OFFSET
0,4
COMMENTS
Divisibility sequence; that is, if n divides m, then a(n) divides a(m).
In general, cubing the terms of a Horadam sequence with signature (c,d) will result in a fourth-order recurrence with signature (c^3+2*c*d, c^4*d+3*(c*d)^2+2*d^3, -(c*d)^3-2*c*d^4, -d^6). - Gary Detlefs, Nov 12 2021
a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using (1/3,2/3)-fences and third-squares (1/3 X 1 pieces, always placed so that the shorter sides are horizontal). A (w,g)-fence is a tile composed of two w X 1 pieces separated by a gap of width g. a(n+1) also equals the number of tilings of an n-board using (1/6,1/3)-fences and (1/6,5/6)-fences. - Michael A. Allen, Jan 11 2022
REFERENCES
D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, 1969, Vol. 1, p. 85, (exercise 1.2.8. Nr. 30) and p. 492 (solution).
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..173
Feryal Alayont and Evan Henning, Edge Covers of Caterpillars, Cycles with Pendants, and Spider Graphs, J. Int. Seq. (2023) Vol. 26, Art. 23.9.4.
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 38, 2012, pp. 1871-1876.
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums II, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 42, 2012, pp. 2053-2059.
A. Brousseau, A sequence of power formulas, Fib. Quart., 6 (1968), 81-83.
Andrej Dujella, A bijective proof of Riordan's theorem on powers of Fibonacci numbers, Discrete Math. 199 (1999), no. 1-3, 217--220. MR1675924 (99k:05016).
Kenneth Edwards and Michael A. Allen, A new combinatorial interpretation of the Fibonacci numbers cubed, Fib. Q. 58:5 (2020) 128-134.
D. Foata and G.-N. Han, Nombres de Fibonacci et polynomes orthogonaux
Mariana Nagy, Simon R. Cowell and Valeriu Beiu, Survey of Cubic Fibonacci Identities - When Cuboids Carry Weight, arXiv:1902.05944 [math.HO], 2019.
Hilary I. Okagbue, Muminu O. Adamu, Sheila A. Bishop and Abiodun A. Opanuga, Digit and Iterative Digit Sum of Fibonacci numbers, their identities and powers, International Journal of Applied Engineering Research ISSN 0973-4562 Volume 11, Number 6 (2016) pp 4623-4627.
J. Riordan, Generating functions for powers of Fibonacci numbers, Duke. Math. J. 29 (1962) 5-12.
E. L. Roettger and H. C. Williams, Appearance of Primes in Fourth-Order Odd Divisibility Sequences, J. Int. Seq., Vol. 24 (2021), Article 21.7.5.
H. C. Williams and R. K. Guy, Odd and even linear divisibility sequences of order 4, INTEGERS, 2015, #A33.
Index entries for linear recurrences with constant coefficients, signature (3,6,-3,-1).
FORMULA
a(n) = A000045(n)^3.
G.f.: x*p(3, x)/q(3, x) with p(3, x) = Sum_{m=0..2} A056588(2, m)*x^m = 1 -2*x -x^2 and q(3, x) = Sum_{m=0..4} A055870(4, m)*x^m = 1 -3*x -6*x^2 +3*x^3 +x^4 = (1+x-x^2)*(1-4*x-x^2) (factorization deduced from Riordan result).
Recursion (cf. Knuth's exercise): 1*a(n) -3*a(n-1) -6*a(n-2) +3*a(n-3) +1*a(n-4) = 0, n >= 4, a(0)=0, a(1)=a(2)=1, a(3)=2^3. See 5th row of signed Fibonomial triangle for coefficients: A055870(4, m), m=0..4
a(n) = (Fibonacci(3n) - 3(-1)^n*Fibonacci(n))/5. - Ralf Stephan, May 14 2004
a(n) and a(n+1) are found as rightmost and leftmost terms (respectively) in M^n * [1 0 0 0] where M = the 4 X 4 upper triangular Pascal's triangle matrix [1 3 3 1 / 1 2 1 0 / 1 1 0 0 / 1 0 0 0]. E.g., Ma(4) = 27, a(5) = 125. M^4 * [1 0 0 0] = [125 75 45 27]; where 75 = A066259(4) and 45 = A066258(3). The characteristic polynomial of M = x^4 - 3x^3 - 6x^2 + 3x + 1. a(n)/a(n-1) of the sequence and companions tend to 2+sqrt(5) = 4.2360679..., an eigenvalue of M and a root of the polynomial. - Gary W. Adamson, Oct 31 2004
From R. J. Mathar, Oct 16 2006: (Start)
G.f.: x*(1-2*x-x^2)/((1+x-x^2)*(1-4*x-x^2)). - Colin Barker, Feb 28 2012
a(n) = F(n-2)*F(n+1)^2 + F(n-1)*(-1)^n. - J. M. Bergot, Mar 17 2016
a(n) = ((-3*(1/2*(-1-sqrt(5)))^n-(2-sqrt(5))^n+3*(1/2*(-1+sqrt(5)))^n+(2+sqrt(5))^n)) / (5*sqrt(5)). - Colin Barker, Jun 04 2016
a(n) = F(n-1)*F(n)*F(n+1) + F(n)*(-1)^(n-1). - Tony Foster III, Apr 11 2018
5*a(n) = L(2*n-1)*F(n+2) - L(2*n+1)*F(n-2) - 7*(-1)^n*F(n), where L(n) = A000032(n). - Peter Bala, Nov 12 2019
F(n+1)*F(n)*F(n-1) = 2*Sum_{j=1..n-1} P(j)*a(n-j) for n>0, where Pell number P(n) = A000129(n). - Michael A. Allen, Jan 11 2022
EXAMPLE
a(4) = 27 because the fourth Fibonacci number is 3 and 3^3 = 27.
a(5) = 125 because the fifth Fibonacci number is 5 and 5^3 = 125.
MATHEMATICA
Table[Fibonacci[n]^3, {n, 0, 30}] (* Vladimir Joseph Stephan Orlovsky, Jul 21 2008 *)
PROG
(Magma) [Fibonacci(n)^3: n in [0..30]]; // Vincenzo Librandi, Jun 04 2011
(PARI) a(n)=fibonacci(n)^3 \\ Charles R Greathouse IV, Sep 24 2015
(PARI) concat(0, Vec(x*(1-2*x-x^2)/((1+x-x^2)*(1-4*x-x^2)) + O(x^30))) \\ Colin Barker, Jun 04 2016
(Sage) [fibonacci(n)^3 for n in (0..30)] # G. C. Greubel, Feb 20 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Jul 10 2000
STATUS
approved