|
%I #12 Dec 13 2015 01:35:56
%S 1,0,1,0,2,1,0,3,3,1,0,4,7,4,1,0,5,13,11,5,1,0,6,22,25,16,6,1,0,7,34,
%T 50,41,22,7,1,0,8,50,91,92,63,29,8,1,0,9,70,155,187,155,92,37,9,1,0,
%U 10,95,250,353,343,247,129,46,10,1,0,11,125,386,628,701,590,376,175,56,11,1
%N Triangular array with g.f. Product_{n >= 1} 1/(1 - x*z^n/(1 - z)).
%C A refinement of A227682.
%C A colored composition of n is defined as a composition of n where each part p comes in one of p colors (denoted by an integer from 1 to p) and the color numbers are nondecreasing through the composition. The color numbers thus form a partition, called the color partition, of some integer.
%C For example, the composition 1 + 3 + 2 of 6 gives rise to three colored compositions of 6, namely, 1(c1) + 3(c1) + 2(c1), 1(c1) + 3(c1) + 2(c2) and 1(c1) + 3(c2) + 2(c2), where the color number of a part is shown after the part prefaced by the letter c.
%C T(n,k) equals the number of colored compositions of n into k parts.
%C See A253830 for the enumeration of colored compositions having parts with distinct colors.
%H P. Bala, <a href="/A253829/a253829.pdf">Colored Compositions</a>
%F G.f.: G(x,z) := Product_{n >= 1} (1 - z)/(1 - z - x*z^n) = exp( Sum_{n >= 1} (x*z)^n/(n*(1 - z)^n*(1 - z^n)) ) =
%F 1 + Sum_{n >= 1} (x*z/(1 - z))^n/( Product_{i = 1..n} 1 - z^i ) = 1 + x*z + (2*x + x^2)*z^2 + (3*x + 3*x^2 + x^3)*z^3 + ....
%F Note, G(x*(1 - z),z) is the generating function of A008284.
%F T(n,k) = Sum_{i = k..n} binomial(i-1,k-1)*A008284(n+k-i,k).
%F Recurrence equation: T(n,k) = T(n-1,k) + T(n-1,k-1) + T(n-k,k) - T(n-k-1,k) with boundary conditions T(n,n) = 1, T(n,0) = 0 for n >= 1 and T(n,k) = 0 for n < k.
%F Row sums are A227682.
%e Triangle begins
%e n\k| 0 1 2 3 4 5 6 7
%e = = = = = = = = = = = = = = = = =
%e 0 | 1
%e 1 | 0 1
%e 2 | 0 2 1
%e 3 | 0 3 3 1
%e 4 | 0 4 7 4 1
%e 5 | 0 5 13 11 5 1
%e 6 | 0 6 22 25 16 6 1
%e 7 | 0 7 34 50 41 22 7 1
%e ...
%e T(4,2) = 7: The compositions of 4 into two parts are 2 + 2, 1 + 3 and 3 + 1. Coloring the parts as described above produces seven colored compositions of 4 into two parts:
%e 2(c1) + 2(c1), 2(c1) + 2(c2), 2(c2) + 2(c2),
%e 1(c1) + 3(c1), 1(c1) + 3(c2), 1(c1) + 3(c3),
%e 3(c1) + 1(c1).
%p G := 1/(product(1-x*z^j/(1-z), j = 1 .. 12)): Gser := simplify(series(G, z = 0, 14)): for n to 12 do P[n] := coeff(Gser, z^n) end do: for n to 12 do seq(coeff(P[n], x^j), j = 1 .. n) end do;
%Y Cf. A008284, A227682 (row sums), A253830.
%K nonn,easy,tabl
%O 0,5
%A _Peter Bala_, Jan 19 2015
|
