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

6, 11, 14, 16, 17, 20, 23, 24, 34, 38, 39, 42, 48, 49, 58, 66, 74, 77, 83, 87, 102, 104, 110, 120, 136, 145, 156, 158, 172, 177, 178, 183, 195, 201, 233, 235, 250, 256, 260, 273, 277, 282, 288, 293, 306, 319, 325, 329, 373, 389, 399, 415, 458, 512, 589, 609, 695, 862, 989, 1063

Offset: 1

Author

Jon E. Schoenfield, Nov 28 2015

Keywords

• nonn
• base

Comments

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).

a(60) = 1063 is almost certainly the last term in the sequence.

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.

Examples

			The first several terms and their corresponding Fibonacci numbers, number of digits D, digit sum S, and average digit values are as follows:
.
   k | Fibonacci(k) |  D |  S | avg. digit value
  ---+--------------+----+----+-----------------
   6 |            8 |  1 |  8 | 8.00000000000000
  11 |           89 |  2 | 17 | 8.50000000000000
  14 |          377 |  3 | 17 | 5.66666666666667
  16 |          987 |  3 | 24 | 8.00000000000000
  17 |         1597 |  4 | 22 | 5.50000000000000
  20 |         6765 |  4 | 24 | 6.00000000000000
  23 |        28657 |  5 | 28 | 5.60000000000000
  24 |        46368 |  5 | 27 | 5.40000000000000
  34 |      5702887 |  7 | 37 | 5.28571428571429
  38 |     39088169 |  8 | 44 | 5.50000000000000
  39 |     63245986 |  8 | 43 | 5.37500000000000
  42 |    267914296 |  9 | 46 | 5.11111111111111
  48 |   4807526976 | 10 | 54 | 5.40000000000000
  49 |   7778742049 | 10 | 55 | 5.50000000000000
  58 | 591286729879 | 12 | 73 | 6.08333333333333
.
(Fibonacci(58) is almost certainly the last Fibonacci number whose average digit exceeds 98/17 = 5.764705...)
		

Crossrefs

Cf. A000045.

Programs

Mathematica

Select[Range@ 1200, Mean@ IntegerDigits@ Fibonacci@ # > 5 &] (* Michael De Vlieger, Nov 28 2015 *)