A276920 Number of solutions to x^3 + y^3 + z^3 + t^3 == 0 (mod n) for 1 <= x, y, z, t <= n.

1, 8, 27, 72, 125, 216, 595, 704, 1539, 1000, 1331, 1944, 3133, 4760, 3375, 5632, 4913, 12312, 8911, 9000, 16065, 10648, 12167, 19008, 16125, 25064, 45927, 42840, 24389, 27000, 35371, 47104, 35937, 39304, 74375, 110808, 58645, 71288, 84591, 88000

Offset

1,2

Comments

a(n) = n^3 if n is in A074243. - Robert Israel, Oct 13 2016

Links

Chai Wah Wu, Table of n, a(n) for n = 1..10000 (terms n = 1..242 from Robert Israel)

Example

For n = 3, we see that all nondecreasing solutions {x, y, z, t} are in {{1, 1, 1, 3}, {1, 1, 2, 2}, {1, 2, 3, 3}, {2, 2, 2, 3}, {3, 3, 3, 3}}. The numbers in the sets can be ordered in 4, 6, 12, 4 and 1 ways respectively. Therefore, a(3) = 4 + 6 + 12 + 4 + 1 = 27. - David A. Corneth, Oct 11 2016

Maple

CF:= table([[false, false, true] = 12, [true, false, false] = 12, [true, false, true] = 6, [false, false, false] = 24, [true, true, true] = 1, [false, true, true] = 4, [false, true, false] = 12, [true, true, false] = 4]):
f1:= proc(n)
  option remember;
  local count, t, x, y, z, signature;
  if isprime(n) and n mod 3 = 2 then return n^3 fi;
  count:= 0;
  for t from 1 to n do
    for x from 1 to t do
      for y from 1 to x do
        for z from 1 to y do
          if t^3 + x^3 + y^3 + z^3 mod n = 0 then
            signature:= map(evalb, [z=y, y=x, x=t]);
            count:= count + CF[signature];
          fi
  od od od od;
  count
end proc:
f:= proc(n) local t;
    mul(f1(t[1]^t[2]), t=ifactors(n)[2])
end proc:
map(f, [$1..40]); # Robert Israel, Oct 13 2016

Mathematica

JJJ[4, n, lam] = Sum[If[Mod[a^3 + b^3 + c^3 + d^3, n] == Mod[lam, n], 1, 0], {d, 0, n - 1}, {a, 0, n - 1}, {b, 0, n - 1}, {c, 0 , n - 1}]; Table[JJJ[4, n, 0], {n, 1, 50}]

Prog

(PARI) a(n) = sum(x=1, n, sum(y=1, n, sum(z=1, n, sum(t=1, n, Mod(x, n)^3 + Mod(y, n)^3 + Mod(z, n)^3 + Mod(t, n)^3 == 0)))); \\ Michel Marcus, Oct 11 2016
(PARI) qperms(v) = {my(r=1, t); v = vecsort(v); for(i=1, #v-1, if(v[i]==v[i+1], t++, r*=binomial(i, t+1); t=0)); r*=binomial(#v, t+1)}
a(n) = {my(t=0); forvec(v=vector(4, i, [1, n]), if(sum(i=1, 4, Mod(v[i], n)^3)==0, t+=qperms(v)), 1); t} \\ David A. Corneth, Oct 11 2016
(Python)
def A276920(n):
    ndict = {}
    for i in range(n):
        i3 = pow(i, 3, n)
        for j in range(i+1):
            j3 = pow(j, 3, n)
            m = (i3+j3) % n
            if m in ndict:
                if i == j:
                    ndict[m] += 1
                else:
                    ndict[m] += 2
            else:
                if i == j:
                    ndict[m] = 1
                else:
                    ndict[m] = 2
    count = 0
    for i in ndict:
        j = (-i) % n
        if j in ndict:
            count += ndict[i]*ndict[j]
    return count # Chai Wah Wu, Jun 06 2017

Crossrefs

Cf. A000189, A047726, A060839, A063454, A074243, A087412, A087786, A254073, A276919.

Sequence in context: A372227 A246992 A351601 * A193006 A212745 A229423

Adjacent sequences: A276917 A276918 A276919 * A276921 A276922 A276923

Keyword

nonn,mult

Author

José María Grau Ribas, Sep 22 2016

Status

approved