A228609 Partial sums of the cubes of the tribonacci sequence A000073.
0, 1, 2, 10, 74, 417, 2614, 16438, 101622, 633063, 3941012, 24511836, 152535900, 949133883, 5905611508, 36746590964, 228646935796, 1422699232325, 8852413871022, 55082039340022, 342734883853750, 2132586518002125
Offset: 0
References
- R. Schumacher, Explicit formulas for sums involving the squares of the first n Tribonacci numbers, Fib. Q., 58:3 (2020), 194-202.
Links
- Michael De Vlieger, Table of n, a(n) for n = 0..1261
- H. Ohtsuka, Advanced Problems and Solutions, Fib. Quart. 51 (2) (2013) 186, H-736.
- Index entries for linear recurrences with constant coefficients, signature (5,5,25,-58,26,-42,54,-13,1,-3,1).
Programs
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Mathematica
CoefficientList[Series[x (-1 + 3 x + 11 x^3 - 5 x^4 + x^5 - 3 x^6 + x^7 + 5 x^2)/((x^3 - 5 x^2 + 7 x - 1) (x^6 + 4 x^5 + 11 x^4 + 12 x^3 + 11 x^2 + 4 x + 1) (x - 1)^2), {x, 0, 21}], x] (* Michael De Vlieger, Jan 12 2022 *) Accumulate[LinearRecurrence[{1,1,1},{0,1,1},30]^3] (* or *) LinearRecurrence[ {5,5,25,-58,26,-42,54,-13,1,-3,1},{0,1,2,10,74,417,2614,16438,101622,633063,3941012},30] (* Harvey P. Dale, Sep 11 2022 *)
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PARI
T(n)=([0, 1, 0; 0, 0, 1; 1, 1, 1]^n)[1, 3]; \\ A000073 a(n) = sum(k=1, n, T(k)^3); \\ Michel Marcus, Jan 12 2022
Formula
a(n) = a(n-1) + (A000073(n))^3.
G.f.: x*(-1+3*x+11*x^3-5*x^4+x^5-3*x^6+x^7+5*x^2) / ( (x^3-5*x^2+7*x-1) *(x^6+4*x^5+11*x^4+12*x^3+11*x^2+4*x+1) *(x-1)^2 )