A180178 - OEIS

%I #15 Nov 15 2019 21:30:10

%S 1,1,1,0,2,1,1,1,3,1,1,2,3,4,1,1,4,4,6,5,1,1,4,9,8,10,6,1,1,5,12,17,

%T 15,15,7,1,1,6,15,28,30,26,21,8,1,1,7,21,38,56,51,42,28,9,1,1,8,27,56,

%U 85,102,84,64,36,10,1,1,9,34,80,130,172,175,134,93,45,11,1,1,10,42,108,200

%N Triangle read by rows: T(n,k) is the number of compositions of n without 3's and having k parts; 1 <= k <= n.

%D P. Chinn and S. Heubach, Compositions of n with no occurrence of k, Congressus Numerantium, 164 (2003), pp. 33-51 (see Table 5).

%D R.P. Grimaldi, Compositions without the summand 1, Congressus Numerantium, 152, 2001, 33-43.

%H P. Chinn and S. Heubach, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL6/Heubach/heubach5.html">Integer Sequences Related to Compositions without 2's</a>, J. Integer Seqs., Vol. 6, 2003.

%F Number of compositions of n without p's and having k parts = Sum_{j=(pk-n)/(p-1)..k} (-1)^(k-j)*binomial(k,j)*binomial(n-pk+pj-1, j-1).

%F For a given p, the g.f. of the number of compositions without p's is G(t,z) = t*g(z)/(1-t*g(z)), where g(z) = z/(1-z) - z^p; here z marks sum of parts and t marks number of parts.

%e T(7,4)=8 because we have (1,2,2,2), (2,1,2,2), (2,2,1,2), (2,2,2,1), (1,1,1,4), (1,1,4,1), (1,4,1,1), and (4,1,1,1).

%e Triangle starts:

%e   1;

%e   1, 1;

%e   0, 2, 1;

%e   1, 1, 3, 1;

%e   1, 2, 3, 4, 1;

%e   1, 4, 4, 6, 5, 1;

%p p := 3: T := proc (n, k) options operator, arrow: sum((-1)^(k-j)*binomial(k, j)*binomial(n-p*k+p*j-1, j-1), j = (p*k-n)/(p-1) .. k) end proc: for n to 13 do seq(T(n, k), k = 1 .. n) end do; # yields sequence in triangular form

%p p := 3: g := z/(1-z)-z^p: G := t*g/(1-t*g): Gser := simplify(series(G, z = 0, 15)): for n to 13 do P[n] := sort(coeff(Gser, z, n)) end do: for n to 13 do seq(coeff(P[n], t, k), k = 1 .. n) end do; # yields sequence in triangular form

%p with(combinat): m := 3: T := proc (n, k) local ct, i: ct := 0: for i to numbcomp(n, k) do if member(m, composition(n, k)[i]) = false then ct := ct+1 else end if end do: ct end proc: for n to 13 do seq(T(n, k), k = 1 .. n) end do; # yields sequence in triangular form

%t p = 3; max = 14; g = z/(1-z) - z^p; G = t*g/(1-t*g); Gser = Series[G, {z, 0, max+1}]; t[n_, k_] := SeriesCoefficient[Gser, {z, 0, n}, {t, 0, k}]; Table[t[n, k], {n, 1, max}, {k, 1, n}] // Flatten (* _Jean-François Alcover_, Jan 28 2014, after Maple *)

%Y Cf. A011973, A180177, A180179, A180180, A180181, A180182, A180183.

%K nonn,tabl

%O 1,5

%A _Emeric Deutsch_, Aug 15 2010