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A341216
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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.
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0
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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, 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, 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
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OFFSET
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1,3
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COMMENTS
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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).
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.
That makes the bottom triangle of the array unchanging, so it is omitted.
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LINKS
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FORMULA
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T(1,k) = 1, for k > 1;
T(n,k) = T(n-1,k) if k mod n > T(n-1,k) or k mod n = 0;
T(n,k) = T(n-1,k) + 1 otherwise.
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EXAMPLE
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n\k 2 3 4 5 6 7 8 9 10 11 12 13
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1 1 1 1 1 1 1 1 1 1 1 1 1
2 2 1 2 1 2 1 2 1 2 1 2
3 2 3 1 3 1 2 2 3 1 3
5 4 1 4 1 3 3 4 1 4
6 2 5 1 3 3 5 1 5
7 6 1 4 3 6 1 6
8 2 5 4 7 1 7
9 6 5 8 1 8
10 6 9 1 9
11 10 1 10
12 2 11
13 12
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CROSSREFS
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The last entry in each column is A128982.
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KEYWORD
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AUTHOR
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STATUS
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approved
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