“…Note that k(n + 1)/2 = 154. Then the above π is given like π = (7,11,19,20 . We may read this configuration from left to right of the first row, and to the second row, and so on.…”
Section: Proof Of Theorem 11: the Case N ≡ 1 (Mod K)mentioning
confidence: 99%
“…Note that k(n + 1)/2 = 147. Then the above π ′ is given like π ′ = (6,10,18,19,25,36,37,5,11,16,20,26,35,38,4,12,14,21,27,34,39,3,7,17,22,28,33,40,2,8,15,23,29,32,41,1,9,13,24,30,31 . We see that the maximal 7-consecutive sum is 151.…”
Section: Proof Of Theorem 11: the Case N ≡ −1 (Mod K)mentioning
confidence: 99%
“…Note that k(n + 1)/2 = 125/2. Then the above π ′ is given like π ′ = (5,8,15,16,21,4,9,13,17,22,3,10,11,18,23,2,6,14,19,24,1,7,12,20).…”
Section: Proof Of Theorem 11: the Case N ≡ −1 (Mod K)mentioning
confidence: 99%
“…We must remark that msum(21, 3) = 2 is not our result. This comes from [7]. Moreover, msum(6, 3), msum(9, 3) = msum(9, 6) and msum(15, 3) can be obtained only from Theorem 2.4 (i) and (1.1).…”
Let n and k be positive integers with n > k. Given a permutation (π 1 , . . . , πn) of integers 1, . . . , n, we consider k-consecutive sums of π, i.e., s i := k−1 j=0 π i+j for i = 1, . . . , n, where we let π n+j = π j . What we want to do in this paper is to know the exact value of msum(n, k) := min max{s i : i = 1, . . . , n} − k(n + 1) 2 : π ∈ Sn ,where Sn denotes the set of all permutations of 1, . . . , n. In this paper, we determine the exact values of msum(n, k) for some particular cases of n and k. As a corollary of the results, we obtain msum(n, 3), msum(n, 4) and msum(n, 6) for any n.
“…Note that k(n + 1)/2 = 154. Then the above π is given like π = (7,11,19,20 . We may read this configuration from left to right of the first row, and to the second row, and so on.…”
Section: Proof Of Theorem 11: the Case N ≡ 1 (Mod K)mentioning
confidence: 99%
“…Note that k(n + 1)/2 = 147. Then the above π ′ is given like π ′ = (6,10,18,19,25,36,37,5,11,16,20,26,35,38,4,12,14,21,27,34,39,3,7,17,22,28,33,40,2,8,15,23,29,32,41,1,9,13,24,30,31 . We see that the maximal 7-consecutive sum is 151.…”
Section: Proof Of Theorem 11: the Case N ≡ −1 (Mod K)mentioning
confidence: 99%
“…Note that k(n + 1)/2 = 125/2. Then the above π ′ is given like π ′ = (5,8,15,16,21,4,9,13,17,22,3,10,11,18,23,2,6,14,19,24,1,7,12,20).…”
Section: Proof Of Theorem 11: the Case N ≡ −1 (Mod K)mentioning
confidence: 99%
“…We must remark that msum(21, 3) = 2 is not our result. This comes from [7]. Moreover, msum(6, 3), msum(9, 3) = msum(9, 6) and msum(15, 3) can be obtained only from Theorem 2.4 (i) and (1.1).…”
Let n and k be positive integers with n > k. Given a permutation (π 1 , . . . , πn) of integers 1, . . . , n, we consider k-consecutive sums of π, i.e., s i := k−1 j=0 π i+j for i = 1, . . . , n, where we let π n+j = π j . What we want to do in this paper is to know the exact value of msum(n, k) := min max{s i : i = 1, . . . , n} − k(n + 1) 2 : π ∈ Sn ,where Sn denotes the set of all permutations of 1, . . . , n. In this paper, we determine the exact values of msum(n, k) for some particular cases of n and k. As a corollary of the results, we obtain msum(n, 3), msum(n, 4) and msum(n, 6) for any n.
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