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a(n) = number m such that the golden mean phi's m-th convergent F(m+1)/F(m) matches its actual value to n decimal places (here F(m) is the m-th Fibonacci number A000045(m)).
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%I #16 Aug 21 2017 17:33:15

%S 4,7,10,12,14,17,19,21,23,26,29,31,33,36,37,40,43,46,48,50,52,55,57,

%T 59,62,64,66,69,71,74,76,79,80,84,85,89,90,92,96,98,101,102,105,107,

%U 109,112,115,117,119,122,125,127,128,132,134,136,138,141,143,145,148

%N a(n) = number m such that the golden mean phi's m-th convergent F(m+1)/F(m) matches its actual value to n decimal places (here F(m) is the m-th Fibonacci number A000045(m)).

%H David Consiglio, Jr., <a href="/A104036/b104036.txt">Table of n, a(n) for n = 1..1000</a>

%H David Consiglio, Jr., <a href="/A104036/a104036.py.txt">Python program</a>

%H F. Richman, <a href="http://math.fau.edu/Richman/contfrac.htm">Continued fractions</a>

%e a(1) = 4 because F(5) / F(4) = 5/3 = 1.666... which is the lowest Fibonacci ratio that matches Phi (1.618...) to 1 decimal place.

%e a(2) = 7 because F(7) / F(6) = 21 / 13 = 1.615... which is the lowest Fibonacci ratio that matches Phi to 2 decimal places.

%Y Cf. A001622.

%K nonn,base

%O 1,1

%A _Lekraj Beedassy_, Mar 31 2005

%E a(7)-a(8) corrected and extended by _David Consiglio, Jr._, Oct 13 2015