cp's OEIS Frontend

Numbers n such that n^3+1 is expressible as the sum of two nonzero cubes (both greater than 1).

Values of z associated with A050794.

Sequence is infinite. One subsequence is (from x = 1 + 9 m^3, y = 9 m^4, z = 3*m*(3*m^3 + 1), x^3 + y^3 = z^3 + 1): z(m) = 3*m*(3*m^3 + 1) = {12, 150, 738, 2316, 5640, 11682, 21630, 36888, 59076, 90030, ...} = a (1, 3, 6, 10, 13, 17, 20, 23, 30, 37, ...). - Zak Seidov, Sep 16 2013

Numbers n such that n^3+1 is a member of A001235. - Altug Alkan, May 09 2016

From Fred W. Helenius (fredh(AT)ix.netcom.com), Jul 22 2008: (Start)

There is an infinite family of solutions to c^3+1=a^3+b^3 given by

(a,b,c) = (9n^3 + 1, 9n^4, 9n^4 + 3n). The present sequence actually asks about

x^3 + y^3 = z^3 - 1 with x < y < z; for that we can take

(x,y,z) = (9n^3 - 1, 9n^4 - 3n, 9n^4) for n > 1.

I extracted these solutions from Theorem 235 in Hardy & Wright; the result shown there is that all nontrivial rational solutions of

x^3 + y^3 = u^3 + v^3 are given by

x = r(1 - (a - 3b)(a^2 + 3b^2))

y = r((a + 3b)(a^2 + 3b^2) - 1)

u = r((a + 3b) - (a^2 + 3b^2)^2)

v = r((a^2 + 3b^2)^2 - (a - 3b))

where r,a,b are rational and r is not zero.

Specializing to r = 1, b = n/2 and a = 3n/2 gives

x = 1, y = 9n^3 - 1, u = 3n - 9n^4, v = 9n^4.

The solutions given above are obtained by changing signs and moving cubes from one side of the equation to the other as necessary.

Unfortunately, not all integral solutions are found so easily: the third value in A050788 corresponds to 135^3 + 138^3 = 172^3 - 1; this is not produced by such simple choices of r,a,b. (End)

It is easy to check that with x = 6n^2, y = 6n^3 - 1, and this z = 6n^3 + 1, it satisfies the Diophantine equation x^3 + y^3 = z^3 - 2. Thus these are near-misses for Fermat equation.

For n>2, it seems to be the only solution of x^n + y^n = z^n - 2 (or even that differ by 2 from FLT, see A050787 and A050791 for solutions that differ by 1). As 2 is not a cube, these solutions are not included in the theory for x^3 + y^3 = u^3 + v^3.

Values of a^3 + b^3 such that (a^3 + b^3)^3 - 1 is of the form x^3 + y^3 where a, b, x, y > 0.

38134 = 2*23*829 is the first term that is nonsquare. What are the next square terms of this sequence?

n is a member of A007412 and n^3 is a member of A003072, obviously.

Some results coming from the Alarcon and Duval reference.

For n = 0, there are infinitely many solutions because every triple (k,k,k) with k >= 0 satisfies the equation.

a(n) = 0 iff 3 divides n and 9 doesn't divide n (equivalent to n is in A016051).

When n belongs to A074232 (complement of A016051), a(n) is always a multiple of 3 because

1) if (a,a,b) [resp. (a,b,b)] with a < b is a primitive solution, then these triples generate 3 solutions with the permutations (a,a,b), (a,b,a), (b,a,a), [resp. (a,b,b), (b,b,a), (b,a,b)] and,

2) if (a,b,c) with a < b < c is a primitive solution, then this triple generates 6 solutions with the permutations (a,b,c), (b,c,a), (c,a,b), (a,c,b), (c,b,a), (b,a,c).

For prime p <> 3, a(p) = a(2*p) = 3.

An inequality: (n/4)^(1/3) <= max(x, y, z) <= (n+2)/3.

This sequence is unbounded.

A261029 gives the number of triples without counting the permutations and, in link, a list of primitive triples up to n = 2000.