A075155 Cubes of Lucas numbers.
8, 1, 27, 64, 343, 1331, 5832, 24389, 103823, 438976, 1860867, 7880599, 33386248, 141420761, 599077107, 2537716544, 10749963743, 45537538411, 192900170952, 817138135549, 3461452853383, 14662949322176, 62113250509227, 263115950765039, 1114577054530568
Offset: 0
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1000
- Mohammad K. Azarian, Identities Involving Lucas or Fibonacci and Lucas Numbers as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 45, 2012, pp. 2221-2227.
- Index entries for linear recurrences with constant coefficients, signature (3,6,-3,-1).
Programs
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Magma
[ Lucas(n)^3 : n in [0..120]]; // Vincenzo Librandi, Apr 14 2011
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Mathematica
CoefficientList[Series[(8 - 23*x - 24*x^2 + x^3)/((x^2 + 4*x - 1)*(x^2 - x - 1)), {x,0,50}], x] (* or *) Table[LucasL[n]^3, {n,0,30}] (* or *) LinearRecurrence[{3,6,-3,-1}, {8, 1, 27, 64}, 30] (* G. C. Greubel, Dec 21 2017 *) -
PARI
a(n)=(fibonacci(n-1)+fibonacci(n+1))^3 \\ Charles R Greathouse IV, Feb 09 2016
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Python
from sympy import lucas def a(n): return lucas(n)**3 print([a(n) for n in range(25)]) # Michael S. Branicky, Aug 01 2021
Formula
a(n) = 3*(-1)^n*L(n) + L(3*n).
a(n) = (-1)^n*A075151(n).
a(n) = L(n)*C(n)^2, L(n) = Lucas numbers (A000032), C(n) = reflected Lucas numbers (comment to A061084).
a(n) = 3*a(n-1) + 6*a(n-2) - 3*a(n-3) - a(n-4), n>=4.
G.f.: ( 8-23*x-24*x^2+x^3 )/( (x^2+4*x-1)*(x^2-x-1) ).
a(n) = L(3*n) + (F(n+4) - F(n-4))*(-1)^n, n>3 and F(n)=A000045(n). - J. M. Bergot, Feb 09 2016
a(n) + Sum_{i=0..n+1} a(i) = 19/2 + (5/2)*L(3*n+2). - Greg Dresden, Feb 24 2025
Extensions
Simpler definition from Ralf Stephan, Nov 01 2004