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 A187799 Decimal expansion of 20/phi^2, where phi is the golden ratio. Also (with a different offset), decimal expansion of 3 - sqrt(5). 5
 7, 6, 3, 9, 3, 2, 0, 2, 2, 5, 0, 0, 2, 1, 0, 3, 0, 3, 5, 9, 0, 8, 2, 6, 3, 3, 1, 2, 6, 8, 7, 2, 3, 7, 6, 4, 5, 5, 9, 3, 8, 1, 6, 4, 0, 3, 8, 8, 4, 7, 4, 2, 7, 5, 7, 2, 9, 1, 0, 2, 7, 5, 4, 5, 8, 9, 4, 7, 9, 0, 7, 4, 3, 6, 2, 1, 9, 5, 1, 0, 0, 5, 8, 5, 5, 8, 5, 5, 9, 1, 6, 2, 1, 2, 1, 7, 7, 2, 5, 0, 3, 0, 4, 9, 1, 8, 2, 3, 8, 4, 9 (list; constant; graph; refs; listen; history; text; internal format)
 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 Cf. A187426, A187798, A094874. Sequence in context: A021571 A013675 A198878 * A288935 A132714 A230327 Adjacent sequences:  A187796 A187797 A187798 * A187800 A187801 A187802 KEYWORD nonn,cons AUTHOR Joost Gielen, Aug 30 2013 EXTENSIONS Extended by Charles R Greathouse IV, Aug 31 2013 STATUS approved

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Last modified September 21 07:08 EDT 2019. Contains 327253 sequences. (Running on oeis4.)