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 A185173 Minimum number of distinct sums from consecutive terms in a circular permutation. 0
 1, 3, 6, 9, 13, 17, 22, 28, 35, 41, 49, 57, 65, 73, 82, 93 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS a(n) <= n(n+1)/2, but this apparently is impossible for n >= 4. - N. J. A. Sloane, Mar 14 2012 LINKS 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) PROG (Sage) # a(n)=distinct_sum_count(n) def distinct_sum_count(n):     min_sum_count=n*(n+1)/2     for p in Permutations(n=n):         if p[0]==1 and p[1]

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Last modified October 17 16:42 EDT 2019. Contains 328120 sequences. (Running on oeis4.)