%I
%S 0,1,0,1,0,0,1,1,0,0,1,2,0,0,0,1,2,1,0,0,0,1,3,2,0,0,0,0,1,4,3,1,0,0,
%T 0,0,1,5,3,2,0,0,0,0,0,1,7,4,3,1,0,0,0,0,0,1,9,6,4,2,0,0,0,0,0,0,1,12,
%U 8,4,3,1,0,0,0,0,0,0
%N Table read by antidiagonals: T(l,L) is the number of all possible covers of Llength line segment by llength line segments with allowed gaps < l.
%C Table starts:
%C 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
%C 0, 0, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 21, 28, 37, ...
%C 0, 0, 0, 1, 2, 3, 3, 4, 6, 8, 10, 13, 18, 24, 31, ...
%C 0, 0, 0, 0, 1, 2, 3, 4, 4, 5, 7, 10, 13, 16, 20, ...
%C 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 5, 6, 8, 11, 15, ...
%C 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 6, 7, 9, ...
%C 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 7, 7, ...
%C 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 7, ...
%C 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, ...
%C 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, ...
%C 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, ...
%C 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, ...
%C 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, ...
%C 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, ...
%C .....................................................
%C Second row is A228361 which is also correspond to Padovan's spiral numbers A134816 for n>1.
%C Third row is A228362.
%C T(l,L) is also the number of compositions of L where parts do not exceeds l and where are no two adjacent parts less than l.
%C T(2,5) = 3: [2,2,1], [2,1,2], [1,2,2]
%C T(2,9) = 9: [2,2,2,2,1], [2,2,2,1,2], [2,2,1,2,2], [2,1,2,2,2], [1,2,2,2,2], [2,1,2,1,2,1], [1,2,2,1,2,1], [1,2,1,2,2,1], [1,2,1,2,1,2]
%C T(3,8) = 6: [3,3,2], [3,1,3,1], [3,2,3], [1,3,3,1], [1,3,1,3], [2,3,3]
%F For all l>=1:
%F G.f.: (1  Sum[x^i, {i, l, 2 l  1}])^1*Sum[x^i, {i, 0, l  1}]^2*x^l.
%F G.f. for l=1: x/(1x).
%F G.f. for l=2: x^2*(1+x)^2/(1x^2x^3).
%F G.f. for l=3: x^3*(1 + x + x^2)^2/(1  x^3  x^4  x^5).
%F For l>1, L>=0:
%F c[k, l, m] = Sum[(1)^i binomial[k  1  i*l, m  1] binomial[m, i], {i, 0, floor[(k  m)/l]}] // number of compositions of k into exactly m parts which do not exceed l.
%F a[L, l, m] = Sum[ binomial[m + 1, m + 1  j]*c[L  l*m, l  1, j], {j, 0, m + 1}] //the number of all possible covers of Llength line segment by m llength line segments.
%F T[l, L] := Sum[a[L, l, j], {j, 1, ceiling[L/l]}].
%t Gf[l_, z] := (1  Sum[z^i, {i, l, 2 l  1}])^1*Sum[z^i, {i, 0, l  1}]^2*z^l
%t T[l_, L_] := CoefficientList[Series[Gf[l, z], {z, 0, 100}], z][[L + 1]]
%t Table[T[n  b + 1, b  1], {n, 1, 30}, {b, n, 1, 1}] // Flatten
%K nonn,tabl
%O 1,12
%A _Philipp O. Tsvetkov_, Aug 21 2013
