OFFSET
1,1
LINKS
Ivan Panchenko, Table of n, a(n) for n = 1..1000
Mohammad K. Azarian, The Value of a Series of Reciprocal Fibonacci Numbers, Problem B-1133, Fibonacci Quarterly, Vol. 51, No. 3, August 2013, p. 275; Solution published in Vol. 52, No. 3, August 2014, pp. 277-278.
FORMULA
10*(3 - sqrt(5)) = 30 - 10*sqrt(5) = (5 - sqrt(5))^2 = 20/phi^2.
2 * Sum_{i > 1} (-1)^i/(F(i)F(i + 1)) = 3 - sqrt(5), where F(i) is the i-th Fibonacci number. This formula comes from John D. Watson, Jr.'s solution to Azarian's Problem B-1133 in the Fibonacci Quarterly. Azarian originally posed the problem as an infinite alternating sum explicitly written out for the first dozen terms or so. See the Azarian links above. - Alonso del Arte, Aug 25 2016
EXAMPLE
20/phi^2 = 7.6393202250021030359082633...
3 - sqrt(5) = 0.76393202250021030359082633... (with offset 0).
MATHEMATICA
First@ RealDigits[N[5*(Sqrt[5] - 1)^2, 111]] (* Michael De Vlieger, Feb 25 2015 *)
PROG
(PARI) 5*(sqrt(5)-1)^2 \\ Charles R Greathouse IV, Aug 31 2013
(Magma) 5*(Sqrt(5)-1)^2; // Vincenzo Librandi, Feb 24 2015
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
Joost Gielen, Aug 30 2013
EXTENSIONS
Extended by Charles R Greathouse IV, Aug 31 2013
STATUS
approved