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.

1, 1, 2, 4, 42, 504, 7492, 172480, 8639632

Offset

1,3

Comments

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.

Links

Table of n, a(n) for n=1..9.

Sunil K. Chebolu and Papa A. Sissokho, Zero-sum-free tuples and hyperplane arrangements, Integers 22 (2022), #A13.

Sean A. Irvine, Java program (github)

Example

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.
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.

Prog

(SageMath)
n = 3 #the index for the sequence a(n)
orbits = {} #dictionary of permutations that are consecutive zero-sum-free
seen = [] #list of seen permutations that are consecutive zero-sum-free
a = 0 #the value of a(n)
for labeling in Permutations(range(1, 2*n)):
    if labeling not in seen:
        sums = [labeling[0]]
        for i in range(1, 2*n-1):
            nextsum = (labeling[i] + sums[i-1]) % (2*n)
            if any([nextsum == 0, nextsum in sums]):
                break
            sums.append(nextsum)
        if len(sums) == (2*n)-1:
            a += 1
            orbits[a] = []
            for m in [x for x in range(1, 2*n) if gcd(x, 2*n) == 1]:
                equiv = [(m*labeling[i]) % (2*n) for i in range(2*n-1)]
                if equiv not in orbits[a]:
                    orbits[a].append(equiv)
                seen.append(equiv)
                equiv = [equiv[2*n-2-i] for i in range(2*n-1)]
                if equiv not in orbits[a]:
                    orbits[a].append(equiv)
                seen.append(equiv)
print(f"a({n}) = {a}\n")
print("Equivalencies:")
for i in range(1, a+1):
    print(f"{i}.")
    for x in orbits[i]:
        print(x)
    print('\n')

Crossrefs

Cf. A141599.

Sequence in context: A009547 A215921 A009811 * A009591 A009717 A018317

Adjacent sequences: A364234 A364235 A364236 * A364238 A364239 A364240

Keyword

nonn,hard,more

Author

Anna Coleman, Joshua Harrington, Maggie X. Lai, Philip Thomas, and Wing Hong Tony Wong, Jul 18 2023

Extensions

a(8)-a(9) from Sean A. Irvine, Aug 15 2023

Status

approved