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%I #7 Oct 07 2015 04:11:41
%S 2,4,6,20,12,34,132,60,68,238,996,396,340,476,1858,8132,2988,2244,
%T 2380,3716,15510,69940,24396,16932,15708,18580,31020,135490,624132,
%U 209820,138244,118524,122628,155100,270980,1223134,5725124,1872396,1188980
%N Triangle read by rows: T(n,k) is number of paths from (0,0) to (3n,0) that stay in the first quadrant (but may touch the horizontal axis), consisting of steps u=(2,1), U=(1,2), or d=(1,-1) and having abscissa of first return equal to 3k.
%C Row sums yield A027307. T(n,n) = A108424(n).
%H Emeric Deutsch, <a href="http://www.jstor.org/stable/2589192">Problem 10658: Another Type of Lattice Path</a>, American Math. Monthly, 107, 2000, 368-370.
%F G.f.: tzA(z)A(tz)+tzA(z)A^2(tz), where A=1+zA^2+zA^3=(2/3)*sqrt((z+3)/z)*sin((1/3)*arcsin(sqrt(z)*(z+18)/(z+3)^(3/2)))-1/3 (the g.f. of A027307).
%F T(n, k) = A108424(k)*A027307(n-k) (there are explicit formulas in A108424 and A027307).
%e T(2,1)=4 because we have u(d)ud, u(d)Udd, Ud(d)ud and Ud(d)Udd, the d step of the first return being shown between parentheses.
%e Triangle begins:
%e 2;
%e 4,6;
%e 20,12,34;
%e 132,60,68,238;
%e ...
%p a:=n->sum(2^(i+1)*binomial(2*n,i)*binomial(n,i+1),i=0..n-1)/n: b:=proc(n) if n=1 then 2 else (n*2^n*binomial(2*n,n)/((2*n-1)*(n+1)))*sum(binomial(n-1,j)^2/2^j/binomial(n+j+1,j),j=0..n-1): fi end: T:=proc(n,k) if k=n then b(n) else b(k)*a(n-k) fi end:for n from 1 to 9 do seq(T(n,k),k=1..n) od; > # yields sequence in triangular form
%Y Cf. A027307, A108424, A108435.
%K nonn,tabl
%O 1,1
%A _Emeric Deutsch_, Jun 05 2005
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