A180180 - OEIS

%I #16 Nov 15 2019 21:30:28

%S 1,1,1,1,2,1,1,3,3,1,0,4,6,4,1,1,3,10,10,5,1,1,4,12,20,15,6,1,1,5,15,

%T 31,35,21,7,1,1,6,19,44,65,56,28,8,1,1,8,24,60,106,120,84,36,9,1,1,8,

%U 33,80,160,222,203,120,45,10,1,1,9,40,111,230,372,420,322,165,55,11,1

%N Triangle read by rows: T(n,k) is the number of compositions of n without 5'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 7).

%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(8,2)=5 because we have (1,7),(7,1),(2,6),(6,2),and (4,4).

%e Triangle starts:

%e  1;

%e  1,  1;

%e  1,  2,  1;

%e  1,  3,  3,  1;

%e  0,  4,  6,  4,  1;

%e  1,  3, 10, 10,  5,  1;

%p p:= 5: 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:=5: g:=z/(1-z)-z^(p): G:=t*g/(1-t*g): Gser:=simplify(series(G,z=0,15)): for n from 1 to 13 do P[n]:=sort(coeff(Gser,z,n)); # yields sequence in triangular form

%p with(combinat): m := 5: 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 12 do seq(T(n, k), k = 1 .. n) end do; # yields sequence in triangular form

%t p = 5; 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, A180178, A180179, A180181, A180182, A180183.

%K nonn,tabl

%O 1,5

%A _Emeric Deutsch_, Aug 15 2010