
EXAMPLE

a(4)=9 because the circular permutation 1243 has no way to get 5 as a sum of consecutive terms.
a(5)=13 because the circular permutation 12534 has no way to get 6 or 9 as a sum of consecutive terms.
From Bert Dobbelaere, Jun 24 2019: (Start)
Permutations achieving the minimum number of distinct sums:
a(1) = 1: {1}
a(2) = 3: {1, 2}
a(3) = 6: {1, 2, 3}
a(4) = 9: {1, 2, 4, 3}
a(5) = 13: {1, 2, 5, 3, 4}
a(6) = 17: {1, 3, 2, 4, 6, 5}
a(7) = 22: {1, 3, 2, 5, 7, 4, 6}
a(8) = 28: {1, 4, 3, 7, 6, 2, 8, 5}
a(9) = 35: {1, 3, 2, 4, 5, 8, 9, 6, 7}
a(10) = 41: {1, 3, 10, 9, 4, 6, 7, 2, 8, 5}
a(11) = 49: {1, 3, 5, 2, 8, 7, 4, 11, 10, 9, 6}
a(12) = 57: {1, 2, 6, 11, 10, 7, 4, 8, 9, 12, 5, 3}
a(13) = 65: {1, 2, 10, 12, 11, 13, 7, 5, 8, 3, 9, 4, 6}
a(14) = 73: {1, 4, 7, 2, 9, 14, 13, 11, 12, 10, 3, 6, 5, 8}
a(15) = 82: {1, 4, 5, 2, 3, 8, 14, 11, 12, 10, 15, 7, 6, 9, 13}
a(16) = 93: {1, 3, 8, 5, 10, 13, 4, 11, 16, 12, 14, 7, 6, 15, 2, 9}
(End)
