A254073 Number of solutions to x^3 + y^3 + z^3 == 1 (mod n) for 1 <= x, y, z <= n.

1, 4, 9, 16, 25, 36, 90, 64, 162, 100, 121, 144, 252, 360, 225, 256, 289, 648, 468, 400, 810, 484, 529, 576, 625, 1008, 1458, 1440, 841, 900, 1143, 1024, 1089, 1156, 2250, 2592, 1602, 1872, 2268, 1600, 1681, 3240, 2115, 1936, 4050, 2116, 2209, 2304, 4410

Offset: 1

Author

José María Grau Ribas, Jan 25 2015

Keywords

• nonn
• mult

Comments

It appears that a(n) = n^2 for n in A088232 (numbers n such that 3 does not divide phi(n)) and that a(n) != n^2 for n in A066498 (numbers n such that 3 divides phi(n)). - Michel Marcus, Mar 13 2015

It appears that a(p) != p^2 for primes in A002476 (primes of form 6m + 1). - Michel Marcus, Mar 13 2015

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A087412.

Programs

Mathematica

a[n_] := Sum[ If[ Mod[x^3 + y^3 + z^3, n] == 1, 1, 0], {x, n}, {y, n}, {z, n}]; a[1]=1; Table[a[n], {n, 2,22}]

PARI

a(n) = {nb = 0; for (x=1, n, for (y=1, n, for (z=1, n, if ((Mod(x^3,n) + Mod(y^3,n) + Mod(z^3,n)) % n == Mod(1, n), nb ++);););); nb;} \\ Michel Marcus, Mar 11 2015

PARI

a(n)={my(p=Mod(sum(i=0, n-1, x^(i^3%n)), 1-x^n)^3); polcoeff(lift(p), 1%n)} \\ Andrew Howroyd, Jul 18 2018

Python

def A254073(n):
    ndict = {}
    for i in range(n):
        m = pow(i,3,n)
        if m in ndict:
            ndict[m] += 1
        else:
            ndict[m] = 1
    count = 0
    for i in ndict:
        ni = ndict[i]
        for j in ndict:
            k = (1-i-j) % n
            if k in ndict:
                count += ni*ndict[j]*ndict[k]
    return count # Chai Wah Wu, Jun 06 2017