OFFSET
0,2
COMMENTS
Partial sums of A033888, i.e., a(n) = Sum_{k=0..n} Fibonacci(4*k). - Vladeta Jovovic, Jun 09 2002
From Paul Weisenhorn, May 17 2009: (Start)
a(n) is the solution of the 2 equations a(n)+1=A^2 and 5*a(n)+1=B^2
which are equivalent to the Pell equation (10*a(n)+3)^2-5*(A*B)^2=4.
(End)
Numbers a(n) such as a(n)+1 and 5*a(n)+1 are perfect squares. - Sture Sjöstedt, Nov 03 2011
REFERENCES
A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 29.
H. J. H. Tuenter, Fibonacci summation identities arising from Catalan's identity, Fib. Q., 60:4 (2022), 312-319.
LINKS
Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).
FORMULA
a(n) = -3/5 + (1/5*sqrt(5)+3/5)*(2*1/(7+3*sqrt(5)))^n/(7+3*sqrt(5)) + (1/5*sqrt(5)-3/5)*(-2*1/(-7+3*sqrt(5)))^n/(-7+3*sqrt(5)). Recurrence: a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3). G.f.: 3*x/(1-7*x+x^2)/(1-x). - Vladeta Jovovic, Jun 09 2002
a(n) = A081068(n) - 1.
a(n) is the next integer from ((3+sqrt(5))*((7+3*sqrt(5))/2)^(n-1)-6)/10. - Paul Weisenhorn, May 17 2009
a(n) = 7*a(n-1) - a(n-2) + 3, n>1. - Gary Detlefs, Dec 07 2010
a(n) = sum_{k=0..n} Fibonacci(4k). - Gary Detlefs, Dec 07 2010
a(n) = (Lucas(4n+2)-3)/5, where Lucas(n)= A000032(n). - Gary Detlefs, Dec 07 2010
a(n) = (1/5)*(Fibonacci(4n+4) - Fibonacci(4n)-3). - Gary Detlefs, Dec 08 2010
a(n) = 3*A092521(n). - R. J. Mathar, Nov 03 2011
a(0)=0, a(1)=3, a(2)=24, a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3). - Harvey P. Dale, Jul 25 2013
Sum_{n>=1} 1/a(n) = 2/(3 + sqrt(5)) = A094874 - 1. - Amiram Eldar, Oct 05 2020
MAPLE
fs4:=n->sum(fibonacci(4*k), k=0..n):seq(fs4(n), n=0..21); # Gary Detlefs, Dec 07 2010
MATHEMATICA
Table[Fibonacci[2 n]*Fibonacci[2 n + 2], {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, Jul 01 2011 *)
Accumulate[Fibonacci[4*Range[0, 30]]] (* or *) LinearRecurrence[{8, -8, 1}, {0, 3, 24}, 30] (* Harvey P. Dale, Jul 25 2013 *)
PROG
(Magma) [Fibonacci(2*n)*Fibonacci(2*n+2): n in [0..30]]; // Vincenzo Librandi, Apr 18 2011
(PARI) a(n)=fibonacci(2*n)*fibonacci(2*n+2) \\ Charles R Greathouse IV, Jul 02 2013
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
N. J. A. Sloane, Jun 09 2002
STATUS
approved