cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A328204 Numbers of quadruples contained in the divisors of A328149(n).

Original entry on oeis.org

1, 1, 2, 2, 2, 2, 3, 3, 2, 6, 2, 4, 4, 2, 3, 4, 4, 3, 2, 10, 2, 2, 4, 6, 4, 2, 5, 4, 2, 10, 2, 5, 6, 2, 4, 6, 4, 2, 2, 14, 3, 4, 4, 4, 4, 2, 6, 1, 8, 2, 11, 2, 4, 6, 4, 4, 6, 4, 2, 4, 17, 2, 2, 4, 6, 4, 4, 1, 8, 4, 2, 12, 2, 9, 6, 2, 6, 4, 4, 4, 2, 18, 3, 2, 6
Offset: 1

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Author

Michel Lagneau, Jun 07 2020

Keywords

Comments

A quadruple (x, y, z, w) of A328149 is a set of positive integers that satisfy x^3 + y^3 + z^3 = w^3.

Examples

			a(7) = 3 because the set of divisors of A328149(7) = 240: {1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30, 40, 48, 60, 80, 120, 240} contains the three quadruples {3, 4, 5, 6}, {6, 8, 10, 12} and {12, 16, 20, 24}. The first quadruple is primitive.
		

Crossrefs

Programs

  • Maple
    with(numtheory):
    for n from 3 to 3000 do :
       d:=divisors(n):n0:=nops(d):it:=0:
        for i from 1 to n0-3 do:
         for j from i+1 to n0-2 do :
          for k from j+1 to n0-1 do:
          for m from k+1 to n0 do:
           if d[i]^3 + d[j]^3 + d[k]^3 = d[m]^3
            then
            it:=it+1:
            else
           fi:
          od:
         od:
        od:
        od:
        if it>0 then
        printf(`%d, `,it):
        else fi:
       od:
  • Mathematica
    nq[n_] := If[Mod[n, 6] > 0, 0, Block[{t, u, v, c = 0, d = Divisors[n], m}, m = Length@ d; Do[t = d[[i]]^3 + d[[j]]^3; Do[u = t + d[[h]]^2; If[u > n^3, Break[]]; If[Mod[n^3, u] == 0 && IntegerQ[v = u^(1/3)] && Mod[n, v] == 0, c++], {h, j+1, m-1}], {i, m-3}, {j, i+1, m - 2}]; c]]; Select[Array[nq, 1638], # > 0 &] (* program from Giovanni Resta adapted for the sequence. See A330894 *)
  • PARI
    isok(n) = {my(d=divisors(n), nb=0, m); if (#d > 3, for (i=1, #d-3, for (j=i+1, #d-2, for (k=j+1, #d-1, if (ispower(d[i]^3+d[j]^3+d[k]^3, 3, &m) && !(n%m), nb++););););); nb;}
    lista(nn) = my(m); for (n=1, nn, if (m=isok(n), print1(m, ", "))); \\ Michel Marcus, Nov 15 2020