A107239 Sum of squares of tribonacci numbers (A000073).

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

Offset: 0

Author

Jonathan Vos Post, May 17 2005

Keywords

• easy
• nonn

Examples

			a(7) = 71 = 0^2 + 0^2 + 1^2 + 1^2 + 2^2 + 4^2 + 7^2
		

References

• R. Schumacher, Explicit formulas for sums involving the squares of the first n Tribonacci numbers, Fib. Q., 58:3 (2020), 194-202.

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• M. Feinberg, Fibonacci-Tribonacci, Fib. Quart. 1(3) (1963), 71-74.
• Z. Jakubczyk, Advanced Problems and Solutions, Fib. Quart. 51 (3) (2013) 185, H-715.
• Eric Weisstein's World of Mathematics, Tribonacci Number
• Index entries for linear recurrences with constant coefficients, signature (3,1,3,-7,1,-1,1).

Crossrefs

Cf. A000073, A000213, A085697.

Cf. A107240, A107241, A107242, A107243, A107244, A107245, A107246, A107247.

Programs

Magma

R:=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

Maple

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]:
seq(a(n), n=0..30);  # Alois P. Heinz, Nov 22 2021

Mathematica

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 *)

Sage

@CachedFunction
def T(n): # A000073
    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))
[A107239(n) for n in (0..40)] # G. C. Greubel, Nov 20 2021

Formula

a(n) = T(0)^2 + T(1)^2 + ... + T(n)^2 where T(n) = A000073(n).

From R. J. Mathar, Aug 19 2008: (Start)

a(n) = Sum_{i=0..n} A085697(i).

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

a(n+1) = A000073(n)*A000073(n+1) + ( (A000073(n+1) - A000073(n-1))^2 - 1 )/4 for n>0 [Jakubczyk]. - R. J. Mathar, Dec 19 2013