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A228609
Partial sums of the cubes of the tribonacci sequence A000073.
1
0, 1, 2, 10, 74, 417, 2614, 16438, 101622, 633063, 3941012, 24511836, 152535900, 949133883, 5905611508, 36746590964, 228646935796, 1422699232325, 8852413871022, 55082039340022, 342734883853750, 2132586518002125
OFFSET
0,3
REFERENCES
R. Schumacher, Explicit formulas for sums involving the squares of the first n Tribonacci numbers, Fib. Q., 58:3 (2020), 194-202.
LINKS
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).
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 )
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 *)
PROG
(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
CROSSREFS
Sequence in context: A275550 A243250 A088189 * A001395 A047853 A151387
KEYWORD
easy,nonn
AUTHOR
R. J. Mathar, Dec 18 2013
STATUS
approved