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A296559
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).
1
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, 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, 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
OFFSET
0,9
COMMENTS
Sum of entries in row n = 2^{n-1} = A011782(n) (n>=1).
Sum(kT(n,k), k>=0) = (3n+5)*2^{n-4} = A106472(n-1) (n>=3).
FORMULA
G.f.: G(t,x) = (1-x)/(1 - (1 + t)x - (1 - t)x^3).
EXAMPLE
T(3,2) = 2 because we have [1,2],[2,1].
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].
Triangle begins:
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, 10, 6, 1,
4, 13, 18, 28, 26, 16, 15, 7, 1,
...
MAPLE
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
MATHEMATICA
nmax = 12;
s = Series[(1-x)/(1 - (1+t) x - (1-t) x^3), {x, 0, nmax}, {t, 0, nmax}];
T[n_, k_] := SeriesCoefficient[s, {x, 0, n}, {t, 0, k}];
Table[T[n, k], {n, 0, nmax}, {k, 0, n}] // Flatten (* Jean-François Alcover, Dec 16 2017 *)
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Emeric Deutsch, Dec 15 2017
STATUS
approved