%I #18 Feb 27 2021 22:13:10

%S 1,1,2,1,1,2,1,2,3,4,1,1,1,1,2,1,2,3,4,5,6,1,1,1,1,1,1,2,1,2,2,3,3,4,

%T 5,6,1,1,2,3,3,3,4,5,6,1,2,3,4,5,6,7,8,9,10,1,1,1,1,1,1,1,1,1,1,2,1,2,

%U 3,4,5,6,7,8,9,10,11,12,1,1,1,1,1,1,1,1,1,1,1,1,2,1,2,2,2

%N Triangle read by columns T(n,k) k > n >= 1: Last survivor positions in a modified Josephus problem for n numbers, where after each deletion the counting starts over at the lowest existing number n, rather than continuing from the current position.

%C Arrange 1,2,3,...,n clockwise in a circle. Start the count at the lowest surviving value and delete the k-th value counting clockwise around the circle. Repeat this procedure until one number remains, which is T(n,k).

%C Note: In the complete n X k array with n >= 1 and k >= 1, T(n,k) = T(k-1,k) for all n >= k > 1 and T(n,1)=n.

%C That makes the bottom triangle of the array unchanging, so it is omitted.

%H <a href="/index/J#Josephus">Index entries for sequences related to the Josephus Problem</a>

%F T(1,k) = 1, for k > 1;

%F T(n,k) = T(n-1,k) if k mod n > T(n-1,k) or k mod n = 0;

%F T(n,k) = T(n-1,k) + 1 otherwise.

%e n\k 2 3 4 5 6 7 8 9 10 11 12 13

%e _______________________________________________________________

%e 1 1 1 1 1 1 1 1 1 1 1 1 1

%e 2 2 1 2 1 2 1 2 1 2 1 2

%e 3 2 3 1 3 1 2 2 3 1 3

%e 5 4 1 4 1 3 3 4 1 4

%e 6 2 5 1 3 3 5 1 5

%e 7 6 1 4 3 6 1 6

%e 8 2 5 4 7 1 7

%e 9 6 5 8 1 8

%e 10 6 9 1 9

%e 11 10 1 10

%e 12 2 11

%e 13 12

%Y The last entry in each column is A128982.

%K nonn,tabl

%O 1,3

%A _Gary Yane_, Feb 06 2021