A260707 a(n) is the percentage of digits equal to the n-th digit, up to that digit.

100, 50, 66, 25, 60, 16, 28, 12, 22, 30, 36, 16, 30, 14, 6, 18, 17, 22, 26, 5, 23, 9, 21, 16, 24, 11, 22, 17, 3, 23, 19, 6, 21, 2, 17, 19, 21, 21, 7, 22, 9, 2, 23, 18, 20, 19, 23, 4, 20, 22, 23, 25, 22, 3, 9, 25, 10, 22, 3, 16, 24, 22, 25, 23, 4, 24, 4, 25, 24, 25, 25, 5, 26, 27, 5, 27, 28, 8, 24, 3, 28, 8, 24, 5, 28, 9, 3, 28, 8, 28, 29, 30, 9, 30, 4, 31

Offset

1,1

Comments

The percentages are rounded down to give an integer. For example, 2/3 becomes 66%. - N. J. A. Sloane, Nov 17 2015

Eric Angelini asks "Does the sequence enter some kind of loop at some point?"

Comment from Jack Brennen, Nov 18 2015: I wrote a Python program to explore the sequence and computed over 10^7 terms without the sequence entering a loop. It exhibits some short term looping, but the loops don't last long.

Links

Table of n, a(n) for n=1..96.

E. Angelini, Percentage of digits used so far, SeqFan list, Nov. 17, 2015

Futility Closet, A New Pangram, Nov. 16, 2015

Example

The third digit in the string of terms 1005066025... is 0. When we reach that digit we have seen one 1 and two 0's, so 0 has appeared 2/3 = 66% of the time, and therefore a(3)=66. - N. J. A. Sloane, Nov 17 2015

Mathematica

P = {100}; Do[AppendTo[P, Floor[100 Count[Take[#, n], #[[n]]]/n &@ Flatten@ Map[IntegerDigits, P]]], {n, 2, 120}]; P (* Michael De Vlieger, Nov 17 2015 *)

Prog

(PARI) {c=vector(10); c[2]++; print1(100); a=[1, 0, 0]; for(n=2, 99, print1(", "p=100*c[a[n]+1]++\n); a=concat(a, digits(p)))}

Crossrefs

Sequence in context: A008903 A033420 A234322 * A180102 A242615 A090292

Adjacent sequences: A260704 A260705 A260706 * A260708 A260709 A260710

Keyword

nonn,base

Author

M. F. Hasler, Nov 17 2015

Status

approved