Eric Angelini wrote:
Hello Math-fun fans,
I've just submitted this to Sloane's OEIS:
1 2 3 4 5 6 7 8 9 11 13 15 17 19 22 31 33 35 37
What about 0? <snip>
"Jump-my-digits" numbers.
Take any integer of the sequence and repeat it as many times as you wish -- like this (for 258): 258,258,258,258,258,258,258,...
Choose now any digit of 258, "2", for instance, and jump over the next 2 digits: you'll land on another "2". The same can be done with "5" and with "8": jumping respectively over 5 and 8 digits will see you land on another "5" or another "8".
If a_{0 mod 3} = 2, a_{1 mod 3}, and a_{2 mod 3}=8, this doesn't amount to a_{i+a_i}=a_i, which is what I expected on the first reading. Your definition gives a_{i+a_i+1}=a_i. If you use the definition I expected, then 20 can be included, since there's a period 2 signal and 0 goes back to itself. For that sequence, I get 0,1,2,3,4,5,6,7,8,9,11,20,22,24,26,28,33,...
What could be the smallest such number containing all 10 different digits? (0->9) If it doesn't exist, the smallest one containing 9 different digits, etc. Best, É.
Well, with your definition, 0 can't show up except in 0. In mine, 1 can't show up except in repdigits of 1. In mine, 1<gcd(d,d') for all pairs of digits d,d' (where gcd(x,0)=oo) and their indices have to take different values modulo their gcd. Also, 1<gcd(d,period) unless period=1. That means 02428242 is my winner. In yours, 1<gcd(d+1,d'+1) and 1<gcd(d+1,period) unless period=1. So you can have at most three different digits. That means 258 is your winner. Unless I've made a mistake, you can't get a number with four different digits. (Unless, of course, you switch to a different base.) -- Mike Stay staym@clear.net.nz http://www.cs.auckland.ac.nz/~msta039