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Search: id:A141523
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| A141523 |
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A "Lucas" type version of the tribonacci sequence in that the zeroth term is 3 in the sequence and a Binet version: a(0)=3;a(1)=1;a(2)=1; a(n)=a(n-1)+a(n-2)+a(n-3). ( result differs from A001644). |
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| 3, 1, 1, 5, 7, 13, 25, 45, 83, 153, 281, 517, 951, 1749, 3217, 5917, 10883, 20017, 36817, 67717, 124551, 229085, 421353, 774989, 1425427, 2621769, 4822185, 8869381, 16313335, 30004901, 55187617
(list; graph; listen)
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OFFSET
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0,1
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COMMENT
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The sequence is "designed" to give a zeroth term as 3:
The Mathematica to generate the Binet function:
f[n_Integer] = Module[{a}, a[n] /. RSolve[{a[
n] == a[n - 1] + a[n - 2] + a[n - 3], a[0] == 3, a[1] == 1.a[2] == 1}, a[n], n][[1]] // FullSimplify]
Code is equivalent to:
a = Table[x /. NSolve[x^3 - x^2 - x - 1 == 0, x][[n]], {n, 1, 3}]
g[n_] = (0.634392934063409) a[[3]]^n + (1.1828035329682955 + 0.6810930616541588*I)a[[1]]^n + (1.1828035329682955- 0.6810930616541588*I) a[[2]]^n
Rationalize[N[Table[Re[g[n]], {n, 0, 25}], 100], 0]
Round[%]
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FORMULA
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a(0)=3;a(1)=1;a(2)=1; a(n)=a(n-1)+a(n-2)+a(n-3).
O.g.f.: (3-2x-3x^2)/(1-x-x^2-x^3). a(n)= A001644(n)-2*A000073(n). [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Aug 22 2008]
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MATHEMATICA
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Clear[f, g, n, a] a[0] = 3; a[1] = 1; a[2] = 1; a[n_] := a[n] = a[n - 1] + a[n - 2] + a[n - 3]; Table[a[n], {n, 0, 30}]
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CROSSREFS
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Cf. A001644.
Sequence in context: A047812 A129392 A118538 this_sequence A086385 A123162 A083075
Adjacent sequences: A141520 A141521 A141522 this_sequence A141524 A141525 A141526
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KEYWORD
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nonn,uned
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AUTHOR
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Roger L. Bagula and Gary W. Adamson (rlbagulatftn(AT)yahoo.com), Aug 11 2008
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