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A108428
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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 doubledescents (i.e. dd's).
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0
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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,5
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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)
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REFERENCES
| Problem 10658, American Math. Monthly, 107, 2000, 368-370.
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FORMULA
| T(n, k)=(1/n)sum(binomial(n, j)binomial(n, k-j)binomial(n+j, k+1), j=0..k). G.f.=G=G(t, z) satisfies t^2*zG^3-t^2*zG^2-(1+z-tz)G+1=0.
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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
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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
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CROSSREFS
| Cf. A027307.
Sequence in context: A046539 A198929 A172347 * A174126 A075613 A189150
Adjacent sequences: A108425 A108426 A108427 * A108429 A108430 A108431
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KEYWORD
| nonn,tabf
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AUTHOR
| Emeric Deutsch (deutsch(AT)duke.poly.edu), Jun 03 2005
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