cp's OEIS Frontend

A345117 a(n) is the index (in Z/nZ) of the first already visited element in the process of moving around Z/nZ, starting at 0 with stride 1 and increasing stride by 1 after each step.

%I A345117 #20 Jun 11 2024 15:53:06
%S A345117 0,1,0,2,1,0,3,4,1,0,10,3,2,1,0,8,11,3,17,1,0,6,9,21,3,10,1,0,4,6,12,
%T A345117 16,3,21,1,0,5,28,6,15,26,3,38,1,0,45,18,9,6,28,15,3,7,1,0,10,21,20,
%U A345117 52,6,8,29,3,32,1,0,59,10,36,21,27,6,46,62,3,15,1
%N A345117 a(n) is the index (in Z/nZ) of the first already visited element in the process of moving around Z/nZ, starting at 0 with stride 1 and increasing stride by 1 after each step.
%H A345117 Sean A. Irvine, <a href="https://github.com/archmageirvine/joeis/blob/master/src/irvine/oeis/a345/A345117.java">Java program</a> (github)
%F A345117 a(n) = A000217(A212652(n)) mod n. - _Rémy Sigrist_, Jun 08 2021
%e A345117 For n = 3 we start with Z/3Z represented as [0,1,2]. In the first step we mark the zero to obtain [0*,1,2] and move one step (to 1). Then we mark the 1 to obtain [0*,1*,2] and move two steps (to 0*). We have landed on a number already visited, so the process ends here, and as we have landed on 0 last, a(3) = 0.
%e A345117 For n = 4 we start with [0,1,2,3]. After the first step we get [0*,1,2,3] and we land at 1. After the second step we have [0*,1*,2,3] and we have landed at 3. In the penultimate step we mark the 3 to get [0*,1*,2,3*] and move 3 steps (to 2). We mark the 2 and move 4 steps to the 2*, which we have already visited. Therefore, a(4) = 2.
%e A345117 For n = 5 the list of steps is as follows: [0,1,2,3,4] -> [0*,1,2,3,4] -> [0*,1*,2,3,4] -> [0*,1*,2,3*,4] -> we land on 1 again, therefore a(5) = 1.
%e A345117 For n = 7 the list of steps is as follows: [0,1,2,3,4,5,6] -> [0*,1,2,3,4,5,6] -> [0*,1*,2,3,4,5,6] -> [0*,1*,2,3*,4,5,6] -> [0*,1*,2,3*,4,5,6*] -> we land on 3 again, therefore a(7) = 3.
%e A345117 Note: the '*' after a number means that this number was already visited.
%o A345117 (Python)
%o A345117 def a(n):
%o A345117     row = ['x' for i in range(n)]
%o A345117     free = True
%o A345117     count = index = 0
%o A345117     while(free):
%o A345117         row[index] = count
%o A345117         count += 1
%o A345117         index = (index + count) % n
%o A345117         if row[index] != 'x':
%o A345117             free = False
%o A345117     return index
%Y A345117 Cf. A000217, A212652.
%K A345117 nonn,easy
%O A345117 1,4
%A A345117 _Sophie Weber_, Jun 08 2021