A025397 Numbers that are the sum of 3 positive cubes in exactly 3 ways.

5104, 9729, 12104, 12221, 12384, 14175, 17604, 17928, 19034, 20691, 21412, 21888, 24480, 28792, 29457, 30528, 31221, 32850, 34497, 35216, 36288, 38259, 39339, 39376, 40060, 40097, 40832, 40851, 41033, 41040, 41364, 41966, 42056, 42687, 43408, 45144

Offset: 1

Author

David W. Wilson

Keywords

• nonn

Links

• Donovan Johnson, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A008917, A025323, A025396, A025398, A025405, A025456, A343969, A344240, A344804.

Programs

Maple

N:= 10^5: # to get all terms <= N
Reps:= Matrix(N,3,(i,j) -> {}):
for i from 1 to floor(N^(1/3)) do
  Reps[i^3,1]:= {[i]}
od:
for j from 2 to 3 do
for i from 1 to floor(N^(1/3)) do
  for x from i^3+1 to N do
    Reps[x,j]:= Reps[x,j] union
      map(t -> if t[-1] <= i then [op(t),i] fi, Reps[x-i^3,j-1]);
  od
od
od:
select(t -> nops(Reps[t,3])=3, [$1..N]); # Robert Israel, Aug 28 2015

Mathematica

Reap[ For[ n = 1, n <= 50000, n++, pr = Select[ PowersRepresentations[n, 3, 3], Times @@ # != 0 &]; If[pr != {} && Length[pr] == 3, Print[n, pr]; Sow[n]]]][[2, 1]] (* Jean-François Alcover, Jul 31 2013 *)

PARI

is(n)=k=ceil((n-2)^(1/3)); d=0; for(a=1, k, for(b=a, k, for(c=b, k, if(a^3+b^3+c^3==n, d++)))); d
n=3; while(n<50000, if(is(n)==3, print1(n, ", ")); n++) \\ Derek Orr, Aug 27 2015

Formula

n such that A025456(n) = 3. - Robert Israel, Aug 28 2015