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A364237 a(n) is the number of non-equivalent permutations of {1,2,...,2n-1} such that no subset of consecutive terms from the permutation sums to 0 modulo 2n, where two permutations are equivalent if one can be obtained from the other by multiplying every entry with an integer relatively prime to 2n and/or reversing the permutation.

%I A364237 #14 Sep 11 2023 23:02:43
%S A364237 1,1,2,4,42,504,7492,172480,8639632
%N A364237 a(n) is the number of non-equivalent permutations of {1,2,...,2n-1} such that no subset of consecutive terms from the permutation sums to 0 modulo 2n, where two permutations are equivalent if one can be obtained from the other by multiplying every entry with an integer relatively prime to 2n and/or reversing the permutation.
%C A364237 If we consider all permutations of {1,2,...,2n-1} such that no subset of consecutive terms from the permutation sums to 0 modulo 2n, then the number of such permutations is given by the number of constructive orderings mentioned in A141599. For example, given the permutation 14325 that satisfies the given conditions, observe that the partial sums modulo 6, namely 1=1, 1+4=5, 1+4+3=2, 1+4+3+2=4, and 1+4+3+2+5=3, are distinct.
%H A364237 Sunil K. Chebolu and Papa A. Sissokho, <a href="http://math.colgate.edu/~integers/w13/w13.pdf">Zero-sum-free tuples and hyperplane arrangements</a>, Integers 22 (2022), #A13.
%H A364237 Sean A. Irvine, <a href="https://github.com/archmageirvine/joeis/blob/master/src/irvine/oeis/a364/A364237.java">Java program</a> (github)
%e A364237 When n=3, there are four permutations of {1,2,3,4,5} such that no subset of consecutive terms from the permutation sums to 0 modulo 6, namely 14325, 25314, 41352, and 52341. Note that 14325 and 52341 are equivalent by reversing the permutations. Furthermore multiplication by 5 on every entry also yields the same equivalence. Additionally, 25314 and 41352 are analogously equivalent. Hence a(3)=2.
%e A364237 When n=4, 6142573 and 3752416 are equivalent by reversing the permutations but not by multiplying any integer relatively prime to 8, whereas 6142573 and 2346751 are equivalent by multiplication of 3 on every entry.
%o A364237 (SageMath)
%o A364237 n = 3 #the index for the sequence a(n)
%o A364237 orbits = {} #dictionary of permutations that are consecutive zero-sum-free
%o A364237 seen = [] #list of seen permutations that are consecutive zero-sum-free
%o A364237 a = 0 #the value of a(n)
%o A364237 for labeling in Permutations(range(1,2*n)):
%o A364237     if labeling not in seen:
%o A364237         sums = [labeling[0]]
%o A364237         for i in range(1,2*n-1):
%o A364237             nextsum = (labeling[i] + sums[i-1]) % (2*n)
%o A364237             if any([nextsum == 0, nextsum in sums]):
%o A364237                 break
%o A364237             sums.append(nextsum)
%o A364237         if len(sums) == (2*n)-1:
%o A364237             a += 1
%o A364237             orbits[a] = []
%o A364237             for m in [x for x in range(1,2*n) if gcd(x,2*n) == 1]:
%o A364237                 equiv = [(m*labeling[i]) % (2*n) for i in range(2*n-1)]
%o A364237                 if equiv not in orbits[a]:
%o A364237                     orbits[a].append(equiv)
%o A364237                 seen.append(equiv)
%o A364237                 equiv = [equiv[2*n-2-i] for i in range(2*n-1)]
%o A364237                 if equiv not in orbits[a]:
%o A364237                     orbits[a].append(equiv)
%o A364237                 seen.append(equiv)
%o A364237 print(f"a({n}) = {a}\n")
%o A364237 print("Equivalencies:")
%o A364237 for i in range(1,a+1):
%o A364237     print(f"{i}.")
%o A364237     for x in orbits[i]:
%o A364237         print(x)
%o A364237     print('\n')
%Y A364237 Cf. A141599.
%K A364237 nonn,hard,more
%O A364237 1,3
%A A364237 _Anna Coleman_, _Joshua Harrington_, _Maggie X. Lai_, _Philip Thomas_, and _Wing Hong Tony Wong_, Jul 18 2023
%E A364237 a(8)-a(9) from _Sean A. Irvine_, Aug 15 2023