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A106789
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Sum of two consecutive squares of Lucas 3-step numbers (A001644).
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2
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10, 10, 58, 170, 562, 1962, 6562, 22202, 75242, 254330, 860474, 2911226, 9848050, 33316090, 112707970, 381286954, 1289885834, 4363653034, 14762129274, 49939929610, 168945571442, 571538767370, 1933501811618, 6540989771354
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OFFSET
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0,1
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COMMENTS
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A106729 is sum of two consecutive squares of Lucas numbers (A001254), for which L(n)^2 + L(n+1)^2 = 5*{F(n)^2 + F(n+1)^2} = 5*A001519(n). Sum of two consecutive squares of Lucas 3-step numbers can be expressed in terms of tribonacci numbers, but not quite as neatly, as derived from the identity A001644(n) = T(n) + 2*T(n-1) + 3*T(n-2) = 3*T(n+1) - 2*T(n) - T(n-1) where the tribonacci numbers T(n) = A000073(n).
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LINKS
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FORMULA
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G.f.: 2*(5 - 5*x + 4*x^2 - 18*x^3 - x^4 - 5*x^5)/((1+x+x^2-x^3)*(1-3*x-x^2-x^3)).
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EXAMPLE
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a(3) = A001644(3)^2 + A001644(4)^2 = 7^2 + 11^2 = 49 + 121 = 170 = 13^2 + 1.
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MATHEMATICA
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CoefficientList[Series[2*(5-5*x+4*x^2-18*x^3-x^4-5*x^5)/(1-2*x-3*x^2 -6*x^3+x^4+x^6), {x, 0, 40}], x] (* G. C. Greubel, Apr 21 2019 *)
Total/@Partition[LinearRecurrence[{1, 1, 1}, {3, 1, 3}, 40]^2, 2, 1] (* Harvey P. Dale, Apr 03 2022 *)
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PROG
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(PARI) my(x='x+O('x^40)); Vec(2*(5-5*x+4*x^2-18*x^3-x^4-5*x^5)/(1-2*x-3*x^2-6*x^3+x^4+x^6)) \\ G. C. Greubel, Apr 21 2019
(Magma) R<x>:=PowerSeriesRing(Integers(), 40); Coefficients(R!( 2*(5-5*x+4*x^2-18*x^3-x^4-5*x^5)/(1-2*x-3*x^2-6*x^3+x^4+x^6) )); // G. C. Greubel, Apr 21 2019
(Sage) (2*(5-5*x+4*x^2-18*x^3-x^4-5*x^5)/(1-2*x-3*x^2-6*x^3+x^4+x^6)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Apr 21 2019
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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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