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A264935 Numbers k such that the average of the digits of the k-th Fibonacci number is greater than 5.

%I A264935 #7 Nov 28 2015 22:08:52
%S A264935 6,11,14,16,17,20,23,24,34,38,39,42,48,49,58,66,74,77,83,87,102,104,
%T A264935 110,120,136,145,156,158,172,177,178,183,195,201,233,235,250,256,260,
%U A264935 273,277,282,288,293,306,319,325,329,373,389,399,415,458,512,589,609,695,862,989,1063
%N A264935 Numbers k such that the average of the digits of the k-th Fibonacci number is greater than 5.
%C A264935 As k increases, it appears that the average of the digits of the k-th Fibonacci number approaches 9/2 (the same as would be expected with increasingly longer strings of random decimal digits).
%C A264935 a(60) = 1063 is almost certainly the last term in the sequence.
%C A264935 It seems nearly certain that there are only 11 Fibonacci numbers whose average digit is exactly 5; their indices are k = 5, 10, 35, 78, 97, 138, 184, 189, 300, 437, and 550.
%e A264935 The first several terms and their corresponding Fibonacci numbers, number of digits D, digit sum S, and average digit values are as follows:
%e A264935 .
%e A264935    k | Fibonacci(k) |  D |  S | avg. digit value
%e A264935   ---+--------------+----+----+-----------------
%e A264935    6 |            8 |  1 |  8 | 8.00000000000000
%e A264935   11 |           89 |  2 | 17 | 8.50000000000000
%e A264935   14 |          377 |  3 | 17 | 5.66666666666667
%e A264935   16 |          987 |  3 | 24 | 8.00000000000000
%e A264935   17 |         1597 |  4 | 22 | 5.50000000000000
%e A264935   20 |         6765 |  4 | 24 | 6.00000000000000
%e A264935   23 |        28657 |  5 | 28 | 5.60000000000000
%e A264935   24 |        46368 |  5 | 27 | 5.40000000000000
%e A264935   34 |      5702887 |  7 | 37 | 5.28571428571429
%e A264935   38 |     39088169 |  8 | 44 | 5.50000000000000
%e A264935   39 |     63245986 |  8 | 43 | 5.37500000000000
%e A264935   42 |    267914296 |  9 | 46 | 5.11111111111111
%e A264935   48 |   4807526976 | 10 | 54 | 5.40000000000000
%e A264935   49 |   7778742049 | 10 | 55 | 5.50000000000000
%e A264935   58 | 591286729879 | 12 | 73 | 6.08333333333333
%e A264935 .
%e A264935 (Fibonacci(58) is almost certainly the last Fibonacci number whose average digit exceeds 98/17 = 5.764705...)
%t A264935 Select[Range@ 1200, Mean@ IntegerDigits@ Fibonacci@ # > 5 &] (* _Michael De Vlieger_, Nov 28 2015 *)
%Y A264935 Cf. A000045.
%K A264935 nonn,base
%O A264935 1,1
%A A264935 _Jon E. Schoenfield_, Nov 28 2015