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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).
0
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, 1167, 1167, 650, 190, 25, 1, 1, 36, 400, 2045, 5685, 9318, 9318, 5685, 2045, 400, 36, 1, 1, 49, 749, 5341, 21133, 50813, 77947, 77947, 50813, 21133, 5341, 749, 49, 1, 1, 64, 1288
OFFSET
0,5
COMMENTS
Row n contains 2n terms (n > 0).
Row sums yield A027307.
T(n,1) = T(n,2n-2) = n^2*T(n,k) = T(n,2n-k-1) (mirror symmetry).
LINKS
Emeric Deutsch, Problem 10658: Another Type of Lattice Path, American Math. Monthly, 107, 2000, 368-370.
FORMULA
T(n, k) = (1/n)*Sum_{j=0..k} binomial(n, j)*binomial(n, k-j)*binomial(n+j, k+1).
G.f.: G = G(t, z) satisfies t^2*zG^3 - t^2*zG^2 - (1 + z - tz)G + 1 = 0.
EXAMPLE
T(2,1)=4 because we have udUdd, uudd, Uddud and Ududd.
Triangle begins:
1;
1, 1;
1, 4, 4, 1;
1, 9, 23, 23, 9, 1;
1, 16, 76, 156, 156, 76, 16, 1;
...
MAPLE
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
CROSSREFS
Cf. A027307.
Sequence in context: A198929 A172347 A375863 * A373431 A174126 A075613
KEYWORD
nonn,tabf
AUTHOR
Emeric Deutsch, Jun 03 2005
STATUS
approved