OFFSET
0,1
COMMENTS
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).
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..1000
FORMULA
EXAMPLE
MATHEMATICA
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 *)
PROG
(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
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Jonathan Vos Post, May 16 2005
STATUS
approved