OFFSET
0,3
LINKS
Robert Israel, Table of n, a(n) for n = 0..2381
Florian Luca (editor), Problem H-755, The Fibonacci Quarterly, Volume 54, Number 1 (February 2016), p. 92 (see solution by Adnan Ali).
Index entries for linear recurrences with constant coefficients, signature (6,-9,-6,20,-6,-9,6,-1).
FORMULA
a(n) = n*Sum_{k=1..n} Lucas(k)^2 (by Adnan Ali, see References).
From Colin Barker, Sep 15 2017: (Start)
G.f.: x*(1 + 14*x - 33*x^2 + 18*x^3 - 18*x^4 + 14*x^5 - 4*x^6)/((1 - x)^2*(1 + x)^2*(1 - 3*x + x^2)^2).
a(n) = 6*a(n-1) - 9*a(n-2) - 6*a(n-3) + 20*a(n-4) - 6*a(n-5) - 9*a(n-6) + 6*a(n-7) - a(n-8) for n > 7. (End)
a(n) = n*(Lucas(2*n+1)-1) for n even, otherwise a(n) = n*(Lucas(2*n+1)-3). - Bruno Berselli, Sep 15 2017
MAPLE
lucas:= gfun:-rectoproc({a(n)=a(n-1)+a(n-2), a(0)=2, a(1)=1}, a(n), remember):
f:= n -> n*(lucas(n)*lucas(n+1)-2):
map(f, [$0..100]); # Robert Israel, Sep 17 2017
MATHEMATICA
Table[n (LucasL[n] LucasL[n+1] - 2), {n, 0, 40}]
PROG
(Magma) [n*(Lucas(n)*Lucas(n+1)-2): n in [0..35]];
(PARI) concat(0, Vec(x*(1 + 14*x - 33*x^2 + 18*x^3 - 18*x^4 + 14*x^5 - 4*x^6) / ((1 - x)^2*(1 + x)^2*(1 - 3*x + x^2)^2) + O(x^30))) \\ Colin Barker, Sep 15 2017
(PARI) a000032(n) = fibonacci(n+1)+fibonacci(n-1); a(n) = n*(a000032(n)*a000032(n+1)-2); \\ Altug Alkan, Sep 15 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Sep 15 2017
STATUS
approved