A296559 - OEIS

%I #19 Dec 16 2017 05:31:14

%S 1,0,1,0,1,1,1,0,2,1,1,2,1,3,1,1,4,3,3,4,1,2,4,9,5,6,5,1,3,7,12,16,9,

%T 10,6,1,4,13,18,28,26,16,15,7,1,6,19,36,42,55,41,27,21,8,1,9,29,60,82,

%U 90,97,64,43,28,9,1,13,47,94,152,170,177,160,99,65,36,10,1,19,73,158,252,335,333,323,253,151,94,45,11,1

%N Triangle read by rows: T(n,k) is the number of compositions of n having k parts equal to 1 or 2 (0<=k<=n).

%C Sum of entries in row n = 2^{n-1} = A011782(n) (n>=1).

%C Sum(kT(n,k), k>=0) = (3n+5)*2^{n-4} = A106472(n-1) (n>=3).

%F G.f.: G(t,x) = (1-x)/(1 - (1 + t)x - (1 - t)x^3).

%e T(3,2) = 2 because we have [1,2],[2,1].

%e T(6,3) = 5 because we have [2,2,2],[1,1,1,3],[1,1,3,1],[1,3,1,1],[3,1,1,1].

%e Triangle begins:

%e   1,

%e   0, 1,

%e   0, 1, 1,

%e   1, 0, 2, 1,

%e   1, 2, 1, 3, 1,

%e   1, 4, 3, 3, 4, 1,

%e   2, 4, 9, 5, 6, 5, 1,

%e   3, 7, 12, 16, 9, 10, 6, 1,

%e   4, 13, 18, 28, 26, 16, 15, 7, 1,

%e   ...

%p g := (1-x)/(1-(1+t)*x-(1-t)*x^3): gser := simplify(series(g, x = 0, 17)): for n from 0 to 15 do p[n] := sort(expand(coeff(gser, x, n))) end do: for n from 0 to 15 do seq(coeff(p[n], t, j), j = 0 .. n) end do; # yields sequence in triangular form

%t nmax = 12;

%t s = Series[(1-x)/(1 - (1+t) x - (1-t) x^3), {x, 0, nmax}, {t, 0, nmax}];

%t T[n_, k_] := SeriesCoefficient[s, {x, 0, n}, {t, 0, k}];

%t Table[T[n, k], {n, 0, nmax}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Dec 16 2017 *)

%Y Cf. A011782, A105114, A105422, A106472.

%K nonn,tabl

%O 0,9

%A _Emeric Deutsch_, Dec 15 2017