A106789 Sum of two consecutive squares of Lucas 3-step numbers (A001644).

10, 10, 58, 170, 562, 1962, 6562, 22202, 75242, 254330, 860474, 2911226, 9848050, 33316090, 112707970, 381286954, 1289885834, 4363653034, 14762129274, 49939929610, 168945571442, 571538767370, 1933501811618, 6540989771354

Offset: 0

Author

Jonathan Vos Post, May 16 2005

Keywords

• easy
• nonn

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).

Examples

			a(0) = A001644(0)^2 + A001644(1)^2 = 3^2 + 1^2 = 9 + 1 = 10.
a(1) = A001644(1)^2 + A001644(2)^2 = 1^2 + 3^2 = 1 + 9 = 10.
a(2) = A001644(2)^2 + A001644(3)^2 = 3^2 + 7^2 = 9 + 49 = 58.
a(3) = A001644(3)^2 + A001644(4)^2 = 7^2 + 11^2 = 49 + 121 = 170 = 13^2 + 1.
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A000073, A001644.

Cf. A001254, A001519, A106729.

Programs

Magma

R:=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

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 *)

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

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

Formula

a(n) = A001644(n)^2 + A001644(n+1)^2.

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)).