A087412 a(n) is the number of solutions to x^3 + y^3 == 1 (mod n).

1, 2, 3, 4, 5, 6, 6, 8, 18, 10, 11, 12, 6, 12, 15, 16, 17, 36, 24, 20, 18, 22, 23, 24, 25, 12, 54, 24, 29, 30, 33, 32, 33, 34, 30, 72, 24, 48, 18, 40, 41, 36, 33, 44, 90, 46, 47, 48, 42, 50, 51, 24, 53, 108, 55, 48, 72, 58, 59, 60

Offset: 1

Author

Yuval Dekel (dekelyuval(AT)hotmail.com), Oct 21 2003

Keywords

• mult
• nonn

Links

• Andrew Howroyd, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A087786.

Programs

Mathematica

a[n_] := Module[{v = Table[0, {n}]}, For[i = 0, i <= n-1, i++, v[[Mod[i^3, n] + 1]]++]; Sum[v[[i+1]] v[[Mod[1-i, n] + 1]], {i, 0, n-1}]];
a /@ Range[1, 60] (* Jean-François Alcover, Sep 17 2019, after Andrew Howroyd *)

PARI

a(n) = {nb = 0; for (x = 0, n-1, for (y = 0, n-1, if (Mod(x^3,n) + Mod(y^3,n) == Mod(1, n), nb++););); nb;} \\ Michel Marcus, Aug 06 2013

PARI

a(n)={my(v=vector(n)); for(i=0, n-1, v[i^3%n + 1]++); sum(i=0, n-1, v[i+1]*v[(1-i)%n + 1])} \\ Andrew Howroyd, Jul 17 2018

Formula

From Andrew Howroyd, Jul 17 2018: (Start)

a(p^e) = p^e for p prime and p mod 3 = 2.

Conjecture: a(3^e) = 2*3^e for e > 1.

a(p^e) = p^(e-1)*(p - 1 + Sum_{b=1..p-1} Legendre(12*b^(-1) - 3*b^2, p)) for p prime and p <> 3.

The final formula arises from factoring x^3 + y^3 as (x + y)*(x*2 - x*y + y^2), then substituting b = x + y and counting the solutions to the resulting quadratic equation 3*x^2 - 3*b*x + b^2 == b^(-1) (mod p) for each nonzero value of b. (End)

Extensions

More terms from Michel Marcus, Aug 06 2013