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Search: id:A108425
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| A108425 |
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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 (i.e. ud and Ud's). |
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+0
4
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| 2, 4, 6, 8, 36, 22, 16, 144, 248, 90, 32, 480, 1600, 1560, 394, 64, 1440, 7840, 14400, 9420, 1806, 128, 4032, 32480, 95760, 115416, 55692, 8558, 256, 10752, 120064, 517440, 986272, 860832, 325360, 41586, 512, 27648, 408576, 2419200, 6668928
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OFFSET
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1,1
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COMMENT
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Row sums yield A027307. T(n,n)=A006318(n) (the large Schroeder numbers; asks for a bijective proof). T(n,1)=2^n.
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REFERENCES
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Problem 10658, American Math. Monthly, 107, 2000, 368-370.
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FORMULA
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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.
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EXAMPLE
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Example T(2,1)=4 because we have uudd, uUddd, Uuddd and UUdddd.
Triangle begins:
2;
4,6;
8,36,22;
16,144,248,90;
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MAPLE
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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
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CROSSREFS
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Cf. A027307, A006318, A108426.
Sequence in context: A057809 A135632 A068541 this_sequence A059569 A083433 A083435
Adjacent sequences: A108422 A108423 A108424 this_sequence A108426 A108427 A108428
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KEYWORD
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nonn,tabl
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AUTHOR
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Emeric Deutsch (deutsch(AT)duke.poly.edu), Jun 03 2005
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