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A104856
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Triangle read by rows: T(n,k)=binomial(n,k)binomial(k,floor(k/2))binomial(n-k,floor((n-k)/2)) (0<=k<=n).
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0
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1, 1, 1, 2, 2, 2, 3, 6, 6, 3, 6, 12, 24, 12, 6, 10, 30, 60, 60, 30, 10, 20, 60, 180, 180, 180, 60, 20, 35, 140, 420, 630, 630, 420, 140, 35, 70, 280, 1120, 1680, 2520, 1680, 1120, 280, 70, 126, 630, 2520, 5040, 7560, 7560, 5040, 2520, 630, 126, 252, 1260, 6300
(list; table; graph; refs; listen; history; internal format)
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OFFSET
| 0,4
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COMMENTS
| T(n,k) is the number of paths in the first quadrant, starting from the origin, with unit steps up, down, right, or left, having a total of n steps, exactly k of which are vertical (up or down). Example: T(3,2)=6 because we have NNE, NEN, ENN, NSE, ENS and NES. [From Emeric Deutsch (deutsch(AT)duke.poly.edu), Nov 22 2008]
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REFERENCES
| David M. Bloom, Problem 10921, Amer. Math. Monthly 110, (2003), 958-959.
E. Deutsch and D. Lovit, Math. Magazine, vol. 80, No. 1, 2007, p. 80, Problem 1739. [From Emeric Deutsch (deutsch(AT)duke.poly.edu), Nov 22 2008]
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FORMULA
| T(n, k)=binomial(n, k)binomial(k, floor(k/2))binomial(n-k, floor((n-k)/2)) (0<=k<=n).
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MAPLE
| T:=(n, k)->binomial(n, k)*binomial(k, floor(k/2))*binomial(n-k, floor((n-k)/2)): for n from 0 to 10 do seq(T(n, k), k=0..n) od; # yields sequence in triangular form
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CROSSREFS
| Row sums yield A005566. T(n, 0)=T(n, n)=A001405(n).
Cf. A005566, A001405.
Sequence in context: A104346 A193450 A109906 * A038715 A057040 A096235
Adjacent sequences: A104853 A104854 A104855 * A104857 A104858 A104859
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KEYWORD
| nonn,tabl
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AUTHOR
| Emeric Deutsch (deutsch(AT)duke.poly.edu), Apr 23 2005
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