|
|
A110391
|
|
a(n) = L(3*n)/L(n), where L(n) = Lucas number.
|
|
9
|
|
|
1, 4, 6, 19, 46, 124, 321, 844, 2206, 5779, 15126, 39604, 103681, 271444, 710646, 1860499, 4870846, 12752044, 33385281, 87403804, 228826126, 599074579, 1568397606, 4106118244, 10749957121, 28143753124, 73681302246, 192900153619, 505019158606, 1322157322204
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,2
|
|
COMMENTS
|
Subsidiary sequences: a(n) = L((2k+1)*n)/L(n) for k = 2,3, etc. This is the sequence for k = 1.
|
|
LINKS
|
|
|
FORMULA
|
a(n) = +2*a(n-1) +2*a(n-2) -a(n-3).
G.f.: ( 1+2*x-4*x^2 ) / ( (1+x)*(x^2-3*x+1) ). (End)
Exp( Sum_{n >= 1} a(n)*t^n/n ) = 1 + 4*t + 11*t^2 + 29*t^3 + ... is the o.g.f. for A002878. This is the case x = 1 of the general result exp( Sum_{n >= 1} L(3*n,x)/L(n,x)*t^n/n ) = Sum_{n >= 0} L(2*n + 1,x)*t^n, where L(n,x) is the n-th Lucas polynomial of A114525. - Peter Bala, Mar 18 2015
a(n) = 2^(-n)*(-(-2)^n+(3-sqrt(5))^n+(3+sqrt(5))^n). - Colin Barker, Jun 03 2016
|
|
EXAMPLE
|
a(1) = L(3)/L(1) = 4/1 = 4.
|
|
MAPLE
|
with(combinat): L:=n->fibonacci(n+2)-fibonacci(n-2): seq(L(3*n)/L(n), n=0..30); # Emeric Deutsch, Jul 31 2005
|
|
MATHEMATICA
|
LinearRecurrence[{2, 2, -1}, {1, 4, 6}, 40] (* Harvey P. Dale, Aug 20 2020 *)
|
|
PROG
|
(PARI) Vec((1+2*x-4*x^2)/((1+x)*(x^2-3*x+1)) + O(x^30)) \\ Colin Barker, Jun 03 2016
(Magma) [Lucas(3*n)/Lucas(n): n in [0..30]]; // G. C. Greubel, Dec 17 2017
(PARI) for(n=0, 30, print1((fibonacci(3*n+1) + fibonacci(3*n-1))/( fibonacci(n+1) + fibonacci(n-1)), ", ")) \\ G. C. Greubel, Dec 17 2017
|
|
CROSSREFS
|
|
|
KEYWORD
|
easy,nonn
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|