

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
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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 B1133, Fibonacci Quarterly, Vol. 51, No. 3, August 2013, p. 275; Solution published in Vol. 52, No. 3, August 2014, pp. 277278.


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 ith Fibonacci number. This formula comes from John D. Watson, Jr.'s solution to Azarian's Problem B1133 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



