A381623 Triangle read by rows: T(n,k) is the number of the k-th eliminated person in the variation of the Josephus elimination process for n people, where the first person is eliminated, then two people are skipped, and then the process repeats.
1, 1, 2, 1, 2, 3, 1, 4, 2, 3, 1, 4, 3, 5, 2, 1, 4, 2, 6, 3, 5, 1, 4, 7, 5, 3, 6, 2, 1, 4, 7, 3, 8, 6, 2, 5, 1, 4, 7, 2, 6, 3, 9, 5, 8, 1, 4, 7, 10, 5, 9, 6, 3, 8, 2, 1, 4, 7, 10, 3, 8, 2, 9, 6, 11, 5, 1, 4, 7, 10, 2, 6, 11, 5, 12, 9, 3, 8, 1, 4, 7, 10, 13, 5, 9, 2, 8, 3, 12, 6, 11
Offset: 1
Examples
Consider 4 people in a circle. Initially, person number 1 is eliminated, and persons 2 and 3 are skipped. The remaining people are now in order 4, 2, 3. Then, person 4 is eliminated, and 2 and 3 are skipped. The remaining people are in order 2, 3. Now, person 2 is eliminated. Person 3 is eliminated last. Thus, the fourth row of the triangle is 1, 4, 2, 3. Triangle begins: 1; 1, 2; 1, 2, 3,; 1, 4, 2, 3; 1, 4, 3, 5, 2; 1, 4, 2, 6, 3, 5; 1, 4, 7, 5, 3, 6, 2; 1, 4, 7, 3, 8, 6, 2, 5; ...
Programs
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Python
def J(n,A): l=[] for i in range(n): l.append(i+1) index = 0 P=[] for i in range(n): index+=A[i] index=index%len(l) P.append(l[index]) l.pop(index) return P def invPerm(p): inv = [] for i in range(len(p)): inv.append(None) for i in range(len(p)): inv[p[i]-1]=i+1 return inv def DUU(n): return [0] + [2 for i in range(n)] seq = [] for i in range(1,20): seq += J(i, DUU(i)) print(", ".join([str(v) for v in seq])) -
Python
def row(n): i, J, out = 0, list(range(1, n+1)), [] while len(J) > 1: i = i%len(J) out.append(J.pop(i)) i = (i + 2)%len(J) return out + [J[0]] print([e for n in range(1, 14) for e in row(n)]) # Michael S. Branicky, Mar 27 2025
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