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A364319
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a(n) = (A077446(n) + 1)/2 for n >= 0.
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0
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0, 1, 3, 6, 16, 33, 91, 190, 528, 1105, 3075, 6438, 17920, 37521, 104443, 218686, 608736, 1274593, 3547971, 7428870, 20679088, 43298625, 120526555, 252362878, 702480240, 1470878641, 4094354883, 8572908966, 23863649056, 49966575153, 139087539451
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OFFSET
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0,3
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COMMENTS
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a(n) and b(n) = A006452(n+1), for n >= 0, give the nonnegative solution of the equation binomial(a(n), 2) = b(n)^2 - 1.
This shows that the number of independent elements of an antisymmetric a(n) X a(n) matrix coincides with the number of independent elements of a traceless b(n) X b(n) matrix. The n = 0 case is trivial: 0 = 0. (The question about this coincidence was posed to W. L. by Martin Bordemann, Mar 03 1991.)
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LINKS
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FORMULA
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a(n) = 6*a(n-2) - a(n-4) - 2, for n >= 0, with a(-4) = -32, a(-3) = -15, a(-2) = -5, a(-1) = -2.
O.g.f.: G(x) = x*(1 + 2*x - 3*x^2 - 2*x^3)/((1 - x)*(1 - 6*x^2 + x^4)) = x*(1 + 2*x - 3*x^2 - 2*x^3)/((1 - x)*(1 - 2*x - x^2)*(1 + 2*x - x^2)).
Bisection: a(2*k) = (5*S(k-1, 6) + S(k-2, 6) + 1)/2 and a(2*k+1) = (S(k, 6) + 5*S(k-1, 6) + 1)/2, for k >= 0, with the Chebyshev polynomials S(n, x) (A049310) with S(-2, x) = -1, S(-1, x) = 0, evaluated at x = 6. S(n, 6) = A001109(n+1).
Bisection: a(2*k) = (1 + 8*q(k) - p(k))/2 and a(2*k+1) = (1 + 8*q(k) + p(k))/2, for k >= 0, with p(k) = A001541(k) = S(k, 6) - 3*S(k-1, 6) and q(k) = A001109(k) = S(k-1, 6).
E.g.f.: (cosh(x) - cosh(sqrt(2)*x)*(cosh(x) - 3*sinh(x)) + sinh(x) - sqrt(2)*(cosh(x) - 2*sinh(x))*sinh(sqrt(2)*x))/2. - Stefano Spezia, Aug 29 2023
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EXAMPLE
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The solutions (a(n), b(n)) begin:
n: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ...
------------------------------------------------------------------------
a: 0 1 3 6 16 33 91 190 528 1105 3075 6438 17920 37521 104443 218686 ...
b: 1 1 2 4 11 23 64 134 373 781 2174 4552 12671 26531 73852 154634 ...
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MATHEMATICA
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LinearRecurrence[{1, 6, -6, -1, 1}, {0, 1, 3, 6, 16}, 31] (* Robert P. P. McKone, Aug 29 2023 *)
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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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STATUS
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approved
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