|
|
A334964
|
|
Numbers that are the sum of three coprime positive cubes
|
|
1
|
|
|
3, 10, 17, 29, 36, 43, 55, 62, 66, 73, 92, 99, 118, 127, 129, 134, 141, 153, 155, 160, 179, 190, 197, 216, 218, 225, 244, 251, 253, 258, 277, 281, 307, 314, 342, 345, 349, 352, 359, 368, 371, 378, 397, 405, 408, 415, 433, 434, 466, 469, 471, 476, 495, 514, 521, 532, 540, 547, 557, 560, 566, 567
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
The greatest common divisor of the three cubes must be 1, but they need not be pairwise coprime.
|
|
LINKS
|
|
|
EXAMPLE
|
a(3)=17 is in the sequence because 17 = 1^3 + 2^3 + 2^3 with gcd(1,2,2)=1.
|
|
MAPLE
|
N:= 1000: # for all terms <= N
S:= {seq(seq(seq(x^3+y^3+z^3, z=select(t -> igcd(x, y, t)=1, [$y..floor((N-x^3-y^3)^(1/3))])), y=x..floor(((N-x^3)/2)^(1/3))), x=1..floor((N/3)^(1/3)))}:
sort(convert(S, list));
|
|
PROG
|
(PARI) list(lim)=my(v=List(), s, g, x3); lim\=1; if(lim<3, return([])); for(x=1, sqrtnint(lim\3, 3), x3=x^3; for(y=x, sqrtnint((lim-x3)\2, 3), s=x3+y^3; g=gcd(x, y); if(g>1, for(z=y, sqrtnint(lim-s, 3), if(gcd(g, z)==1, listput(v, s+z^3))), for(z=y, sqrtnint(lim-s, 3), listput(v, s+z^3))))); Set(v) \\ Charles R Greathouse IV, May 18 2020
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|