A228860 -id:A228860 - cp's OEIS frontend

A228917 Number of undirected circular permutations i_0, i_1, ..., i_n of 0, 1, ..., n such that i_0+i_1, i_1+i_2, ...,i_{n-1}+i_n, i_n+i_0 are among those k with 6k-1 and 6k+1 twin primes.

1, 1, 1, 2, 2, 2, 5, 2, 12, 39, 98, 526, 2117, 6663, 15043, 68403, 791581, 4826577, 19592777, 102551299, 739788968, 4449585790, 36547266589, 324446266072, 2743681178070

Offset: 1

Zhi-Wei Sun, Sep 08 2013

Conjecture: a(n) > 0 for all n > 0.
This implies the twin prime conjecture, and it is similar to the prime circle problem mentioned in A051252.
For each n = 2,3,... construct an undirected simple graph T(n) with vertices 0,1,...,n which has an edge connecting two distinct vertices i and j if and only if 6*(i+j)-1 and 6*(i+j)+1 are twin primes. Then a(n) is just the number of Hamiltonian cycles contained in T(n). Thus a(n) > 0 if and only if T(n) is a Hamilton graph.
Zhi-Wei Sun also made the following similar conjectures for odd primes, Sophie Germain primes, cousin primes and sexy primes:
(1) For any integer n > 0, there is a permutation i_0, i_1, ..., i_n of 0, 1, ..., n such that i_0+i_1, i_1+i_2, ..., i_{n-1}+i_n, i_n+i_0 are integers of the form (p-1)/2, where p is an odd prime. Also, we may replace the above (p-1)/2 by (p+1)/4 or (p-1)/6; when n > 4 we may substitute (p-1)/4 for (p-1)/2.
(2) For any integer n > 2, there is a permutation i_0, i_1, ..., i_n of 0, 1,..., n  such that i_0+i_1, i_1+i_2, ..., i_{n-1}+i_n, i_n+i_0 are integers of the form (p+1)/6, where p is a Sophie Germain prime.
(3) For any integer n > 3, there is a permutation i_0, i_1, ..., i_n of 0, 1,..., n  such that i_0+i_1, i_1+i_2, ..., i_{n-1}+i_n, i_n+i_0 are among those integers k with 6*k+1 and 6*k+5 both prime.
(4) For any integer n > 4, there is a permutation i_0, i_1, ..., i_n of 0, 1,..., n  such that i_0+i_1, i_1+i_2, ..., i_{n-1}+i_n, i_n+i_0 are among those integers k with 2*k-3 and 2*k+3 both prime.

			a(n) = 1 for n = 1,2,3 due to the permutation (0,...,n).
a(4) = 2 due to the permutations (0,1,4,3,2) and (0,2,1,4,3).
a(5) = 2 due to the permutations (0,1,4,3,2,5), (0,3,4,1,2,5).
a(6) = 2 due to the permutations
  (0,1,6,4,3,2,5) and (0,3,4,6,1,2,5).
a(7) = 5 due to the permutations
  (0,1,6,4,3,2,5,7), (0,1,6,4,3,7,5,2), (0,2,1,6,4,3,7,5),
  (0,3,4,6,1,2,5,7), (0,5,2,1,6,4,3,7).
a(8) = 2 due to the permutations
  (0,1,6,4,8,2,3,7,5) and (0,1,6,4,8,2,5,7,3).
a(9) = 12 due to the permutations
  (0,1,6,4,3,9,8,2,5,7), (0,1,6,4,8,9,3,2,5,7),
  (0,1,6,4,8,9,3,7,5,2), (0,2,1,6,4,8,9,3,7,5),
  (0,2,8,9,1,6,4,3,7,5), (0,3,4,6,1,9,8,2,5,7),
  (0,3,9,1,6,4,8,2,5,7), (0,3,9,8,4,6,1,2,5,7),
  (0,5,2,1,6,4,8,9,3,7), (0,5,2,8,4,6,1,9,3,7),
  (0,5,2,8,9,1,6,4,3,7), (0,5,7,3,9,1,6,4,8,2).
a(10) > 0 due to the permutation (0,5,2,3,9,1,6,4,8,10,7).
a(11) > 0 due to the permutation (0,10,8,9,3,7,11,6,4,1,2,5).
a(12) > 0 due to the permutation
        (0, 5, 2, 1, 6, 4, 3, 9, 8, 10, 7, 11, 12).

Zhi-Wei Sun, Twin primes and circular permutations, a message to Number Theory List, Sept. 8, 2013.
Z.-W. Sun, Some new problems in additive combinatorics, arXiv preprint arXiv:1309.1679 [math.NT], 2013-2014.

Cf. A000040, A001359, A006512, A023200, A046132, A023201, A046117, A005384, A051252, A228766, A228860, A228886.

(* A program to compute required circular permutations for n = 7. To get "undirected" circular permutations, we should identify a circular permutation with the one of the opposite direction; for example, (0,7,5,2,3,4,6,1) is identical to (0,1,6,4,3,2,5,7) if we ignore direction. Thus a(7) is half of the number of circular permutations yielded by this program. *)
tp[n_]:=tp[n]=PrimeQ[6n-1]&&PrimeQ[6n+1]
V[i_]:=Part[Permutations[{1,2,3,4,5,6,7}],i]
m=0
Do[Do[If[tp[If[j==0,0,Part[V[i],j]]+If[j<7,Part[V[i],j+1],0]]==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]];Label[aa];Continue,{i,1,7!}]

a(10)-a(25) from Max Alekseyev, Sep 12 2013

A228886 Number of permutations i_0,i_1,...,i_n of 0,1,...,n with i_0 = 0 and i_n = n, and with i_0+i_1, i_1+i_2, ..., i_{n-1}+i_n, i_n+i_0 all coprime to both n-1 and n+1.

Original entry on oeis.org

1, 0, 1, 0, 1, 8, 3, 24, 78, 1164, 34, 4021156, 400, 87180, 2499480, 7509358, 1700352, 39982182134232, 1427688, 212987263960, 9533487948, 36638961135462, 29317847040, 30258969747586970112, 1655088666624

Offset: 1

Zhi-Wei Sun, Sep 07 2013

Conjecture: a(n) > 0 except for n = 2, 4.
Note that this conjecture is stronger than the one stated in the comments in A228860. Also, there is no permutation i_0, ..., i_7 of 0, ..., 7 with i_0+i_1, i_1+i_2, ..., i_6+i_7, i_7+i_0 all coprime to 7*13 - 1 = 90.
For n > 1 let S(n) be an undirected simple graph with vertices 0, 1, ..., n which has an edge connecting two distinct vertices i and j if and only if i + j is relatively prime to both n-1 and n+1. Then a(n) is just the number of those Hamiltonian cycles in S(n) on which the vertices 0 and n are adjacent.
We have some other similar conjectures. For example, for any positive integer n there is a permutation i_0,i_1,...,i_n of 0,1,...,n with i_0 = 0 and i_n = n such that i_0 + i_1, i_1 + i_2, ..., i_{n-1} + i_n, i_n + i_0 are all relatively prime to both 2*n-1 and 2*n+1.
On Sep 10 2013, Zhi-Wei Sun proved the conjecture for any positive odd integer n. In fact, if n == 1 or 3 (mod 6) then we may take the permutation 0,n-2,2,n-4,4,...,1,n-1,n which meets the requirement; if n == 3 or 5 (mod 6) then the permutation 0,1,n-1,3,n-3,...,n-2,2,n suffices for our purpose.

			a(2) = 0 since 1+2 = 2^2-1.
a(3) = 1 due to the identical permutation (0,1,2,3).
a(4) = 0 since 1+2 divides 4^2-1.
a(5) = 1 due to the permutation (0,1,4,3,2,5).
a(6) = 8 due to the permutations
  (0,1,2,4,5,3,6), (0,1,3,5,4,2,6),
  (0,2,4,5,1,3,6), (0,3,1,2,4,5,6),
  (0,3,1,5,4,2,6), (0,4,2,1,3,5,6),
  (0,4,2,1,5,3,6), (0,4,5,3,1,2,6).
a(7) = 3 due to the permutations(0,1,4,3,2,5,6,7), (0,1,6,5,2,3,4,7), (0,5,2,3,4,1,6,7).
a(8) > 0 due to the permutation (0,1,4,7,3,2,6,5,8).
a(9) > 0 due to the permutation (0,1,2,5,4,7,6,3,8,9).
a(10) > 0 due to the permutation (0,1,3,2,5,8,6,4,9,7,10).

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. A228860, A228917, A228956.

(*A program to compute the 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]]+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(11)-a(25) from Max Alekseyev, Sep 13 2013

A228885 Determinant of the n X n matrix with (i,j)-entry equal to 1 or 0 according as i + j is coprime to n or not.

Original entry on oeis.org

1, -1, -2, 0, 4, -4, -6, 0, 0, -16, -10, 0, 12, -36, -2048, 0, 16, 0, -18, 0, 27648, -100, -22, 0, 0, -144, 0, 0, 28, -4194304, -30, 0, 2048000, -256, -127401984, 0, 36, -324, -14155776, 0, 40, -764411904, -42, 0, 0, -484, -46, 0, 0, 0, -536870912, 0, 52, 0, -419430400000, 0, 3057647616, -784, -58, 0

Offset: 1

Zhi-Wei Sun, Sep 06 2013

sign

Conjecture: If n is squarefree, then (-1)^(n*(n-1)/2)*a(n) > 0.
When p^2 divides n with p prime, (i + n/p) + j is coprime to n if and only if i + j is coprime to n. So a(n) = 0 if n is not squarefree.
It is easy to show that Phi(n) divides a(n) for any n > 0, where Phi(n) is Euler's totient function. Also, a(p) = (-1)^((p-1)/2)*(p-1) for any odd prime p.

			a(1) = 1 since 1 + 1 = 2 is relatively prime to 1.

Zhi-Wei Sun, Table of n, a(n) for n = 1..500
Zhi-Wei Sun, On some determinants with Legendre symbol entries, preprint, arXiv:1308.2900 [math.NT], 2013-2019.

Cf. A228860, A228884.

Cf. A057077, A008966, A349741, A000010.

a[n_]:=Det[Table[If[GCD[i+j,n]==1,1,0],{i,1,n},{j,1,n}]]
Table[a[n],{n,1,60}]

a(n) = matdet(matrix(n, n, i, j, gcd(n, i+j)==1)); \\ Michel Marcus, Aug 25 2021

Conjectures from Ridouane Oudra, Mar 13 2025: (Start)
a(n) = (-1)^floor(n/2)*mu(n)^2*Product_{k=1..n} phi(gcd(n,k)).
a(n) = (-1)^floor(n/2)*mu(n)^2*Product_{d|n} phi(d)^phi(n/d).
a(n) = A057077(n)*A008966(n)*A349741(n)*A000010(n). (End)

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A228917 Number of undirected circular permutations i_0, i_1, ..., i_n of 0, 1, ..., n such that i_0+i_1, i_1+i_2, ...,i_{n-1}+i_n, i_n+i_0 are among those k with 6k-1 and 6k+1 twin primes.

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A228886 Number of permutations i_0,i_1,...,i_n of 0,1,...,n with i_0 = 0 and i_n = n, and with i_0+i_1, i_1+i_2, ..., i_{n-1}+i_n, i_n+i_0 all coprime to both n-1 and n+1.

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A228885 Determinant of the n X n matrix with (i,j)-entry equal to 1 or 0 according as i + j is coprime to n or not.

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cp's OEIS Frontend

A228917 Number of undirected circular permutations i_0, i_1, ..., i_n of 0, 1, ..., n such that i_0+i_1, i_1+i_2, ...,i_{n-1}+i_n, i_n+i_0 are among those k with 6*k-1 and 6*k+1 twin primes.

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A228886 Number of permutations i_0,i_1,...,i_n of 0,1,...,n with i_0 = 0 and i_n = n, and with i_0+i_1, i_1+i_2, ..., i_{n-1}+i_n, i_n+i_0 all coprime to both n-1 and n+1.

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Programs

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A228885 Determinant of the n X n matrix with (i,j)-entry equal to 1 or 0 according as i + j is coprime to n or not.

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Examples

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Mathematica

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Formula

A228917 Number of undirected circular permutations i_0, i_1, ..., i_n of 0, 1, ..., n such that i_0+i_1, i_1+i_2, ...,i_{n-1}+i_n, i_n+i_0 are among those k with 6k-1 and 6k+1 twin primes.