A158649 Number of integral solutions to the equation (x_1)^3 + ... + (x_n)^3 = (x_1 + ... + x_n)^2 with 1 <= x_1 <= ... <= x_n.

1, 1, 2, 2, 4, 5, 18, 30, 94, 226, 715, 2024, 6546, 20622, 69459, 232406, 810943, 2828246, 10133324, 36435912, 133372054, 490864448, 1829474693, 6858293057, 25951243663, 98749175365, 378522780879

Offset: 0

Author

Jens Voß, Mar 23 2009

Keywords

• more
• nonn

Comments

One prominent solution is x_i = i; another obvious one is x_i = n.

It is easy to show that in every solution (x_1, ..., x_n), the sum x_1 + ... + x_n <= n^2 and x_n <= n^(4/3).

There is only one solution with pairwise distinct x_i, it has x_i = i for all i. - Max Alekseyev, Sep 07 2010

x_1 + ... + x_n != 3k + 2. - David A. Corneth, Nov 06 2018

Examples

			a(4) = 4, since there are four solutions of length n=4: (1,2,2,4), (1,2,3,4), (2,2,4,4), and (4,4,4,4).
		

References

• Titu Andreescu and Dorin Andrica, An Introduction To Diophantine Equations, 2002, GIL Publishing House, pp. 38, example 5.
• Peter Giblin, Primes and Programming, 1993, Cambridge University Press. See chapter 9, exercise 1.7.

Links

• Max A. Alekseyev, Problem 3766, Crux Mathematicorum 38(7) (2012), 284-287.
• Edward Barbeau and Samer Seraj, Sum of cubes is square of sum, arXiv:1306.5257 [math.NT], 2013.
• John Mason, Generalising 'sums of cubes equal to squares of sums', The Mathematical Gazette 85:502 (2001), pp. 50-58.
• Alasdair McAndrew, A cute result relating to sums of cubes (2011)
• David Pagni, 82.27 An interesting number fact, The Mathematical Gazette 82:494 (1998), pp. 271-273.
• Carlos Rivera, Puzzle 158. Sum of Cubes equal to Square of Sum, The Prime Puzzles & Problems Connection.
• Greg Ross, Hocus Pocus, Futility Closet, January 2011.
• W. R. Utz, The Diophantine Equation (x_1 + x_2 + ... + x_n)^2 = x_1^3 + x_2^3 + ... + x_n^3, Fibonacci Quarterly 15:1 (1977), pp. 14, 16. Part 1, part 2.

Crossrefs

Cf. A227847.

Programs

Mathematica

a[0] = a[1] = 1;
a[n_] := Module[{x}, cnt = 0; xx = Array[x, n]; m = Floor[n^(4/3)]; x[0] = 1; iter = Table[{x[k], x[k-1], m}, {k, 1, n}]; Do[If[Total[xx] <= n^2, If[Total[xx^3] == Total[xx]^2, cnt++]], Sequence @@ iter // Evaluate]; cnt];
Table[an = a[n]; Print["a(", n, ") = ", an]; an, {n, 0, 10}] (* Jean-François Alcover, Nov 06 2018 *)

Formula

A001055(n) <= a(n) << e^n n^(n/3). - Charles R Greathouse IV, May 24 2013

Extensions

Edited by Max Alekseyev, Aug 18 2010

a(12)-a(13) from Max Alekseyev, Aug 20 2010

a(14) from Max Alekseyev, Sep 07 2010

a(15)-a(17) from Charles R Greathouse IV, Jun 05 2013

a(18)-a(26) from Sela Fried, Jul 10 2025