%I #25 Aug 04 2019 05:53:19
%S 1,1,1,0,2,3,7,11,9,5,41,82,254,2412,9524,13925,85318,220818,1662421,
%T 10496784,20690118,97200566,460358077
%N 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.
%C Conjecture: a(n) > 0 except for n = 4.
%C 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.
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1309.1679">Some new problems in additive combinatorics</a>, preprint, arXiv:1309.1679 [math.NT], 2013-2014.
%e a(1) = 1 due to the circular permutation (0,1).
%e a(2) = 1 due to the circular permutation (0,2,1).
%e a(3) = 1 due to the circular permutation (0,3,2,1).
%e a(5) = 2 due to the circular permutations
%e (0,3,2,4,5,1) and (0,3,5,4,2,1).
%e a(6) = 3 due to the circular permutations
%e (0,3,6,5,4,2,1), (0,6,3,2,4,5,1), (0,6,3,5,4,2,1).
%e a(7) = 7 due to the circular permutations
%e (0,3,6,5,4,2,7,1), (0,3,6,5,4,7,2,1), (0,6,3,2,4,7,5,1),
%e (0,6,3,2,5,4,7,1), (0,6,3,2,7,4,5,1), (0,6,3,5,4,2,7,1),
%e (0,6,3,5,4,7,2,1).
%e a(8) = 11 due to the circular permutations
%e (0,3,6,5,8,4,2,7,1), (0,3,6,5,8,4,7,2,1),
%e (0,3,6,8,4,2,7,5,1), (0,4,6,8,4,7,2,5,1),
%e (0,3,6,8,5,4,2,7,1), (0,3,6,8,5,4,7,2,1),
%e (0,6,3,2,4,7,5,8,1), (0,6,3,2,5,8,4,7,1),
%e (0,6,3,2,7,4,5,8,1), (0,6,3,5,8,4,2,7,1),
%e (0,6,3,5,8,4,7,2,1).
%e a(9) = 9 due to the circular permutations
%e (0,6,3,9,2,4,7,5,8,1), (0,6,3,9,2,5,8,4,7,1),
%e (0,6,3,9,2,7,4,5,8,1), (0,6,3,9,5,8,4,2,7,1),
%e (0,6,3,9,5,8,4,7,2,1), (0,6,3,9,8,4,2,7,5,1),
%e (0,6,3,9,8,4,7,2,5,1), (0,6,3,9,8,5,4,2,7,1),
%e (0,6,3,9,8,5,4,7,2,1).
%e a(20) > 0 due to the circular permutation
%e (0,3,12,9,15,18,6,20,19,14,13,4,2,7,16,17,11,10,5,8,1).
%t ( * A program to compute required circular permutations for n = 7. *)
%t p[i_,j_]:=tp[i,j]=PrimeQ[2(i^2+j)+1]
%t V[i_]:=Part[Permutations[{1,2,3,4,5,6,7}],i]
%t m=0
%t 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}];
%t 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!}]
%Y Cf. A228917, A228956, A229005, A229038.
%K nonn,more,hard
%O 1,5
%A _Zhi-Wei Sun_, Sep 13 2013
%E a(10)-a(23) from _Alois P. Heinz_, Sep 13 2013