[seqfan] Percentage of digits used so far

[Eric Angelini]

Hello SeqFans,
a recent post by Lee Sallows (link at
the end of this msg) gave me the idea of P;

P=100,50,66,25,60,16,28,12,22,30,36,
16,46,14,6,6,17,22,26,5,23,9,21,16,24,
3,25,17,6,23,29,18,3,...

The p(n)th integer of P is the fraction A/B (in percents, decimals erased) with:
A = the number of times we have
used the (n)th digit of P so far;
B = the overall quantity of digits used
so far to write P.

Example:
100 means that we have used one "1"
to write 1 digit so far --> 1/1 is 100%
50 means that we have used one "0"
to write 2 digits so far --> 1/2 is 50%
66 means that we have used two "0"
to write 3 digits so far --> 2/3 is 66%
25 means that we have used one "5"
to write 4 digits so far --> 1/4 is 25%
60 means that we have used three "0"
to write 5 digits so far --> 3/5 is 60%
16 means that we have used one "6"
to write 6 digits so far --> 1/6 is 16%
28 means that we have used two "6"
to write 7 digits so far --> 2/7 is 28%
12 means that we have used one "2"
to write 8 digits so far --> 1/8 is 12%
22 means that we have used two "5"
to write 9 digits so far --> 2/9 is 22%
30 means that we have used three "6"
to write 10 digits so far --> 3/10 is 30%
...

Does P enter into some kind of loop
at some point?
Best,
É.

Lee's link:
http://www.futilitycloset.com/2015/11/16/a-new-pangram/

Catapulté de mon aPhone