A212286 Least k > 0 such that nk = x^3 + y^3 for nonnegative coprime x and y.

1, 3, 7, 7, 21, 4, 19, 1, 37, 31, 39, 5, 2, 57, 67, 73, 7, 7, 103, 6, 133, 133, 147, 157, 18, 7, 1, 211, 237, 7, 259, 273, 301, 1, 13, 10, 4, 9, 403, 421, 3, 8, 487, 19, 541, 553, 579, 11, 637, 651, 9, 703, 31, 757, 26, 9, 853, 871, 903, 13, 27, 2, 1027, 1, 1101, 11, 1159, 1191

Offset: 2

Author

Charles R Greathouse IV, Jun 12 2012

Keywords

• nonn

Comments

Broughan calls this eta(n) and proves that it exists for all n.

Examples

			4 is not the sum of two nonnegative cubes. 8 = 0^3 + 2^3, but 0 and 2 are not coprime. The least multiple of 4 that can be so represented is 28 = 1^3 + 3^3, so a(4) = 28/4.
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 2..1000
• Kevin A. Broughan, Characterizing the Sum of Two Cubes, J. Integer Seqs., Vol. 6, 2003.
• Index entries for sequences related to sums of cubes

Crossrefs

Cf. A004999, A212285, A202679.

Programs

PARI

sumOfTwoCubes(n)=my(k1=ceil((n-1/2)^(1/3)), k2=floor((4*n+1/2)^(1/3)), L); fordiv(n,d,if(d>=k1 && d<=k2 && denominator(L=(d^2-n/d)/3)==1 && issquare(d^2-4*L), return(1))); 0
sumOfTwoRPCubes(n)=if(sumOfTwoCubes(n),if(vecmax(factor(n)[,2])<3,1,for(x=ceil((n\2)^(1/3)),(n+.5)^(1/3),if(gcd(n,x)==1&&ispower(n-x^3,3),return(1)));0),0)
a(n)=forstep(k=n,2*n*(n^2+3),n,if(sumOfTwoRPCubes(k),return(k/n)))

Formula

a(n) <= 2n^2 + 6, a(a(n)) <= n.