A229082 Number of circular permutations 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)/2 with p an odd prime.

1, 1, 1, 0, 2, 3, 7, 11, 9, 5, 41, 82, 254, 2412, 9524, 13925, 85318, 220818, 1662421, 10496784, 20690118, 97200566, 460358077

Offset

1,5

Comments

Conjecture: a(n) > 0 except for n = 4.

Note that if a circular permutation i_0, i_1, ..., i_n of 0, 1, ..., n with i_0 = 0 meets the requirement then we must have i_n = 1. This can be explained as follows: If i_n > 1, then 3 | i_n since 2*(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 2*(i_{n-1}^2+i_n)+1, ..., 2*(i_1^2+i_2)+1 are primes not divisible by 3. Therefore, i_n > 1 would lead to a contradiction.

Links

Table of n, a(n) for n=1..23.

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

Example

a(1) = 1 due to the circular permutation (0,1).
a(2) = 1 due to the circular permutation (0,2,1).
a(3) = 1 due to the circular permutation (0,3,2,1).
a(5) = 2 due to the circular permutations
   (0,3,2,4,5,1) and (0,3,5,4,2,1).
a(6) = 3 due to the circular permutations
   (0,3,6,5,4,2,1), (0,6,3,2,4,5,1), (0,6,3,5,4,2,1).
a(7) = 7 due to the circular permutations
   (0,3,6,5,4,2,7,1), (0,3,6,5,4,7,2,1), (0,6,3,2,4,7,5,1),
   (0,6,3,2,5,4,7,1), (0,6,3,2,7,4,5,1), (0,6,3,5,4,2,7,1),
   (0,6,3,5,4,7,2,1).
a(8) = 11 due to the circular permutations
   (0,3,6,5,8,4,2,7,1), (0,3,6,5,8,4,7,2,1),
   (0,3,6,8,4,2,7,5,1), (0,4,6,8,4,7,2,5,1),
   (0,3,6,8,5,4,2,7,1), (0,3,6,8,5,4,7,2,1),
   (0,6,3,2,4,7,5,8,1), (0,6,3,2,5,8,4,7,1),
   (0,6,3,2,7,4,5,8,1), (0,6,3,5,8,4,2,7,1),
   (0,6,3,5,8,4,7,2,1).
a(9) = 9 due to the circular permutations
   (0,6,3,9,2,4,7,5,8,1), (0,6,3,9,2,5,8,4,7,1),
   (0,6,3,9,2,7,4,5,8,1), (0,6,3,9,5,8,4,2,7,1),
   (0,6,3,9,5,8,4,7,2,1), (0,6,3,9,8,4,2,7,5,1),
   (0,6,3,9,8,4,7,2,5,1), (0,6,3,9,8,5,4,2,7,1),
   (0,6,3,9,8,5,4,7,2,1).
a(20) > 0 due to the circular permutation
  (0,3,12,9,15,18,6,20,19,14,13,4,2,7,16,17,11,10,5,8,1).

Mathematica

( * A program to compute required circular permutations for n = 7. *)
p[i_, j_]:=tp[i, j]=PrimeQ[2(i^2+j)+1]
V[i_]:=Part[Permutations[{1, 2, 3, 4, 5, 6, 7}], i]
m=0
Do[Do[If[p[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!}]

Crossrefs

Cf. A228917, A228956, A229005, A229038.

Sequence in context: A064956 A348745 A110449 * A176363 A067991 A114600

Adjacent sequences: A229079 A229080 A229081 * A229083 A229084 A229085

Keyword

nonn,more,hard

Author

Zhi-Wei Sun, Sep 13 2013

Extensions

a(10)-a(23) from Alois P. Heinz, Sep 13 2013

Status

approved