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A191781
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Triangle read by rows: T(n,k) is the number of length-n left factors of Dyck paths having length of the first ascent equal to k (1 <= k <= n).
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2
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1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 3, 3, 2, 1, 1, 6, 6, 4, 2, 1, 1, 10, 10, 7, 4, 2, 1, 1, 20, 20, 14, 8, 4, 2, 1, 1, 35, 35, 25, 15, 8, 4, 2, 1, 1, 70, 70, 50, 30, 16, 8, 4, 2, 1, 1, 126, 126, 91, 56, 31, 16, 8, 4, 2, 1, 1, 252, 252, 182, 112, 62, 32, 16, 8, 4, 2, 1, 1, 462, 462, 336, 210, 119, 63, 32, 16, 8, 4, 2, 1, 1
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OFFSET
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1,7
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COMMENTS
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Sum of entries in row n is binomial(n, floor(n/2)) = A001405(n).
Sum_{k=1..n} k*T(n,k) = A191782(n).
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LINKS
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FORMULA
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G.f.: G(t,z) = (1 - z*c + t*z^3*c^2)/((1 - z*c)*(1 - t*z)*(1 - t*z^2)), where c = (1-sqrt(1 - 4*z^2))/(2*z^2).
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EXAMPLE
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T(5,2)=3 because we have UUDDU, UUDUD, and UUDUU, where U=(1,1) and D=(1,-1).
Triangle starts:
1;
1, 1;
1, 1, 1;
2, 2, 1, 1;
3, 3, 2, 1, 1;
6, 6, 4, 2, 1, 1;
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MAPLE
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c := ((1-sqrt(1-4*z^2))*1/2)/z^2: G := (1-z*c+t*z^3*c^2)/((1-z*c)*(1-t*z)*(1-t*z^2*c)): Gser := simplify(series(G, z = 0, 17)): 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
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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