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A108428 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 have k doubledescents (i.e., dd's).

%I A108428 #12 Jul 26 2017 03:16:24
%S A108428 1,1,1,1,4,4,1,1,9,23,23,9,1,1,16,76,156,156,76,16,1,1,25,190,650,
%T A108428 1167,1167,650,190,25,1,1,36,400,2045,5685,9318,9318,5685,2045,400,36,
%U A108428 1,1,49,749,5341,21133,50813,77947,77947,50813,21133,5341,749,49,1,1,64,1288
%N A108428 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 have k doubledescents (i.e., dd's).
%C A108428 Row n contains 2n terms (n > 0).
%C A108428 Row sums yield A027307.
%C A108428 T(n,1) = T(n,2n-2) = n^2*T(n,k) = T(n,2n-k-1) (mirror symmetry).
%H A108428 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 A108428 T(n, k) = (1/n)*Sum_{j=0..k} binomial(n, j)*binomial(n, k-j)*binomial(n+j, k+1).
%F A108428 G.f.: G = G(t, z) satisfies t^2*zG^3 - t^2*zG^2 - (1 + z - tz)G + 1 = 0.
%e A108428 T(2,1)=4 because we have udUdd, uudd, Uddud and Ududd.
%e A108428 Triangle begins:
%e A108428   1;
%e A108428   1,  1;
%e A108428   1,  4,  4,   1;
%e A108428   1,  9, 23,  23,   9,  1;
%e A108428   1, 16, 76, 156, 156, 76, 16, 1;
%e A108428   ...
%p A108428 a:=proc(n,k) if n=0 and k=0 then 1 elif n=0 then 0 else (1/n)*sum(binomial(n,j)*binomial(n,k-j)*binomial(n+j,k+1),j=0..k) fi end: print(1); for n from 1 to 8 do seq(a(n,k),k=0..2*n-1) od; # yields sequence in triangular form
%Y A108428 Cf. A027307.
%K A108428 nonn,tabf
%O A108428 0,5
%A A108428 _Emeric Deutsch_, Jun 03 2005