%I A108425
%S A108425 2,4,6,8,36,22,16,144,248,90,32,480,1600,1560,394,64,1440,7840,14400,
%T A108425 9420,1806,128,4032,32480,95760,115416,55692,8558,256,10752,120064,
%U A108425 517440,986272,860832,325360,41586,512,27648,408576,2419200,6668928
%N A108425 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 peaks 
               (i.e. ud and Ud's).
%C A108425 Row sums yield A027307. T(n,n)=A006318(n) (the large Schroeder numbers; 
               asks for a bijective proof). T(n,1)=2^n.
%D A108425 Problem 10658, American Math. Monthly, 107, 2000, 368-370.
%F A108425 T(n, k)=(1/n)binomial(n, k)*sum(2^(n-j)*binomial(n, j)*binomial(n, k-1-j), 
               j=0..k-1). G.f. =G=G(t, z) satisfies zG^3+tzG^2-(1+z-tz)G+1=0.
%e A108425 Example T(2,1)=4 because we have uudd, uUddd, Uuddd and UUdddd.
%e A108425 Triangle begins:
%e A108425 2;
%e A108425 4,6;
%e A108425 8,36,22;
%e A108425 16,144,248,90;
%p A108425 T:=(n,k)->(1/n)*binomial(n,k)*sum(2^(n-j)*binomial(n,j)*binomial(n,k-1-j),
               j=0..k-1): for n from 1 to 10 do seq(T(n,k),k=1..n) od; # yields 
               sequence in triangular form
%Y A108425 Cf. A027307, A006318, A108426.
%Y A108425 Sequence in context: A057809 A135632 A068541 this_sequence A059569 A083433 
               A083435
%Y A108425 Adjacent sequences: A108422 A108423 A108424 this_sequence A108426 A108427 
               A108428
%K A108425 nonn,tabl
%O A108425 1,1
%A A108425 Emeric Deutsch (deutsch(AT)duke.poly.edu), Jun 03 2005