%I #15 Aug 24 2014 10:16:09
%S 1,1,1,1,2,2,4,3,1,7,7,3,13,17,6,1,26,36,16,4,52,77,45,10,1,104,173,
%T 111,30,5,212,387,268,95,15,1,438,857,666,266,50,6,910,1911,1641,714,
%U 175,21,1,1903,4287,3975,1940,546,77,7,4009,9619,9606,5205,1610,294,28,1
%N Number of Dyck paths with no UUU's and no DDD's of semilength n and having k UUDD's (0<=k<=floor(n/2); U=(1,1), D=(1,-1)).
%C T(n,k) is also the number of weighted lattice paths B(n) having k (1,0)-steps of weight 2. B(n) is the set of lattice paths of weight n that start in (0,0), end on the horizontal axis and never go below this axis, whose steps are of the following four kinds: a (1,0)-step of weight 1; a (1,0)-step of weight 2; a (1,1)-step of weight 2; a (1,-1)-step of weight 1. The weight of a path is the sum of the weights of its steps. Example: row 3 is 2,2; indeed, denoting by h (H) the (1,0)-step of weight 1 (2), and u=(1,1), d=(1,-1), the four paths of weight 3 are ud, hH, Hh, and hhh, having 0, 1, 1, and 0 (1,0)-steps of weight 2, respectively. - _Emeric Deutsch_, Aug 23 2014
%C Row n contains 1+floor(n/2) entries.
%C Sum of entries in row n is A004148(n+1) (the secondary structure numbers).
%C T(n,0) = A023431(n).
%C Sum(k*T(n,k), k=0..floor(n/2)) = A110320(n-1).
%H M. Bona and A. Knopfmacher, <a href="http://dx.doi.org/10.1007/s00026-010-0060-7">On the probability that certain compositions have the same number of parts</a>, Ann. Comb., 14 (2010), 291-306. - _Emeric Deutsch_, Aug 23 2014
%F G.f. G=G(t,z) satisfies G = 1 + zG + tz^2*G + z^3*G^2.
%e T(5,2)=3 because we have UDUUDDUUDD, UUDDUDUUDD, and UUDDUUDDUD.
%e Triangle starts:
%e 1;
%e 1;
%e 1,1;
%e 2,2;
%e 4,3,1;
%e 7,7,3;
%e 13,17,6,1;
%e 26,36,16,4;
%p F := RootOf(G = 1+z*G+t*z^2*G+z^3*G^2, G): Fser := series(F, z = 0, 18): for n from 0 to 15 do P[n] := sort(coeff(Fser, z, n)) end do: for n from 0 to 14 do seq(coeff(P[n], t, j), j = 0 .. floor((1/2)*n)) end do; # yields sequence in triangular form
%Y Cf. A004148, A023431, A110320.
%K nonn,tabf
%O 0,5
%A _Emeric Deutsch_, Oct 11 2009