A233308 Number A(n,k) of tilings of a k X k X n box using k*n bricks of shape k X 1 X 1; square array A(n,k), n>=0, k>=1, read by antidiagonals.
1, 1, 1, 1, 2, 1, 1, 2, 9, 1, 1, 2, 4, 32, 1, 1, 2, 4, 21, 121, 1, 1, 2, 4, 8, 92, 450, 1, 1, 2, 4, 8, 45, 320, 1681, 1, 1, 2, 4, 8, 16, 248, 1213, 6272, 1, 1, 2, 4, 8, 16, 93, 1032, 4822, 23409, 1, 1, 2, 4, 8, 16, 32, 668, 3524, 18556, 87362, 1, 1, 2, 4, 8, 16, 32, 189, 3440, 13173, 70929, 326041, 1
Offset: 0
Examples
Square array A(n,k) begins: 1, 1, 1, 1, 1, 1, ... 1, 2, 2, 2, 2, 2, ... 1, 9, 4, 4, 4, 4, ... 1, 32, 21, 8, 8, 8, ... 1, 121, 92, 45, 16, 16, ... 1, 450, 320, 248, 93, 32, ... 1, 1681, 1213, 1032, 668, 189, ... 1, 6272, 4822, 3524, 3440, 1832, ... 1, 23409, 18556, 13173, 13728, 11976, ...
Links
- Alois P. Heinz, Antidiagonals n = 0..17, flattened
Crossrefs
Programs
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Maple
b:= proc(n, l) option remember; local d, t, k; d:= isqrt(nops(l)); if max(l[])>n then 0 elif n=0 then 1 elif min(l[])>0 then t:=min(l[]); b(n-t, map(x->x-t, l)) else for k while l[k]>0 do od; b(n, subsop(k=d, l))+ `if`(irem(k,d)=1 and {seq(l[k+j], j=1..d-1)}={0}, b(n, [seq(`if`(h-k=0, 1, l[h]), h=1..nops(l))]), 0)+ `if`(k<=d and {seq(l[k+d*j], j=1..d-1)}={0}, b(n, [seq(`if`(irem(h-k, d)=0, 1, l[h]), h=1..nops(l))]), 0) fi end: A:= (n, k)-> `if`(k>n, 2^n, b(n, [0$k^2])): seq(seq(A(n, 1+d-n), n=0..d), d=0..11); -
Mathematica
b[n_, l_] := b[n, l] = Module[{d, t, k}, d= Sqrt[Length[l]]; Which[ Max[l]>n, 0, n==0, 1, Min[l]>0, t=Min[l]; b[n-t, l-t], True, k=Position[l, 0, 1][[1, 1]]; b[n, ReplacePart[l, k->d]]+ If[Mod[k, d]==1 && Union[ Table[ l[[k+j]], {j, 1, d-1}]] == {0}, b[n, Table[ If [h-k=0, 1, l[[h]] ], {h, 1, Length[l]}]], 0]+ If[k <= d && Union[ Table[ l[[k+d*j]], {j, 1, d-1}]] == {0}, b[n, Table[ If[ Mod[h-k, d] == 0, 1, l[[h]] ], {h, 1, Length[l]}]], 0] ] ]; a[n_, k_]:= If[k>n, 2^n, b[n, Array[0&, k^2]]]; Table[Table[a[n, 1+d-n], {n, 0, d}], {d, 0, 11}] // Flatten (* Jean-François Alcover, Dec 13 2013, translated from Maple *)
Comments