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A032349
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Number of paths from (0,0) to (3n,0) that stay in first quadrant (but may touch horizontal axis), where each step is (2,1),(1,2) or (1,-1) and start with (2,1).
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13
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1, 4, 24, 172, 1360, 11444, 100520, 911068, 8457504, 80006116, 768464312, 7474561164, 73473471344, 728745517972, 7284188537672, 73301177482172, 742009157612608, 7550599410874820, 77193497566719320, 792498588659426924
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OFFSET
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1,2
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LINKS
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Emeric Deutsch, Problem 10658, American Math. Monthly, 107, 2000, 368-370.
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FORMULA
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G.f.: z*A^2, where A is the g.f. of A027307.
a(n) = 2*Sum_{i=0..n-1} (2*n+i-1)!/(i!*(n-i-1)!*(n+i+1)!). - Vladimir Kruchinin, Oct 18 2011
D-finite with recurrence: n*(2*n-1)*a(n) = (28*n^2-65*n+36)*a(n-1) - (64*n^2-323*n+408)*a(n-2) - 3*(n-4)*(2*n-5)*a(n-3). - Vaclav Kotesovec, Oct 08 2012
a(n) ~ sqrt(45*sqrt(5)-100)*((11+5*sqrt(5))/2)^n/(5*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 08 2012
G.f. A(x) satisfies: A(x) = 1 + x * ( A(x) + sqrt(A(x)) )^2. - Paul D. Hanna, Jun 11 2016
n*(2*n-1)*(5*n-9)*a(n) = 2*(55*n^3-209*n^2+255*n-99)*a(n-1) + (n-3)*(2*n-3)*(5*n-4)*a(n-2) with a(1) = 1 and a(2) = 4.
G.f.: A(x) = series reversion of x*(1 - x)^2/(1 + x)^2. (End)
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MATHEMATICA
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RecurrenceTable[{n*(2*n-1)*a[n] == (28*n^2-65*n+36)*a[n-1] - (64*n^2-323*n+408)*a[n-2] - 3*(n-4)*(2*n-5)*a[n-3], a[1]==1, a[2]==4, a[3]==24}, a, {n, 20}] (* Vaclav Kotesovec, Oct 08 2012 *)
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PROG
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(Maxima)
a(n):=2*sum((2*n+i-1)!/(i!*(n-i-1)!*(n+i+1)!), i, 0, n-1); \\ Vladimir Kruchinin, Oct 18 2011
(PARI) vector(30, n, 2*sum(k=0, n-1, (2*n+k-1)!/(k!*(n-k-1)!*(n+k+1)!))) \\ Altug Alkan, Oct 06 2015
(PARI) {a(n) = my(A=1); for(i=1, n, A = 1 + x*(A + sqrt(A +x*O(x^n)))^2); polcoeff(A, n)}
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CROSSREFS
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Convolution of A027307 with itself.
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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