|
|
A014448
|
|
Even Lucas numbers: L(3n).
|
|
43
|
|
|
2, 4, 18, 76, 322, 1364, 5778, 24476, 103682, 439204, 1860498, 7881196, 33385282, 141422324, 599074578, 2537720636, 10749957122, 45537549124, 192900153618, 817138163596, 3461452808002, 14662949395604, 62113250390418
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,1
|
|
COMMENTS
|
|
|
LINKS
|
H. H. Ferns, Problem B-115, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 5, No. 2 (1967), p. 202; Identities for F_{kn} and L{kn}, Solution to Problem B-115 by Stanley Rabinowitz, ibid., Vol. 6, No. 1 (1968), pp. 92-93.
|
|
FORMULA
|
G.f.: (2-4*x)/(1-4*x-x^2).
a(n) = 4*a(n-1) +a(n-2) with n>1, a(0)=2, a(1)=4.
a(n) = (2+sqrt(5))^n + (2-sqrt(5))^n.
a(n) = Sum_{k=0..n} C(n,k)*Lucas(n+k). - Paul D. Hanna, Oct 19 2010
a(n) = Fibonacci(6*n)/Fibonacci(3*n), n>0. - Gary Detlefs, Dec 26 2010
a(n) = ( Fibonacci(3*n + 2*k) - F(3*n - 2*k) )/Fibonacci(2*k) for nonzero integer k.
a(n) = ( Fibonacci(3*n + 2*k + 1) + F(3*n - 2*k - 1) )/Fibonacci(2*k + 1) for arbitrary integer k. (End)
a(n) = [x^n] ( (1 + 4*x + sqrt(1 + 8*x + 20*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 23 2015
a(n) = L(n)*(L(n-1)*L(n+1) + 2*(-1)^n). - J. M. Bergot, Feb 05 2016
Sum_{n >= 1} 1/( a(n) + (-1)^(n+1)*20/a(n) ) = 3/16.
Sum_{n >= 1} (-1)^(n+1)/( a(n) + (-1)^(n+1)*20/a(n) ) = 1/16. (End)
a(n) = (15*Fibonacci(n)^2*Lucas(n) + Lucas(n)^3)/4 (Ferns, 1967). - Amiram Eldar, Feb 06 2022
|
|
EXAMPLE
|
|
|
MATHEMATICA
|
|
|
PROG
|
(PARI) polsym(x^2-4*x-1, 100)
(PARI) a(n)=sum(k=0, n, binomial(n, k)*(fibonacci(n+k-1)+fibonacci(n+k+1))) \\ Paul D. Hanna, Oct 19 2010
(Sage) [lucas_number2(n, 4, -1) for n in range(0, 23)] # Zerinvary Lajos, May 14 2009
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|