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A229038
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Number of undirected circular permutations i_1, ..., i_n of 1, ..., n such that all the n sums i_1^2+i_2^2, ..., i_{n-1}^2+i_n^2, i_n^2+i_1^2 are among those integers m with the Jacobi symbol (m/(2*n+1)) equal to 1.
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4
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OFFSET
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1,4
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COMMENTS
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Conjecture: a(n) > 0 if 2*n+1 is a prime greater than 13.
When p = 2*n+1 > 13 is a prime, the conjecture indicates that there is a numbering a_1, a_2, ..., a_n of all the (p-1)/2 = n quadratic residues modulo p such that all the n adjacent sums a_1+a_2, a_2+a_3, ..., a_{n-1}+a_n, a_n+a_1 are also quadratic residues modulo p.
Zhi-Wei Sun also made the following conjecture:
(1) If p = 2*n+1 > 13 is a prime, then there is a circular permutation a_1, a_2, ..., a_n of all the (p-1)/2 = n quadratic residues modulo p such that all the n adjacent sums a_1+a_2, a_2+a_3, ..., a_{n-1}+a_n, a_n+a_1 are quadratic nonresidues modulo p.
(2) For any prime p = 2*n+1 > 5, there is a circular permutation a_1, a_2, ..., a_n of all the (p-1)/2 = n quadratic residues modulo p such that all the n adjacent differences a_1-a_2, a_2-a_3, ..., a_{n-1}-a_n, a_n-a_1 are quadratic residues (or quadratic nonresidues) modulo p.
Noga Alon has confirmed this conjecture.
See also A229878 for a stronger conjecture which remains open.
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LINKS
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EXAMPLE
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a(4) = 3 due to the circular permutations
(1,2,3,4), (1,2,4,3), (1,3,2,4).
a(5) = 1 due to the circular permutation (1,2,4,3,5).
a(8) = 1 due to the circular permutation (1,5,8,6,4,3,2,7).
a(9) > 0 due to the circular permutation (1,2,4,3,8,6,5,7,9).
a(10) > 0 due to the circular permutation
(1, 2, 4, 3, 7, 9, 5, 10, 8, 6).
a(11) > 0 due to the circular permutation
(1, 5, 2, 3, 4, 10, 8, 11, 7, 6, 9).
a(12) > 0 due to the circular permutation
(1, 10, 2, 3, 8, 7, 12, 5, 11, 6, 4, 9).
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MATHEMATICA
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(* A program to compute required circular permutations for n = 9. To get "undirected" circular permutations, we should identify a circular permutation with the one of the opposite direction; for example, the circular permutation (1, 9, 7, 5, 6, 8, 3, 4, 2) is identical to (1, 2, 4, 3, 8, 6, 5, 7, 9) if we ignore direction. Thus a(9) is half of the number of circular permutations yielded by this program. *)
f[i_, j_, n_]:=JacobiSymbol[i^2+j^2, n]==1
V[i_]:=Part[Permutations[{2, 3, 4, 5, 6, 7, 8, 9}], i]
m=0
Do[Do[If[f[If[j==0, 1, Part[V[i], j]], If[j<8, Part[V[i], j+1], 1], 19]==False, Goto[aa]], {j, 0, 8}];
m=m+1; Print[m, ":", " ", 1, " ", Part[V[i], 1], " ", Part[V[i], 2], " ", Part[V[i], 3], " ", Part[V[i], 4], " ", Part[V[i], 5], " ", Part[V[i], 6], " ", Part[V[i], 7], " ", Part[V[i], 8]]; Label[aa]; Continue, {i, 1, 8!}]
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CROSSREFS
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KEYWORD
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nonn,more,hard
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AUTHOR
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STATUS
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approved
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