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A319480 Number of ordered pairs (i,j) with 0 < i < j < prime(n)/2 such that R(i^2,prime(n)) > R(j^2,prime(n)), where R(k,p) (with p an odd prime and k an integer) denotes the unique integer r among 0,1,...,(p-1)/2 for which k is congruent to r or -r modulo p. 5

%I #22 Sep 20 2018 06:59:58

%S 0,0,1,3,7,10,14,19,41,42,74,79,85,100,154,163,207,224,245,309,318,

%T 342,449,536,590,553,581,715,738,856,912,1085,1037,1324,1229,1477,

%U 1442,1491,1785,1730,1952,1986,2240,2316,2191,2474,2748,2836,3176

%N Number of ordered pairs (i,j) with 0 < i < j < prime(n)/2 such that R(i^2,prime(n)) > R(j^2,prime(n)), where R(k,p) (with p an odd prime and k an integer) denotes the unique integer r among 0,1,...,(p-1)/2 for which k is congruent to r or -r modulo p.

%C Conjecture: Let p be any odd prime and let N(p) be the number of ordered pairs (i,j) with 0 < i < j < p/2 and R(i^2,p) > R(j^2,p). Then N(p) == floor((p+1)/8) (mod 2).

%C See also A319311 for a similar conjecture.

%H Zhi-Wei Sun, <a href="/A319480/b319480.txt">Table of n, a(n) for n = 2..1000</a>

%e a(3) = 0 since prime(3) = 5 and R(1^2,5) = 1 = R(2^2,5).

%e a(4) = 1 since prime(4) = 7, R(1^2,7) = 1 < R(2^2,7) = 3, R(1^2,7) < R(3^2,7) = 2, and R(2^2,7) = 3 > R(3^2,7) = 2.

%t R[k_,p_]:=R[k,p]=Abs[Mod[k,p,-p/2]];

%t t[p_]:=t[p]=Sum[Boole[R[i^2,p]>R[j^2,p]],{j,2,(p-1)/2},{i,1,j-1}]; Table[t[Prime[n]],{n,2,50}]

%Y Cf. A000040, A000290, A319311.

%K nonn

%O 2,4

%A _Zhi-Wei Sun_, Sep 20 2018

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