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A099251
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Bisection of Motzkin sums (A005043).
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8
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1, 1, 3, 15, 91, 603, 4213, 30537, 227475, 1730787, 13393689, 105089229, 834086421, 6684761125, 54022715451, 439742222071, 3602118427251, 29671013856627, 245613376802185, 2042162142208813, 17047255430494497, 142816973618414817
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OFFSET
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0,3
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COMMENTS
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The Kn4 triangle sums of A175136 lead to the sequence given above (n >= 1). For the definition of the Kn4 and other triangle sums see A180662. - Johannes W. Meijer, May 06 2011
Equals the expected value of trace(O)^(2n), where O is a 3 X 3 orthogonal matrix randomly selected according to Haar measure (see MathOverflow link). - Nathaniel Johnston, Sep 05 2014
In Smith (1985), we apparently have a(n) = P(2*n), where P(n) is the number of linearly independent three-dimensional n-th order isotropic tensors. In the paper, he refers to Smith (1968) for more details. It is not clear why he does not list the values of P(2*n+1). See also the 1978 letter of D. L. Andrews to N. J. A. Sloane.
Eric Weisstein gives some details on how the material in Smith (1968) about isotropic tensors is related to Motzkin sums. (End)
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REFERENCES
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G. F. Smith, On isotropic tensors and rotation tensors of dimension m and order n, Tensor (N.S.), Vol. 19 (1968), 79-88 (MR0224008).
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LINKS
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FORMULA
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Recurrence: n*(2*n + 1)*a(n) = (2*n - 1)*(13*n - 10)*a(n-1) - 3*(26*n^2 - 87*n + 76)*a(n-2) + 27*(n - 2)*(2*n - 5)*a(n-3). - Vaclav Kotesovec, Oct 17 2012
Conjecture: a(n) = (2/Pi)*Integral_{t=0..1} sqrt((1 - t)/t)*(1 - 8*t + 16*t^2)^n. - Benedict W. J. Irwin, Oct 05 2016
a(n) = Sum_{j=0..2*n+1} (C(2*j,j)*(-1)^(j)*C(2*n+1,j+1))/(2*n+1). - Vladimir Kruchinin, Apr 02 2017
a(n) = hypergeom([1/2, -2*n], [2], 4). - Peter Luschny, Jul 25 2020
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MAPLE
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G := (1+x-sqrt(1-2*x-3*x^2))/(2*x*(1+x)): Gser := series(G, x=0, 60):
a := n -> hypergeom([1/2, -2*n], [2], 4):
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MATHEMATICA
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Take[CoefficientList[Series[(1 + x - Sqrt[1 - 2 * x - 3 * x^2])/(2 * x * (1 + x)), {x, 0, 60}], x], {1, -1, 2}] (* Vaclav Kotesovec, Oct 17 2012 *)
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PROG
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(PARI) x='x+O('x^66); v=Vec((1+x-sqrt(1-2*x-3*x^2))/(2*x*(1+x))); vector(#v\2, n, v[2*n-1]) \\ Joerg Arndt, May 12 2013
(Maxima)
a(n):=sum(binomial(2*j, j)*(-1)^(j)*binomial(2*n+1, j+1), j, 0, 2*n+1)/(2*n+1); /*Vladimir Kruchinin, Apr 02 2017*/
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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