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A105509 Numbers m such that 9 is the leading digit of the m-th Fibonacci number in decimal representation.

%I A105509 #22 Feb 08 2022 08:06:04
%S A105509 16,35,59,83,102,126,150,169,193,212,236,260,279,303,327,346,370,394,
%T A105509 413,437,461,480,504,528,547,571,595,614,638,657,681,705,724,748,772,
%U A105509 791,815,839,858,882,906,925,949,973,992,1016,1040,1059,1083,1102,1107
%N A105509 Numbers m such that 9 is the leading digit of the m-th Fibonacci number in decimal representation.
%C A105509 A008963(a(n)) = 9; A105519(a(n)) = A105519(a(n) - 1) + 1.
%C A105509 Comment from _Jonathan Vos Post_, Dec 23 2006: Peterson says: "Calculate 100/89 = 1.1235955056... This fraction generates the first five Fibonacci numbers before blurring into other digits. ... 10000/9899 = 1.0102030508132134559046368... generates the first 10 Fibonacci numbers (using two digits per number). 1000000/998999 generates the first 15 Fibonacci numbers (using three digits per number). ... in successive fractions, two 0's are appended to the numerator and a 9 to the beginning and end of the denominator...."
%H A105509 M. Bicknell-Johnson, <a href="http://www.maa.org/programs/faculty-and-departments/classroom-capsules-and-notes/a-generalized-magic-trick-from-fibonacci-designer-decimals">A generalized magic trick from Fibonacci: Designer decimals</a>, College Mathematics Journal 35(March):125-126, 2004.
%H A105509 O-Y. Chan and J. Smoak, <a href="http://oyeat.com/papers/Decimal5.pdf">More designer decimals: The integers and their geometric extensions</a> College Mathematics Journal 37(November):355-363, 2006.
%H A105509 Ivars Peterson, <a href="http://www.sciencenews.org/articles/20061104/mathtrek.asp">Designer Decimals</a>, Science News, Week of Nov 04 2006; Vol. 170, No. 19.
%H A105509 J. Smoak, and T.J. Osler, <a href="http://www.rowan.edu/math/osler/FibonoacciFractionsFinalVersionCMJ.pdf">A magic trick from Fibonacci</a>. College Mathematics Journal, 34 (2003):58-60.
%F A105509 m such that d(m+5)-d(m) = 2 for d(m) = floor(1 + log_10(F(m))) and F(m) = m-th Fibonacci number = A000045(m). - _Jonathan Vos Post_, Dec 23 2006
%F A105509 a(n) ~ k*n by the equidistribution theorem, where k = 1/(1 - log(9)/log(10)) = 21.8543.... - _Charles R Greathouse IV_, Oct 07 2016
%e A105509 a(10)=21: A008963(212) = A000030(A000045(212)) =
%e A105509 A000030(90343046356137747723758225621187571439538669) = 9.
%t A105509 Select[Range@ 1200, First@ IntegerDigits@ Fibonacci@ # == 9 &] (* _Michael De Vlieger_, Aug 21 2016 *)
%o A105509 (PARI) is(n)=digits(fibonacci(n))[1]==9 \\ _Charles R Greathouse IV_, Oct 07 2016
%Y A105509 Cf. A000030, A000045, A072711, A105501, A105502, A105503, A105504, A105505, A105506, A105507, A105508.
%K A105509 nonn,base
%O A105509 1,1
%A A105509 _Reinhard Zumkeller_, Apr 11 2005