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Search: id:A108439
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| A108439 |
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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 having abscissa of first return equal to 3k. |
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+0
1
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| 2, 4, 6, 20, 12, 34, 132, 60, 68, 238, 996, 396, 340, 476, 1858, 8132, 2988, 2244, 2380, 3716, 15510, 69940, 24396, 16932, 15708, 18580, 31020, 135490, 624132, 209820, 138244, 118524, 122628, 155100, 270980, 1223134, 5725124, 1872396, 1188980
(list; table; graph; listen)
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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)=A108424(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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G.f.= tzA(z)A(tz)+tzA(z)A^2(tz), where A=1+zA^2+zA^3=(2/3)*sqrt((z+3)/z)*sin((1/3)*arcsin(sqrt(z)*(z+18)/(z+3)^(3/2)))-1/3 (the g.f. of A027307). T(n, k)=A108424(k)*A027307(n-k) (there are explicit formulas in A108424 and A027307).
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EXAMPLE
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T(2,1)=4 because we have u(d)ud, u(d)Udd, Ud(d)ud, and Ud(d)Udd, the d step of the first return being shown between parentheses.
Triangle begins:
2;
4,6;
20,12,34;
132,60,68,238;
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MAPLE
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a:=n->sum(2^(i+1)*binomial(2*n, i)*binomial(n, i+1), i=0..n-1)/n: b:=proc(n) if n=1 then 2 else (n*2^n*binomial(2*n, n)/((2*n-1)*(n+1)))*sum(binomial(n-1, j)^2/2^j/binomial(n+j+1, j), j=0..n-1): fi end: T:=proc(n, k) if k=n then b(n) else b(k)*a(n-k) fi end:for n from 1 to 9 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, A108424, A108435.
Adjacent sequences: A108436 A108437 A108438 this_sequence A108440 A108441 A108442
Sequence in context: A066894 A111115 A005227 this_sequence A045960 A028988 A086172
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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 05 2005
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