A229130 -id:A229130 - cp's OEIS frontend

A227456 Number of permutations i_0, i_1, ..., i_n of 0, 1, ..., n with i_0 = 0 and i_n = 1 such that all the n+1 numbers i_0^2+i_1, i_1^2+i_2, ..., i_{n-1}^2+i_n, i_n^2+i_0 are of the form (p+1)/4 with p a prime congruent to 3 modulo 4.

1, 1, 1, 1, 1, 2, 4, 11, 15, 15

Offset: 1

Zhi-Wei Sun, Sep 16 2013

Conjecture: a(n) > 0 for all n > 0. Similarly, for any positive integer n, there is a circular permutation i_0, i_1, ..., i_n of 0, 1, ..., n such that all the n+1 numbers i_0^2+i_1, i_1^2+i_2, ..., i_{n-1}^2+i_n, i_n^2+i_0 are of the form (p-1)/4 with p a prime congruent to 1 modulo 4.

Note that if a circular permutation i_0, i_1, ..., i_n of 0, 1, ..., n with i_0 = 0 makes all the n+1 numbers i_0^2+i_1, i_1^2+i_2, ..., i_{n-1}^2+i_n, i_n^2+i_0 of the form (p+1)/4 with p a prime, then we must have i_n = 1. This can be explained as follows: If i_n > 1, then 3 | i_n since 4*(i_n^2+0)-1 is a prime not divisible by 3, and similarly i_{n-1},...,i_1 are also multiples of 3 since 4*(i_{n-1}^2+i_n)-1, ..., 4*(i_1^2+i_2)-1 are primes not divisible by 3. Therefore, i_n > 1 would lead to a contradiction.

			a(1) = a(2) = a(3) = a(4) = a(5) = 1 due to the permutations (0,1), (0,2,1), (0,3,2,1), (0,3,2,4,1), (0,3,2,4,5,1).
a(6) = 2 due to the permutations
    (0,3,6,2,4,5,1) and (0,3,6,5,2,4,1).
a(7) = 4 due to the permutations
    (0,3,6,2,4,5,7,1), (0,3,6,2,7,4,5,1),
    (0,3,6,5,2,7,4,1), (0,3,6,5,7,4,2,1).
a(8) = 11 due to the permutations
(0,3,6,2,4,5,8,7,1), (0,3,6,2,7,8,4,5,1), (0,3,6,2,8,4,5,7,1),
(0,3,6,2,8,7,4,5,1), (0,3,6,5,2,7,8,4,1), (0,3,6,5,2,8,7,4,1),
(0,3,6,5,7,8,2,4,1), (0,3,6,5,7,8,4,2,1), (0,3,6,5,8,2,7,4,1),
(0,3,6,5,8,4,2,7,1), (0,3,6,5,8,7,4,2,1).
a(9) > 0 due to the permutation (0,3,6,9,2,4,5,8,7,1).
a(10) > 0 due to the permutation (0,3,6,9,2,4,5,10,8,7,1).

Zhi-Wei Sun, Some new problems in additive combinatorics, preprint, arXiv:1309.1679 [math.NT], 2013-2014.

Cf. A229082, A229141, A229130.

(* A program to compute required permutations for n = 8. *)
f[i_,j_]:=f[i,j]=PrimeQ[4(i^2+j)-1]
V[i_]:=V[i]=Part[Permutations[{2,3,4,5,6,7,8}],i]
m=0
Do[Do[If[f[If[j==0,0,Part[V[i],j]],If[j<7,Part[V[i],j+1],1]]==False,Goto[aa]],{j,0,7}];
m=m+1;Print[m,":"," ",0," ",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]," ",1];Label[aa];Continue,{i,1,7!}]

A229141 Number of circular permutations i_1, ..., i_n of 1, ..., n such that all the n sums i_1^2+i_2, ..., i_{n-1}^2+i_n, i_n^2+i_1 are among those integers m with the Jacobi symbol (m/(2n+1)) equal to 1.

Original entry on oeis.org

1, 0, 0, 2, 0, 1, 0, 5, 35, 0

Offset: 1

Zhi-Wei Sun, Sep 15 2013

Conjecture: a(n) > 0 if 2*n+1 is a prime greater than 11.
Zhi-Wei Sun also made the following conjectures:
(1) For any prime p = 2*n+1 > 11, there is a circular permutation i_1, ..., i_n of 1, ..., n such that all the n numbers i_1^2-i_2, i_2^2-i_3, ..., i_{n-1}^2-i_n, i_n^2-i_1 are quadratic residues modulo p.
(2)  Let p = 2*n+1 be an odd prime. If p > 13 (resp., p > 11), then there is a circular permutation i_1, ..., i_n of 1, ..., n such that all the n numbers i_1^2+i_2, i_2^2+i_3, ..., i_{n-1}^2+i_n, i_n^2+i_1 (resp., the n numbers i_1^2-i_2, i_2^2-i_3, ..., i_{n-1}^2-i_n, i_n^2-i_1) are primitive roots modulo p.
(3)  Let p = 2*n+1 be an odd prime. If p > 19 (resp. p > 13), then there is a circular permutation i_1, ..., i_n of 1, ..., n such that all the n numbers i_1^2+i_2^2, i_2^2+i_3^2, ..., i_{n-1}^2+i_n^2, i_n^2+i_1^2 (resp., the n numbers i_1^2-i_2^2, i_2^2-i_3^2, ..., i_{n-1}^2-i_n^2, i_n^2-i_1^2) are primitive roots modulo p.
See also the linked arXiv paper of Sun for more conjectures involving primitive roots modulo primes.

			a(4) = 2 due to the permutations (1,3,2,4) and (1,4,3,2).
a(6) = 1 due to the permutation (1,3,5,2,6,4).
a(8) = 5 due to the permutations
   (1,3,4,2,5,8,6,7), (1,8,3,6,2,4,5,7), (1,8,3,6,7,4,2,5),
   (1,8,3,7,6,2,4,5), (1,8,6,7,3,4,2,5).
a(9) > 0 due to the permutation (1,3,7,6,8,4,9,2,5).

Zhi-Wei Sun, Some new problems in additive combinatorics, preprint, arXiv:1309.1679 [math.NT], 2013-2014.

Cf. A229038, A229082, A229130, A229005.

(* A program to compute the desired circular permutations for n = 8. *)
f[i_,j_,p_]:=f[i,j,p]=JacobiSymbol[i^2+j,p]==1
V[i_]:=Part[Permutations[{2,3,4,5,6,7,8}],i]
m=0
Do[Do[If[f[If[j==0,1,Part[V[i],j]],If[j<7,Part[V[i],j+1],1],17]==False,Goto[aa]],{j,0,7}];
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]];Label[aa];Continue,{i,1,7!}]

a(10) = 0 from R. J. Mathar, Sep 15 2013

cp's OEIS Frontend

A227456 Number of permutations i_0, i_1, ..., i_n of 0, 1, ..., n with i_0 = 0 and i_n = 1 such that all the n+1 numbers i_0^2+i_1, i_1^2+i_2, ..., i_{n-1}^2+i_n, i_n^2+i_0 are of the form (p+1)/4 with p a prime congruent to 3 modulo 4.

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