A320044 - OEIS

%I #17 Dec 12 2018 14:24:33

%S 0,0,1,1,2,4,3,2,4,7,9,10,7,9,12,15,12,14,15,18,19,25,18,19,20,23,28,

%T 22,30,27,34,30,33,37,39,35,40,38,36,51,38,42,50,50,49,53,44,57,62,59,

%U 55,54,49,62,65,62,69,59,65,67,77,69,71,80,80,69,76,78,88,87,87,94,87,87,99,96,87,97,97,94

%N Number of positive integers k < prime(n)/2 with {k^3/prime(n)} > 1/2, where {x} = x - floor(x) is the fractional part of a real number x.

%C Conjecture 1: For any prime p == 5 (mod 6), the difference card{0 < k < p/2: {k^3/p} > 1/2} - (p+1)/6 is nonnegative and even.

%C Conjecture 2: For any prime p not congruent to 1 modulo 5, the number of positive integers k < p/2 with {k^5/p} > 1/2 is even.

%C Conjecture 3: For any prime p == 5 (mod 12), the difference card{0 < k < p/2: {k^6/p} > 1/2} - (p-5)/12 is positive and odd.

%H Zhi-Wei Sun, <a href="/A320044/b320044.txt">Table of n, a(n) for n = 1..10000</a>

%H Zhi-Wei Sun, <a href="https://arxiv.org/abs/1809.07766">Quadratic residues and related permutations and identities</a>, arXiv:1809.07766 [math.NT], 2018.

%e a(3) = 1 since prime(3) = 5 and {0 < k < 5/2: {k^3/5} > 1/2} = {2}.

%e a(4) = 1 since prime(4) = 7 and {0 < k < 7/2: {k^3/7} > 1/2} = {3}.

%e a(5) = 2 since prime(5) = 11 and {0 < k < 11/2: {k^3/11} > 1/2} = {2,4}.

%t s[p_]:=s[p]=Sum[Boole[Mod[k^3,p]>p/2],{k,1,(p-1)/2}];Table[s[Prime[n]],{n,1,80}]

%Y Cf. A000040, A000578, A000584, A001014.

%K nonn

%O 1,5

%A _Zhi-Wei Sun_, Oct 04 2018