A090027 Number of distinct lines through the origin in 5-dimensional cube of side length n.

0, 31, 211, 961, 2851, 7471, 15541, 31471, 55651, 95821, 152041, 239791, 351331, 517831, 723241, 1007041, 1352041, 1821721, 2359051, 3082921, 3904081, 4956901, 6151651, 7677901, 9334261, 11445361, 13746181, 16566691, 19644031, 23432851, 27408331, 32333581

Offset: 0

Author

Joshua Zucker, Nov 25 2003

Keywords

• nonn

Comments

Equivalently, number of lattice points where the GCD of all coordinates = 1.

Examples

			a(2) = 211 because the 211 points with at least one coordinate=2 all make distinct lines and the remaining 31 points and the origin are on those lines.
		

Crossrefs

Cf. A000225, A001047, A060867, A090020, A090021, A090022, A090023, A090024 are for n dimensions with side length 1, 2, 3, 4, 5, 6, 7, 8, respectively. A049691, A090025, A090026, A090027, A090028, A090029 are this sequence for 2, 3, 4, 5, 6, 7 dimensions. A090030 is the table for n dimensions, side length k.

Programs

Mathematica

aux[n_, k_] := If[k == 0, 0, (k + 1)^n - k^n - Sum[aux[n, Divisors[k][[i]]], {i, 1, Length[Divisors[k]] - 1}]];lines[n_, k_] := (k + 1)^n - Sum[Floor[k/i - 1]*aux[n, i], {i, 1, Floor[k/2]}] - 1;Table[lines[5, k], {k, 0, 40}]

Python

from functools import lru_cache
@lru_cache(maxsize=None)
def A090027(n):
    if n == 0:
        return 0
    c, j = 1, 2
    k1 = n//j
    while k1 > 1:
        j2 = n//k1 + 1
        c += (j2-j)*A090027(k1)
        j, k1 = j2, n//j2
    return (n+1)**5-c+31*(j-n-1) # Chai Wah Wu, Mar 30 2021

Formula

a(n) = A090030(5, n).

a(n) = (n+1)^5 - 1 - Sum_{j=2..n+1} a(floor(n/j)). - Chai Wah Wu, Mar 30 2021