
%e | | | | | | |_____| |_____| | | | | | | | | |_|_|_|
%e | | | | | | |_____| | | | | | | | | |_|_|_| |_|_|_|
%e | | | | | | |_____| | | | | |_|_|_| |_____| |_|_|_|
%e |_____| |_|_|_| |_____| |_|_|_| |_____| |_____| |_|_|_|
%e Square array A(n,k) begins:
%e 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
%e 1, 1, 2, 2, 3, 2, 4, 2, 4, ...
%e 1, 2, 4, 6, 9, 11, 21, 24, 43, ...
%e 1, 2, 6, 4, 20, 7, 54, 12, 190, ...
%e 1, 3, 9, 20, 49, 115, 343, 850, 2401, ...
%e 1, 2, 11, 7, 115, 4, 1225, 7, 15242, ...
%e 1, 4, 21, 54, 343, 1225, 7104, 31777, 169952, ...
%e 1, 2, 24, 12, 850, 7, 31777, 4, 1300180, ...
%e 1, 4, 43, 190, 2401, 15242, 169952, 1300180, 13036591, ...
%e ...
%p b:= proc(n, l, d) option remember; local i, k, m, q, s, t;
%p if max(l[])>n then 0 elif n=0 or l=[] then 1
%p elif min(l[])>0 then t:=min(l[]); b(n-t, map(h->h-t, l), d)
%p else for k do if l[k]=0 then break fi od; s, m:=0, nops(l);
%p for i from k to m while l[i]=0 do if irem(d, 1+i-k, 'q')=0
%p and q<=n then s:= s+ b(n, [l[j]$j=1..k-1, q$j=k..i,
%p l[j]$j=i+1..m], d) fi od; s
%p fi
%p end:
%p A:= (n, k)-> `if`(n<k, A(k, n), `if`(k=0, 1,
%p add(b(n, [0$k], d), d=numtheory[divisors](n*k)))):
%p seq(seq(A(n, d-n), n=0..d), d=0..14);
%t $RecursionLimit = 1000; b[n_, l_, d_] := b[n, l, d] = Module[{i, k, m, q, s, t}, Which[ Max[l] > n, 0, n == 0 || l == {}, 1, Min[l] > 0, t = Min[l]; b[n-t, l-t, d], True, k = Position[l, 0, 1][[1, 1]]; {s, m} = {0, Length[l]}; For[i = k, i <= m && l[[i]] == 0, i++, If[(Mod[d, 1+i-k]) == 0 && (q = Quotient[d, 1+i-k]) <= n, s = s + b[n, Join[l[[1 ;; k-1]], Table[q, {j, k, i}], l[[i+1 ;; m]]], d] ] ]; s ] ]; a[n_, k_] := a[n, k] = If[n < k, a[k, n], If[k == 0, 1, Sum[b[n, Array[0&, k], d], {d, Divisors[n*k]}]]]; Table[Table[a[n, d-n], {n, 0, d}], {d, 0, 14}] // Flatten (* _Jean-François Alcover_, Dec 17 2013, translated from Maple *)
%Y Columns (or rows) k=0-10 give: A000012, A000005, A220768, A220769, A220770, A220771, A220772, A220773, A220774, A220775, A220776.
%Y Main diagonal gives: A220778.
%K nonn,tabl
%O 0,8
%A _Alois P. Heinz_, Dec 19 2012
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