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A274761
a(n) = a(n-1) + a(n-2) + a(n-3), with a(0) = 3, a(1) = 2, a(2) = 5.
1
3, 2, 5, 10, 17, 32, 59, 108, 199, 366, 673, 1238, 2277, 4188, 7703, 14168, 26059, 47930, 88157, 162146, 298233, 548536, 1008915, 1855684, 3413135, 6277734, 11546553, 21237422, 39061709, 71845684, 132144815, 243052208, 447042707, 822239730, 1512334645, 2781617082, 5116191457, 9410143184
OFFSET
0,1
COMMENTS
The tetranacci sequence(s) (T(n+3) = T(n+2) + T(n+1) + T(n)) have characteristic roots such that one is real and the other two have imaginary components. The real root is given by r = (1/3)*(1 + x + y), where x = (19 + 3*sqrt(33))^(1/3) and y = (19 - 3*sqrt(33))^(1/3). By increasing the power of this root (r^n) it is found that r^n = (1/3)*( A001644(n) + (1/3)*a(n-1)*(x + y) + (1/3)*A000073(n-2)*(x^2 + y^2)). Similar results can be found for the complex roots.
LINKS
Pin-Yen Lin, De Moivre type identities for the Tribonacci numbers, The Fibonacci Quarterly 26, no.2, (1988), 131-134
FORMULA
a(n) = a(n-1) + a(n-2) + a(n-3).
a(n) = 3*A000073(n) - A000073(n-1).
G.f.: (3 - x)/(1 - x - x^2 - x^3).
EXAMPLE
a(4) = a(3) + a(2) + a(1) = 10 + 5 + 2 = 17. - Michael B. Porter, Jul 05 2016
MATHEMATICA
CoefficientList[Series[(3-x)/(1-x-x^2-x^3), {x, 0, 40}], x] (* or *) LinearRecurrence[{1, 1, 1}, {3, 2, 5}, 40]
PROG
(PARI) Vec((3-x)/(1-x-x^2-x^3) + O(x^40)) \\ Altug Alkan, Jul 04 2016
(Magma) R<x>:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (3-x)/(1-x-x^2-x^3) )); // G. C. Greubel, Apr 23 2019
(Sage) ((3-x)/(1-x-x^2-x^3)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Apr 23 2019
(GAP) a:=[3, 2, 5];; for n in [4..40] do a[n]:=a[n-1]+a[n-2]+a[n-3]; od; a; # G. C. Greubel, Apr 23 2019
CROSSREFS
Sequence in context: A257981 A128914 A050063 * A100885 A209199 A215328
KEYWORD
nonn
AUTHOR
G. C. Greubel, Jul 04 2016
STATUS
approved