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A229827
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Number of circular permutations i_1,...,i_n of 1,...,n such that the n numbers i_1^2+i_2, i_2^2+i_3, ..., i_{n-1}^2+i_n, i_n^2+i_1 form a complete system of residues modulo n.
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0
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0, 0, 0, 4, 0, 24, 0, 0, 0, 5308, 0, 123884, 0, 0, 0, 147288372, 0, 7238567052, 0, 0, 0
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OFFSET
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2,4
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COMMENTS
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Conjecture: (i) a(p) > 0 for any prime p > 3. Moreover, for any finite field F(q) with q elements and a polynomial P(x) over F(q) of degree smaller than q-1, if P(x) is not of the form c-x, then there is a circular permutation a_1, ..., a_q of all the elements of F(q) with
{P(a_1)+a_2, P(a_2)+a_3, ..., P(a_{q-1})+a_q, P(a_q)+a_1}
equal to F(q).
(ii) Let F be any field with |F| > 7, and let A be a finite subset of F with |A| = n > 2. Let P(x) be a polynomial over F whose degree is smaller than p-1 if F is of prime characteristic p. If P(x) is not of the form c-x, then there is a circular permutation a_1, ..., a_n of all the elements of A such that the n sums P(a_1)+a_2, P(a_2)+a_3, ..., P(a_{n-1})+a_n, P(a_n)+a_1 are pairwise distinct.
Clearly, if a(n) > 0, then we must have 1^2+2^2+...+n^2 == 0 (mod n), i.e., (n+1)*(2n+1) == 0 (mod 6). Thus, when n is divisible by 2 or 3, we must have a(n) = 0.
It is well known that for any finite field F(q) with q elements we have sum_{x in F(q)} x^k = 0 for every k = 0, ..., q-2.
Verified for n < 256: a(n) > 0 iff n is not divisible by 2 or 3. - Bert Dobbelaere, Apr 23 2021
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LINKS
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EXAMPLE
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a(5) = 4 due to the circular permutations
(1,3,4,5,2), (1,4,2,3,5), (1,5,2,4,3), (1,5,4,2,3).
a(7) > 0 due to the circular permutation (1,2,3,7,4,6,5).
a(11) > 0 due to the circular permutation
(1,2,3,4,6,5,9,11,10,8,7).
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MATHEMATICA
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(* A program to compute desired circular permutations for n = 7. *)
V[i_]:=Part[Permutations[{2, 3, 4, 5, 6, 7}], i]
m=0
Do[If[Length[Union[Table[Mod[If[j==0, 1, Part[V[i], j]]^2+If[j<6, Part[V[i], j+1], 1], 7], {j, 0, 6}]]]<7, Goto[aa]];
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]]; Label[aa]; Continue, {i, 1, 6!}]
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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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EXTENSIONS
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STATUS
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approved
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