A024670 Numbers that are sums of 2 distinct positive cubes.

9, 28, 35, 65, 72, 91, 126, 133, 152, 189, 217, 224, 243, 280, 341, 344, 351, 370, 407, 468, 513, 520, 539, 559, 576, 637, 728, 730, 737, 756, 793, 854, 855, 945, 1001, 1008, 1027, 1064, 1072, 1125, 1216, 1241, 1332, 1339, 1343, 1358, 1395, 1456, 1512, 1547, 1674

Offset: 1

Author

Clark Kimberling

Keywords

• nonn

Comments

This sequence contains no primes since x^3+y^3=(x^2-x*y+y^2)*(x+y). - M. F. Hasler, Apr 12 2008

There are no terms == 3, 4, 5 or 6 mod 9. - Robert Israel, Oct 07 2014

a(n) mod 2: {1,0,1,1,0,1,0,1,0,1,1,0,1,0,1,0,1,0,1,0,1,0,1,1,0,1,0,0,1,0,1,0,1,1,1,0, ...} - Daniel Forgues, Sep 27 2018

Examples

			9 is in the sequence since 2^3 + 1^3 = 9.
35 is in the sequence since 3^3 + 2^3 = 35.
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..10000 (terms 1..902 from M. F. Hasler)
• Index to sequences related to sums of cubes

Crossrefs

See also: Sums of 2 positive cubes (not necessarily distinct): A003325. Sums of 3 distinct positive cubes: A024975. Sums of distinct positive cubes: A003997. Sums of 2 distinct nonnegative cubes: A114090. Sums of 2 nonnegative cubes: A004999. Sums of 2 distinct positive squares: A004431. Cubes: A000578.

Cf. A373971 (characteristic function).

Indices of nonzero terms in A025468.

Programs

Maple

N:= 10000: # to get all terms <= N
S:= select(`<=`,{seq(seq(i^3 + j^3, j = 1 .. i-1), i = 2 .. floor(N^(1/3)))},N);
# if using Maple 11 or earlier, uncomment the next line
# sort(convert(S,list));
# Robert Israel, Oct 07 2014

Mathematica

lst={};Do[Do[x=a^3;Do[y=b^3;If[x+y==n,AppendTo[lst,n]],{b,Floor[(n-x)^(1/3)],a+1,-1}],{a,Floor[n^(1/3)],1,-1}],{n,6!}];lst (* Vladimir Joseph Stephan Orlovsky, Jan 22 2009 *)
Select[Range@ 1700, Total@ Boole@ Map[And[! MemberQ[#, 0], UnsameQ @@ #] &, PowersRepresentations[#, 2, 3]] > 0 &] (* Michael De Vlieger, May 13 2017 *)

PARI

isA024670(n)=for( i=ceil(sqrtn( n\2+1,3)),sqrtn(n-.5,3), isA000578(n-i^3) & return(1)) /* One could also use "for( i=2,sqrtn( n\2-1,3),...)" but this is much slower since there are less cubes in [n/2,n] than in [1,n/2]. Replacing the -1 here by +.5 would yield A003325, allowing for a(n)=x^3+x^3. Replacing -1 by 0 may miss some a(n) of this form due to rounding errors. - M. F. Hasler, Apr 12 2008 */

Python

from itertools import count, takewhile
def aupto(limit):
    cbs = list(takewhile(lambda x: x <= limit, (i**3 for i in count(1))))
    sms = set(c+d for i, c in enumerate(cbs) for d in cbs[i+1:])
    return sorted(s for s in sms if s <= limit)
print(aupto(1674)) # Michael S. Branicky, Sep 28 2021

Extensions

Name edited by Zak Seidov, May 31 2011