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A382355 A version of the Josephus problem: a(n) is the surviving integer under the skip-eliminate-skip version of the elimination process.

%I A382355 #10 Apr 05 2025 23:25:37
%S A382355 1,1,1,4,3,6,3,6,9,3,6,9,12,1,4,7,10,13,16,19,1,4,7,10,13,16,19,22,25,
%T A382355 28,31,3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,1,4,7,10,13,16,19,22,
%U A382355 25,28,31,34,37,40,43,46,49,52,55,58,61,64,67,70,3,6
%N A382355 A version of the Josephus problem: a(n) is the surviving integer under the skip-eliminate-skip version of the elimination process.
%C A382355 This variation of the Josephus problem is related to under-down-under card dealing.
%e A382355 Consider 4 people in a circle in order 1,2,3,4. In the first round, person 1 is skipped, then person 2 is eliminated, then person 3 is skipped. Now people are in order 4,1,3. In the second round, person 4 is skipped, person 1 is eliminated, and person 3 is skipped. Now people are in order 4,3. In the third round person 3 is eliminated. Person 4 is freed. Thus, a(4) = 4.
%o A382355 (Python)
%o A382355 def J(n,A):
%o A382355     l=[]
%o A382355     for i in range(n):
%o A382355         l.append(i+1)
%o A382355     index = 0
%o A382355     P=[]
%o A382355     for i in range(n):
%o A382355         index+=A[i]
%o A382355         index=index%len(l)
%o A382355         P.append(l[index])
%o A382355         l.pop(index)
%o A382355     return P
%o A382355 def invPerm(p):
%o A382355     inv = []
%o A382355     for i in range(len(p)):
%o A382355         inv.append(None)
%o A382355     for i in range(len(p)):
%o A382355         inv[p[i]-1]=i+1
%o A382355     return inv
%o A382355 def survivor(n, A):
%o A382355     return J(n, A)[n-1]
%o A382355 def UDU(n):
%o A382355     return [1] + [2 for i in range(n)]
%o A382355 seq = []
%o A382355 for i in range(1,20):
%o A382355     seq += [survivor(i, UDU(i))]
%o A382355 print(", ".join([str(v) for v in seq]))
%o A382355 (Python)
%o A382355 def a(n):
%o A382355     c, i, J = 1, 0, list(range(1, n+1))
%o A382355     while len(J) > 1:
%o A382355         i = (i + 1)%len(J)
%o A382355         q = J.pop(i)
%o A382355         i = (i + 1)%len(J) # skip the third
%o A382355     return J[0]
%o A382355 print([a(n) for n in range(1, 73)]) # _Michael S. Branicky_, Mar 24 2025
%Y A382355 Cf. A006257, A225381, A321298, A378635, A382354, A382356, A382358.
%K A382355 nonn
%O A382355 1,4
%A A382355 _Tanya Khovanova_, _Nathan Sheffield_, and the MIT PRIMES STEP junior group, Mar 22 2025