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A108446
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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 of the form ud.
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2
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1, 1, 1, 4, 5, 1, 20, 32, 13, 1, 113, 223, 135, 26, 1, 688, 1620, 1300, 412, 45, 1, 4404, 12064, 12050, 5350, 1030, 71, 1, 29219, 91335, 109134, 62450, 17575, 2247, 105, 1, 199140, 699689, 973077, 682234, 254625, 49210, 4438, 148, 1, 1385904, 5407744
(list; table; graph; refs; listen; history; internal format)
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OFFSET
| 0,4
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COMMENTS
| Row sums yield A027307. Column 0 yields A108447. T(n,n-1)=A008778(n-1)=n(n^2+6n-1)/6. Number of ud peaks in all paths from (0,0) to (3n,0) is given by A108448.
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REFERENCES
| Problem 10658, American Math. Monthly, 107, 2000, 368-370.
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FORMULA
| T(n, k)=(1/n)binomial(n, k)*sum(binomial(n-k, j)*binomial(n+2j, k+j-1), j=0..n-k). G.f.=G=G(t, z) satisfies G=1+z(G-1+t)G+zG^3.
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EXAMPLE
| T(2,1)=5 because we have udUdd, uudd, Uddud, Ududd and Uuddd.
Triangle begins:
1;
1,1;
4,5,1;
20,32,13,1;
113,223,135,26,1;
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MAPLE
| T:=proc(n, k) if n=0 and k=0 then 1 elif n=0 then 0 elif k=n then 1 elif k=n then 1 else (1/n)*binomial(n, k)*sum(binomial(n-k, j)*binomial(n+2*j, k+j-1), j=0..n-k) fi end: for n from 0 to 9 do seq(T(n, k), k=0..n) od; # yields sequence in triangular form
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CROSSREFS
| Cf. A027307, A008778, A108447, A108448, A108425, A108426.
Sequence in context: A206282 A082051 A196848 * A109962 A102230 A147724
Adjacent sequences: A108443 A108444 A108445 * A108447 A108448 A108449
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KEYWORD
| nonn,tabl
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AUTHOR
| Emeric Deutsch (deutsch(AT)duke.poly.edu), Jun 10 2005
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