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%I #6 Nov 29 2015 17:43:55
%S 2,2,2,2,2,4,8,4,2,2,6,18,18,18,6,2,2,8,32,48,72,48,32,8,2,2,10,50,
%T 100,200,200,200,100,50,10,2,2,12,72,180,450,600,800,600,450,180,72,
%U 12,2,2,14,98,294,882,1470,2450,2450,2450,1470,882,294,98,14,2,2,16,128,448
%N Triangle read by rows: T(n,k) is the number of lattice paths from (0,0) to (n,n) with steps E=(1,0) and N=(0,1) and having k turns (NE or EN) (1<=k<=2n-1).
%C Row n has 2n-1 entries.
%C Sum of entries of row n = binomial(2n,n) = A000984(n) (the central binomial coefficients).
%C Sum(k*T(n,k),k=0..2n-1) = n*binomial(2n,n) = A005430(n).
%F T(n,2k) = 2*binomial(n-1,k-1)*binomial(n-1,k);
%F T(n,2k-1) = 2*binomial(n-1,k-1)^2.
%F G.f.: [1+t*r(t^2,z)]/[1-t*r(t^2,z)], where r(t,z) is the Narayana function, defined by r = z(1+r)(1+tr).
%e T(3,2)=4 because we have ENNNEE, EENNNE, NEEENN and NNEEEN.
%e Triangle starts:
%e 2;
%e 2,2,2;
%e 2,4,8,4,2;
%e 2,6,18,18,18,6,2;
%e 2,8,32,48,72,48,32,8,2;
%p T := proc (n, k) if `mod`(k, 2) = 0 then 2*binomial(n-1, (1/2)*k-1)*binomial(n-1, (1/2)*k) else 2*binomial(n-1, (1/2)*k-1/2)^2 end if end proc: for n to 9 do seq(T(n, k), k = 1 .. 2*n-1) end do; # yields sequence in triangular form
%Y Cf. A000984, A005430.
%K nonn,tabf
%O 1,1
%A _Emeric Deutsch_, Dec 10 2008
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