A320044 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.

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

Offset

1,5

Comments

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.

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.

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.

Links

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Zhi-Wei Sun, Quadratic residues and related permutations and identities, arXiv:1809.07766 [math.NT], 2018.

Example

a(3) = 1 since prime(3) = 5 and {0 < k < 5/2: {k^3/5} > 1/2} = {2}.
a(4) = 1 since prime(4) = 7 and {0 < k < 7/2: {k^3/7} > 1/2} = {3}.
a(5) = 2 since prime(5) = 11 and {0 < k < 11/2: {k^3/11} > 1/2} = {2,4}.

Mathematica

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

Crossrefs

Cf. A000040, A000578, A000584, A001014.

Sequence in context: A152026 A345928 A060806 * A329876 A128868 A341073

Adjacent sequences: A320041 A320042 A320043 * A320045 A320046 A320047

Keyword

nonn

Author

Zhi-Wei Sun, Oct 04 2018

Status

approved