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A099579
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a(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k-1) * 3^(k-1).
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2
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0, 0, 1, 1, 7, 10, 40, 70, 217, 427, 1159, 2440, 6160, 13480, 32689, 73129, 173383, 392770, 919480, 2097790, 4875913, 11169283, 25856071, 59363920, 137109280, 315201040, 727060321, 1672663441, 3855438727, 8873429050, 20444528200
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OFFSET
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0,5
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COMMENTS
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In general a(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k-1)*r^(k-1) has g.f. x^2/((1-r*x^2)*(1-x-r*x^2)) and satisfies the recurrence a(n) = a(n-1) + 2*r*a(n-2) - r*a(n-3) - r^2*a(n-4).
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LINKS
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FORMULA
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G.f.: x^2/((1-3*x^2)*(1-x-3*x^2)).
a(n) = a(n-1) + 6*a(n-2) - 3*a(n-3) - 9*a(n-4).
a(n) = (i*sqrt(3))^(n-1)*ChebyshevU(n-1, -i/(2*sqrt(3))) - 3^((n-1)/2)*(1 - (-1)^n)/2.
E.g.f.: (1/sqrt(39))*( 2*sqrt(3)*exp(x/2)*sinh(sqrt(13)*x/2) - sqrt(13)*sinh(sqrt(3)*x) ). (End)
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MATHEMATICA
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LinearRecurrence[{1, 6, -3, -9}, {0, 0, 1, 1}, 50] (* G. C. Greubel, Jul 24 2022 *)
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PROG
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(Magma) [n le 4 select Floor((n-1)/2) else Self(n-1) +6*Self(n-2) -3*Self(n-3) -9*Self(n-4): n in [1..41]]; // G. C. Greubel, Jul 24 2022
(SageMath)
@CachedFunction
if (n<4): return (n//2)
else: return a(n-1) +6*a(n-2) -3*a(n-3) -9*a(n-4)
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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