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
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
Programs
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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
Comments