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%I #16 Oct 01 2018 05:55:52
%S 0,0,0,6,3,5,10,22,51,62,58,53,100,146,194,200,185,246,242,310,374,
%T 344,422,497,540,582,652,683,768,946,916,1011,1180,1294,1108,1387,
%U 1592,1656,1829,2050,2048,2386,2365,2186,2184,2770,2902,2890,3296,3292,3754,3063,3562,3650,4184,4391,4164,4506,4812
%N Number of ordered pairs (i,j) with 0 < i < j < prime(n)/2 such that R(i^4,prime(n)) > R(j^4,prime(n)), where R(k,p) (with p an odd prime) denotes the unique integer r among 0,...,(p-1)/2 with k congruent to r or -r modulo p.
%C Conjecture: Let p be an odd prime, and let r(p) be the number of ordered pairs (i,j) with 0 < i < j < p/2 and R(i^4,p) > R(j^4,p), where R(k,p) denotes the unique integer r among 0,...,(p-1)/2 with k congruent to r or -r modulo p. Then r(p) is even if p == 3 (mod 4). Also, r(p) == (p-5)/8 (mod 2) if p == 5 (mod 8). When p == 1 (mod 8), r(p) is even if and only if 2 is a quartic residue modulo p.
%C See also A319311, A319480 and A319882 for similar conjectures.
%H Zhi-Wei Sun, <a href="/A319894/b319894.txt">Table of n, a(n) for n = 2..1000</a>
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1809.07766">Quadratic residues and related permutations</a>, arXiv:1809.07766 [math.NT], 2018.
%e a(6) = 3 since prime(6) = 13, (R(1^4,13),R(2^4,13),...,R(6^4,13)) = (1,3,3,9,1,9), and (2,5), (3,5) and (4,5) are only pairs (i,j) with 0 < i < j < 13/2 and R(i^4,13) > R(j^4,13).
%t f[k_,p_]:=f[k,p]=Abs[Mod[PowerMod[k,4,p],p,-p/2]];Inv[p_]:=Inv[p]=Sum[Boole[f[i,p]>f[j,p]],{j,2,(p-1)/2},{i,1,j-1}];Table[Inv[Prime[n]],{n,2,60}]
%Y Cf. A000040, A000583, A319311, A319480, A319882.
%K nonn
%O 2,4
%A _Zhi-Wei Sun_, Sep 30 2018
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