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A108426
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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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5
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1, 1, 1, 3, 5, 2, 12, 28, 21, 5, 55, 165, 180, 84, 14, 273, 1001, 1430, 990, 330, 42, 1428, 6188, 10920, 10010, 5005, 1287, 132, 7752, 38760, 81396, 92820, 61880, 24024, 5005, 429, 43263, 245157, 596904, 813960, 678300, 352716, 111384, 19448, 1430, 246675
(list; table; graph; refs; listen; history; internal format)
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OFFSET
| 0,4
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COMMENTS
| Row sums yield A027307. T(n,n)=A000108(n) (the Catalan numbers). T(n,0)=A001764(n)=binomial(3n,n)/(2n+1). Number of Ud peaks in all paths from (0,0) to (3n,0) is given by A108427.
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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)*binomial(3n-k, n-1). G.f. =G=G(t, z) satisfies G=1+z(t+G)G^2.
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EXAMPLE
| Example T(2,1)=5 because we have udUdd, uUddd, Uddud, Ududd and UUdddd.
Triangle begins:
1;
1,1;
3,5,2;
12,28,21,5;
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MAPLE
| T:=(n, k)->binomial(n, k)*binomial(3*n-k, n-1)/n: print(1); for n from 1 to 9 do seq(T(n, k), k=0..n) od; # yields sequence in triangular form
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CROSSREFS
| Cf. A027307, A000108, A001764, A108425, A108427.
Sequence in context: A112323 A102507 A076556 * A163237 A053979 A130847
Adjacent sequences: A108423 A108424 A108425 * A108427 A108428 A108429
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KEYWORD
| nonn,tabl
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AUTHOR
| Emeric Deutsch (deutsch(AT)duke.poly.edu), Jun 03 2005
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