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%I #13 Aug 07 2018 16:03:29
%S 1,2,1,1,1,2,2,1,1,2,3,1,3,4,1,1,4,2,5,1,2,6,6,1,2,6,2,2,7,2,8,1,3,8,
%T 2,1,9,10,3,1,10,4,11,3,1,12,12,1,8,4,4,3,13,4,3,2,5,14,15,1,15,16,2,
%U 1,3,6,17,4,6,4,18,1,18,18,2,5,6,6,20,1,4,20
%N a(n) is the minimum number of occurrences of the same value (r0-r1+n) mod n for all pairs (r0,r1) of quadratic residues mod n.
%C Multiplicative by the Chinese remainder theorem since if gcd(m,n) = 1 then r is a quadratic residue mod m*n iff it is a quadratic residue mod m and a quadratic residue mod n. - _Andrew Howroyd_, Aug 07 2018
%H Andrew Howroyd, <a href="/A316975/b316975.txt">Table of n, a(n) for n = 1..1000</a>
%e The quadratic residues mod 10 are 0, 1, 4, 5, 6 and 9. For each pair (r0,r1) of these quadratic residues, we compute (r0-r1+10) mod 10, leading to:
%e 0 1 4 5 6 9
%e +------------
%e 0 | 0 9 6 5 4 1
%e 1 | 1 0 7 6 5 2
%e 4 | 4 3 0 9 8 5
%e 5 | 5 4 1 0 9 6
%e 6 | 6 5 2 1 0 7
%e 9 | 9 8 5 4 3 0
%e The minimum number of occurrences of the same value in the above table is 2 (for 2, 3, 7 and 8). Therefore a(10) = 2.
%t Table[Min@ Tally[#][[All, -1]] &@ Flatten[Mod[#, n] & /@ Outer[Subtract, #, #]] &@ Union@ PowerMod[Range@ n, 2, n], {n, 82}] (* _Michael De Vlieger_, Jul 20 2018 *)
%o (PARI) a(n)={my(r=Set(vector(n,i,i^2%n))); my(v=vector(n)); for(i=1, #r, for(j=1, #r, v[(r[i]-r[j])%n + 1]++)); vecmin(select(t->t>0, v))} \\ _Andrew Howroyd_, Aug 07 2018
%Y Cf. A096008.
%K nonn,mult
%O 1,2
%A _Arnauld Chevallier_, Jul 17 2018
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