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%I #4 Mar 30 2012 17:36:28
%S 1,1,2,3,5,1,8,2,15,4,1,25,8,2,46,19,4,1,79,36,9,2,147,76,24,4,1,256,
%T 146,48,10,2,477,304,109,29,4,1,841,578,224,60,11,2,1570,1180,499,144,
%U 34,4,1,2791,2244,1002,312,72,12,2,5217,4525,2172,731,181,39,4,1,9336,8588,4347,1530,410,84,13,2
%N Triangle read by rows: T(n,k) is the number of length n left factors of Dyck paths having k UDUD's, where U=(1,1) and D=(1,-1).
%C Row n>=2 contains floor(n/2) entries.
%C Sum of entries in row n is binomial(n, floor(n/2)) =A001405(n).
%C T(n,0)=A191792(n).
%C Sum(k*T(n,k), k>=0)=A100071(n-3).
%F G.f.: G(t,z) = C/(1-z*C), where C=C(t,z) is given by z^2*(1+z^2-t*z^2)*C^2 - (1+z^2+z^4- t*z^2-t*z^4)*C + 1 + z^2 - t*z^2 = 0.
%e T(7,2)=2 because we have (UD[UD)UD]U and U(UD[UD)UD], where U=(1,1) and D=(1,-1) (the UDUD's are shown between parentheses).
%e Triangle starts:
%e 1;
%e 1;
%e 2;
%e 3;
%e 5,1;
%e 8,2;
%e 15,4,1;
%e 25,8,2;
%p eq := z^2*(1+z^2-t*z^2)*C^2-(1+z^2+z^4-t*z^2-t*z^4)*C+1+z^2-t*z^2: C := RootOf(eq, C): G := C/(1-z*C): Gser := simplify(series(G, z = 0, 20)): for n from 0 to 17 do P[n] := sort(coeff(Gser, z, n)) end do: 1; 1; for n from 2 to 17 do seq(coeff(P[n], t, k), k = 0 .. floor((1/2)*n)-1) end do; # yields sequence in triangular form
%Y Cf. A001405, A191792, A100071.
%K nonn,tabf
%O 0,3
%A _Emeric Deutsch_, Jun 18 2011
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