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A215100
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a(n) = 3*a(n-1) + 4*a(n-2) + a(n-3) with a(0)=2, a(1)=5, a(2)=22.
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8
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2, 5, 22, 88, 357, 1445, 5851, 23690, 95919, 388368, 1572470, 6366801, 25778651, 104375627, 422608286, 1711106017, 6928126822, 28051412820, 113577851765, 459867333397, 1861964820071, 7538941645566, 30524551550379, 123591386053472, 500411306007498, 2026124013786761
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OFFSET
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0,1
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COMMENTS
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Ramanujan-type sequence number 4 for the argument 2*Pi/7. We have a(n)=bs(3n+2), where the sequence bs(n) and its two conjugate sequences as(n) and cs(n) are defined in the comments to A214683 (see also A215076, A120757, A006053). Since we also have as(3n+2)=cs(3n+2)=0 from the formula for S(n) (see Comments at A214683) we obtain the relation 7^(1/3)*a(n)= (c(1)/c(4))^(n + 2/3) + (c(4)/c(2))^(n + 2/3) + (c(2)/c(1))^(n + 2/3).
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REFERENCES
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R. Witula, E. Hetmaniok and D. Slota, Sums of the powers of any order roots taken from the roots of a given polynomial, Proceedings of the Fifteenth International Conference on Fibonacci Numbers and Their Applications, Eger, Hungary, 2012.
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LINKS
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FORMULA
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G.f.: (2 - x - x^2)/(1 - 3*x - 4*x^2 - x^3).
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EXAMPLE
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From 4*a(2) = a(3) = 88 we get 88*7^(1/3) = 4*((c(1)/c(4))^(8/3) + (c(4)/c(2))^(8/3) + (c(2)/c(1))^(8/3))=(c(1)/c(4))^(11/3) + (c(4)/c(2))^(11/3) + (c(2)/c(1))^(11/3).
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MATHEMATICA
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LinearRecurrence[{3, 4, 1}, {2, 5, 22}, 40]
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PROG
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(PARI) Vec((2-x-x^2)/(1-3*x-4*x^2-x^3) + O(x^40)) \\ Michel Marcus, Apr 20 2016
(Magma) I:=[2, 5, 22]; [n le 3 select I[n] else 3*Self(n-1) +4*Self(n-2) +Self(n-3): n in [1..41]]; // G. C. Greubel, Nov 25 2022
(SageMath)
@CachedFunction
if (n<3): return (2, 5, 22)[n]
else: return 3*a(n-1) + 4*a(n-2) + a(n-3)
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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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