A191788 - OEIS

%I #5 Mar 30 2012 17:36:28

%S 1,1,1,1,2,1,3,2,1,6,3,1,11,5,3,1,21,9,4,1,40,17,8,4,1,76,31,13,5,1,

%T 146,62,26,12,5,1,279,113,45,18,6,1,539,228,94,39,17,6,1,1036,419,165,

%U 64,24,7,1,2011,845,348,140,57,23,7,1,3883,1568,618,237,89,31,8,1,7566,3160,1298,521,205,81,30,8,1

%N Triangle read by rows: T(n,k) is the number of length n left factors of Dyck paths having k base pyramids. A base pyramid is a factor of the form U^j D^j (j>0), starting at the horizontal axis.Here U=(1,1) and D=(1,-1).

%C Row n contains 1+floor(n/2) entries.

%C Sum of entries in row n is binomial(n,floor(n/2))=A001405(n).

%C T(n,0) = A191789(n).

%C Sum(k*T(n,k), k>=0) = A191790(n).

%F G.f.: G(t,z)=(1-z^2)/((1-z*c)*(1-z^2*c+z^4*c-t*z^2)), where c=(1-sqrt(1-4*z^2))/(2*z^2).

%e T(6,2)=3 because we have (UD)(UD)UU, (UD)(UUDD), and (UUDD)(UD) (the base pyramids are shown between parentheses).

%e Triangle starts:

%e 1;

%e 1;

%e 1,1;

%e 2,1;

%e 3,2,1;

%e 6,3,1;

%e 11,5,3,1;

%e 21,9,4,1;

%p c := ((1-sqrt(1-4*z^2))*1/2)/z^2: G := (1-z^2)/((1-z*c)*(1-z^2*c+z^4*c-t*z^2)): Gser := simplify(series(G, z = 0, 20)): for n from 0 to 16 do P[n] := sort(coeff(Gser, z, n)) end do: for n from 0 to 16 do seq(coeff(P[n], t, k), k = 0 .. floor((1/2)*n)) end do; # yields sequence in triangular form

%Y Cf. A001405, A191789, A191790

%K nonn,tabf

%O 0,5

%A _Emeric Deutsch_, Jun 18 2011