|
|
A025397
|
|
Numbers that are the sum of 3 positive cubes in exactly 3 ways.
|
|
9
|
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
LINKS
|
|
|
FORMULA
|
|
|
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 *)
|
|
PROG
|
(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
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|