A229082 -id:A229082 - cp's OEIS frontend

A229130 Number of permutations i_0, i_1, ..., i_n of 0, 1, ..., n with i_0 = 0 and i_n = n such that 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 all relatively prime to both n-1 and n+1.

1, 0, 1, 1, 0, 6, 3, 42, 68, 2794, 0, 5311604, 478, 57009, 2716452, 10778632, 207360, 39187872956340, 106144, 26869397610, 11775466120, 22062519153360, 559350576, 29991180449906858400, 257272815600, 12675330087321600, 52248156883498208

Offset: 1

Zhi-Wei Sun, Sep 15 2013

Conjecture: a(n) > 0 except for n = 2, 5, 11. Similarly, for any positive integer n not equal to 4, there is a permutation i_0, i_1, ..., i_n of 0, 1, ..., n with i_0 = 0 and i_n = n such that 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 all coprime to both n-1 and n+1.
Zhi-Wei Sun also made the following general conjecture:
  For any positive integer k, define E(k) to be the set of those positive integers n for which there is no permutation i_0, i_1, ..., i_n of 0, 1, ..., n with i_0 = 0 and i_n = n such that all the n+1 numbers i_0^k+i_1, i_1^k+i_2, ..., i_{n-1}^k+i_n, i_n^k+i_0 are coprime to both n-1 and n+1. Then E(k) is always finite; in particular, E(1) = {2,4}, E(2) = {2,5,11} and E(3) = {2,4}.

			a(3) = 1 due to the permutation (i_0,i_1,i_2,i_3)=(0,1,2,3).
a(4) = 1 due to the permutation (0,1,3,2,4).
a(6) = 1 due to the permutations
  (0,1,3,2,5,4,6), (0,1,3,4,2,5,6), (0,2,5,1,3,4,6),
  (0,3,2,4,1,5,6), (0,3,4,1,2,5,6), (0,4,1,3,2,5,6).
a(7) = 3 due to the permutations
  (0,1,6,5,4,3,2,7), (0,5,4,3,2,1,6,7), (0,5,6,1,4,3,2,7).
a(8) > 0 due to the permutation (0,2,1,4,6,5,7,3,8).
a(9) > 0 due to the permutation (0,1,2,3,4,5,6,7,8,9).
a(10) > 0 due to the permutation (0,1,3,5,4,7,9,8,6,2,10).
a(11) = 0 since 6 is the unique i among 0,...,11 with i^2+5 coprime to 11^2-1, and it is also the unique j among 1,...,10 with j^2+11 coprime to 11^2-1.

Zhi-Wei Sun, List of required permutations for n = 1..10
Zhi-Wei Sun, Some new problems in additive combinatorics, preprint, arXiv:1309.1679 [math.NT], 2013-2014.

Cf. A228886, A229082, A229038, A229005, A228917, A228956.

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

a(12)-a(17) from Alois P. Heinz, Sep 15 2013
a(19) and a(23) from Alois P. Heinz, Sep 16 2013
a(18), a(20)-a(22) and a(24)-a(27) from Bert Dobbelaere, Feb 18 2020

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.

Original entry on oeis.org

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

A229232 Number of undirected circular permutations pi(1), ..., pi(n) of 1, ..., n with the n numbers pi(1)pi(2)-1, pi(2)pi(3)-1, ..., pi(n-1)pi(n)-1, pi(n)pi(1)-1 all prime.

Original entry on oeis.org

0, 0, 0, 1, 0, 2, 1, 2, 2, 8, 2, 241, 0, 693, 376, 7687, 1082, 127563, 25113, 1353842, 559649

Offset: 1

Zhi-Wei Sun, Sep 16 2013

Conjecture: a(n) > 0 for all n > 5 with n not equal to 13.
Zhi-Wei Sun also made the following conjectures:
(1) For any integer n > 1, there is a permutation pi(1), ..., pi(n) of 1, ..., n such that the n numbers 2*pi(1)*pi(2)-1, ..., 2*pi(n-1)*pi(n)-1, 2*pi(n)*pi(1)-1 are all prime. Also, for any positive integer n not equal to 4, there is a permutation pi(1), ..., pi(n) of 1, ..., n such that the n numbers 2*pi(1)*pi(2)+1, ..., 2*pi(n-1)*pi(n)+1, 2*pi(n)*pi(1)+1 are all prime.
(2) Let F be a finite field with q > 7 elements. Then, there is a circular permutation a_1,...,a_{q-1} of the q-1 nonzero elements of F such that all the q-1 elements a_1*a_2-1, a_2*a_3-1, ..., a_{q-2}*a_{q-1}-1, a_{q-1}*a_1-1 are primitive elements of the field F (i.e., generators of the multiplicative group F\{0}). Also, there is a circular permutation b_1,...,b_{q-1} of the q-1 nonzero elements of F such that all the q-1 elements b_1*b_2+1, b_2*b_3+1, ..., b_{q-2}*b_{q-1}+1, b_{q-1}*b_1+1 are primitive elements of the field F.

			a(4) = 1 due to the circular permutation (1,3,2,4).
a(6) = 2 due to the circular permutations
   (1,3,2,4,5,6) and (1,3,2,6,5,4).
a(7) = 1 due to the circular permutation (1,3,2,7,6,5,4).
a(8) = 2 due to the circular permutations
   (1,3,2,7,6,5,4,8) and (1,4,5,6,7,2,3,8).
a(9) = 2 due to the circular permutations
   (1,3,4,5,6,7,2,9,8) and (1,3,8,9,2,7,6,5,4).
a(10) = 8 due to the circular permutations
   (1,3,4,5,6,7,2,9,10,8), (1,3,4,5,6,7,2,10,9,8),
   (1,3,8,9,10,2,7,6,5,4), (1,3,8,10,9,2,7,6,5,4),
   (1,3,10,8,9,2,7,6,5,4), (1,3,10,9,2,7,6,5,4,8),
   (1,4,5,6,7,2,3,10,9,8), (1,4,5,6,7,2,9,10,3,8).
a(13) = 0 since 8 is the unique j among 1, ..., 12 with 13*j-1 prime.

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

Cf. A051252, A227456, A228917, A228956, A229082.

(* A program to compute required circular permutations for n = 8. To get "undirected" circular permutations, we should identify a circular permutation with the one of the opposite direction; for example, (1,8,4,5,6,7,2,3) is identical to (1,3,2,7,6,5,4,8) if we ignore direction. Thus, a(8) is half of the number of circular permutations yielded by this program. *)
V[i_]:=V[i]=Part[Permutations[{2,3,4,5,6,7,8}],i]
f[i_,j_]:=f[i,j]=PrimeQ[i*j-1]
m=0
Do[Do[If[f[If[j==0,1,Part[V[i],j]],If[j<7,Part[V[i],j+1],1]]==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(11)-a(21) from Pontus von Brömssen, Jan 08 2025

A229827 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.

Original entry on oeis.org

0, 0, 0, 4, 0, 24, 0, 0, 0, 5308, 0, 123884, 0, 0, 0, 147288372, 0, 7238567052, 0, 0, 0

Offset: 2

Zhi-Wei Sun, Sep 30 2013

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

			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).

Zhi-Wei Sun, A combinatorial conjecture on finite fields, a message to Number Theory List, Sept. 30, 2013.
Zhi-Wei Sun, Some new problems in additive combinatorics , preprint, arXiv:1309.1679 [math.NT], 2013-2014.

Cf. A227456, A229082, A229141.

(* 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!}]

a(11)-a(22) from Bert Dobbelaere, Apr 23 2021

cp's OEIS Frontend

A229130 Number of permutations i_0, i_1, ..., i_n of 0, 1, ..., n with i_0 = 0 and i_n = n such that 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 all relatively prime to both n-1 and n+1.

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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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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.

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A229232 Number of undirected circular permutations pi(1), ..., pi(n) of 1, ..., n with the n numbers pi(1)*pi(2)-1, pi(2)*pi(3)-1, ..., pi(n-1)*pi(n)-1, pi(n)*pi(1)-1 all prime.

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A229827 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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A229232 Number of undirected circular permutations pi(1), ..., pi(n) of 1, ..., n with the n numbers pi(1)pi(2)-1, pi(2)pi(3)-1, ..., pi(n-1)pi(n)-1, pi(n)pi(1)-1 all prime.