

A180182


Triangle read by rows: T(n,k) is the number of compositions of n without 7's and having k parts; 1<=k<=n.


7



1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 5, 10, 10, 5, 1, 0, 6, 15, 20, 15, 6, 1, 1, 5, 21, 35, 35, 21, 7, 1, 1, 6, 25, 56, 70, 56, 28, 8, 1, 1, 7, 30, 80, 126, 126, 84, 36, 9, 1, 1, 8, 36, 108, 205, 252, 210, 120, 45, 10, 1, 1, 9, 43, 141, 310, 456, 462, 330, 165, 55, 11, 1
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OFFSET

1,5


REFERENCES

P. Chinn and S. Heubach, Compositions of n with no occurrence of k, Congressus Numerantium, 164 (2003), pp. 3351 (see Table 9).
R.P. Grimaldi, Compositions without the summand 1, Congressus Numerantium, 152, 2001, 3343.


LINKS

Table of n, a(n) for n=1..78.
P. Chinn and S. Heubach, Integer Sequences Related to Compositions without 2's, J. Integer Seqs., Vol. 6, 2003.


FORMULA

Number of compositions of n without p's and having k parts = Sum((1)^{kj}*binom(k,j)*binom(npk+pj1,j1), j=(pkn)/(p1)..k).
For a given p, the g.f. of the number of compositions without p's is G(t,z)=tg(z)/[1tg(z)], where g(z)=z/(1z)z^p; here z marks sum of parts and t marks number of parts.


EXAMPLE

T(10,2)=7 because we have (1,9), (9,1), (2,8), (8,2), (6,4), (4,6), and (5,5).
Triangle starts:
1;
1,1;
1,2,1;
1,3,3,1;
1,4,6,4,1;
1,5,10,10,5,1;
0,6,15,20,15,6,1;


MAPLE

p := 7: T := proc (n, k) options operator, arrow: sum((1)^(kj)*binomial(k, j)*binomial(np*k+p*j1, j1), j = (p*kn)/(p1) .. k) end proc: for n to 13 do seq(T(n, k), k = 1 .. n) end do; # yields sequence in triangular form
p := 7: g := z/(1z)z^p: G := t*g/(1t*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
with(combinat): m := 7: 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


MATHEMATICA

p = 7; max = 14; g = z/(1z)  z^p; G = t*g/(1t*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 (* JeanFrançois Alcover, Jan 28 2014, after Maple *)


CROSSREFS

Cf. A011973, A180177, A180178, A180179, A180180, A180181, A180183.
Sequence in context: A223968 A214846 A061676 * A275198 A095145 A095144
Adjacent sequences: A180179 A180180 A180181 * A180183 A180184 A180185


KEYWORD

nonn,tabl


AUTHOR

Emeric Deutsch, Aug 15 2010


STATUS

approved



