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 A319311 Number of ordered pairs (i,j) with 0 < i < j < prime(n)/2 such that (i^2 mod prime(n)) > (j^2 mod prime(n)). 9
 0, 0, 1, 4, 3, 9, 14, 22, 28, 40, 53, 73, 86, 101, 116, 168, 153, 234, 260, 246, 299, 362, 365, 435, 523, 583, 612, 559, 652, 835, 952, 918, 1022, 1154, 1286, 1237, 1486, 1554, 1489, 1730, 1694, 1975, 1889, 2078, 2241, 2520, 2672, 2996, 2784, 2892, 3148, 3058, 3488, 3570, 4023, 3881, 4222, 4087, 4363 (list; graph; refs; listen; history; text; internal format)
 OFFSET 2,4 COMMENTS Conjecture: Let p be an odd prime and let s(p) be the number of ordered pairs (i,j) with 0 < i < j < p/2 and (i^2 mod p) > (j^2 mod p). Then s(p) is even when p == 3 (mod 8). If p == 7 (mod 8), then s(p) == (h(-p)+1)/2 (mod 2), where h(-p) is the class number of the imaginary quadratic field Q(sqrt(-p)). We have verified this conjecture for all primes p < 50000 with p == 3 (mod 4). The conjecture was confirmed by the author in the preprint arXiv:1809.07766v4. - Zhi-Wei Sun, Oct 03 18 LINKS Zhi-Wei Sun, Table of n, a(n) for n = 2..2500 Zhi-Wei Sun, Quadratic residues and related permutations, arXiv:1809.07766 [math.NT], 2018. EXAMPLE a(4) = 1 since prime(4) = 7, and (2,3) is the only ordered pair (i,j) with 0 < i < j < 7/2 and (i^2 mod 7) > (j^2 mod 7). a(5) = 4 since prime(5) = 11, and the only ordered pairs (i,j) with 0 < i < j < 11/2 and (i^2 mod 11) > (j^2 mod 11) are (2,5), (3,4), (3,5) and (4,5). MATHEMATICA s[p_]:=s[p]=Sum[Boole[Mod[i^2, p]>Mod[j^2, p]], {j, 2, (p-1)/2}, {i, 1, j-1}]; Table[s[Prime[n]], {n, 2, 60}] PROG (PARI) a(n) = my(p=prime(n), c=0); for(j=2, p/2, for(i=1, j-1, if((i^2%p) > (j^2%p), c++))); c \\ Felix FrÃ¶hlich, Oct 04 2018 CROSSREFS Cf. A000040, A000290. Sequence in context: A197694 A187770 A103218 * A107381 A242531 A275160 Adjacent sequences:  A319308 A319309 A319310 * A319312 A319313 A319314 KEYWORD nonn AUTHOR Zhi-Wei Sun, Sep 16 2018 STATUS approved

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Last modified February 21 22:10 EST 2020. Contains 332113 sequences. (Running on oeis4.)