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A260707
a(n) is the percentage of digits equal to the n-th digit, up to that digit.
0
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
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
KEYWORD
nonn,base
AUTHOR
M. F. Hasler, Nov 17 2015
STATUS
approved