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A107239
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Sum of squares of tribonacci numbers (A000073).
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15
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0, 0, 1, 2, 6, 22, 71, 240, 816, 2752, 9313, 31514, 106590, 360606, 1219935, 4126960, 13961456, 47231280, 159782161, 540539330, 1828631430, 6186215574, 20927817799, 70798300288, 239508933824, 810252920400, 2741065994769, 9272959837818, 31370198430718
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OFFSET
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0,4
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REFERENCES
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R. Schumacher, Explicit formulas for sums involving the squares of the first n Tribonacci numbers, Fib. Q., 58:3 (2020), 194-202.
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LINKS
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FORMULA
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a(n) = T(0)^2 + T(1)^2 + ... + T(n)^2 where T(n) = A000073(n).
G.f.: x^2*(1-x-x^2-x^3)/((1+x+x^2-x^3)*(1-3*x-x^2-x^3)*(1-x)). (End)
a(n) = 1/4 - (1/11)*Sum_{_R = RootOf(_Z^3-_Z^2-_Z-1)} ((3 + 7*_R + 5*_R^2)/(3*_R^2 - 2*_R - 1)*_R^(-n) - (1/44)*Sum_{_R = RootOf(_Z^3+_Z^2+3*_Z-1)} ((-1 - 2*_R - 9*_R^2)/(3*_R^2 + 2*_R + 3)*_R^(-n). - Robert Israel, Mar 26 2010
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EXAMPLE
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a(7) = 71 = 0^2 + 0^2 + 1^2 + 1^2 + 2^2 + 4^2 + 7^2
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MAPLE
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b:= proc(n) option remember; `if`(n<3, [n*(n-1)/2$2],
(t-> [t, t^2+b(n-1)[2]])(add(b(n-j)[1], j=1..3)))
end:
a:= n-> b(n)[2]:
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MATHEMATICA
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Accumulate[LinearRecurrence[{1, 1, 1}, {0, 0, 1}, 30]^2] (* Harvey P. Dale, Sep 11 2011 *)
LinearRecurrence[{3, 1, 3, -7, 1, -1, 1}, {0, 0, 1, 2, 6, 22, 71}, 30] (* Ray Chandler, Aug 02 2015 *)
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PROG
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(Magma) R<x>:=PowerSeriesRing(Integers(), 40); [0, 0] cat Coefficients(R!( x^2*(1-x-x^2-x^3)/((1+x+x^2-x^3)*(1-3*x-x^2-x^3)*(1-x)) )); // G. C. Greubel, Nov 20 2021
(Sage)
@CachedFunction
if (n<2): return 0
elif (n==2): return 1
else: return T(n-1) +T(n-2) +T(n-3)
def A107231(n): return sum(T(j)^2 for j in (0..n))
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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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