|
%I #60 Nov 11 2018 18:28:41
%S 1,1,2,2,4,5,18,30,94,226,715,2024,6546,20622,69459,232406,810943,
%T 2828246
%N Number of integral solutions to the equation (x_1)^3 + ... + (x_n)^3 = (x_1 + ... + x_n)^2 with 1 <= x_1 <= ... <= x_n.
%C One prominent solution is x_i = i; another obvious one is x_i = n.
%C 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).
%C There is only one solution with pairwise distinct x_i, it has x_i = i for all i. - _Max Alekseyev_, Sep 07 2010
%C x_1 + ... + x_n != 3k + 2. - _David A. Corneth_, Nov 06 2018
%D Titu Andreescu and Dorin Andrica, An Introduction To Diophantine Equations, 2002, GIL Publishing House, pp. 38, example 5.
%D Peter Giblin, Primes and Programming, 1993, Cambridge University Press. See chapter 9, exercise 1.7.
%H Max A. Alekseyev, <a href="https://cms.math.ca/crux/v38/n7/Problems_38_7.pdf">Problem 3766</a>, Crux Mathematicorum 38(7) (2012), 284-287.
%H Edward Barbeau and Samer Seraj, <a href="http://arxiv.org/abs/1306.5257">Sum of cubes is square of sum</a>, arXiv:1306.5257 [math.NT], 2013.
%H John Mason, <a href="http://www.jstor.org/stable/3620469">Generalising 'sums of cubes equal to squares of sums'</a>, The Mathematical Gazette 85:502 (2001), pp. 50-58.
%H Alasdair McAndrew, <a href="http://amca01.wordpress.com/2011/01/10/a-cute-result-relating-to-sums-of-cubes/">A cute result relating to sums of cubes</a> (2011)
%H David Pagni, <a href="http://www.jstor.org/stable/3620410">82.27 An interesting number fact</a>, The Mathematical Gazette 82:494 (1998), pp. 271-273.
%H C. Rivera, <a href="http://www.primepuzzles.net/puzzles/puzz_158.htm">Puzzle 158. Sum of Cubes equal to Square of Sum</a> (2001)
%H Greg Ross, <a href="http://www.futilitycloset.com/2011/01/08/hocus-pocus/">Hocus Pocus</a>
%H 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. <a href="http://www.fq.math.ca/Scanned/15-1/utz-a.pdf">Part 1</a>, <a href="http://www.fq.math.ca/Scanned/15-1/utz-b.pdf">part 2</a>.
%F A001055(n) <= a(n) << e^n n^(n/3). - _Charles R Greathouse IV_, May 24 2013
%e 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).
%t a[0] = a[1] = 1;
%t 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];
%t Table[an = a[n]; Print["a(", n, ") = ", an]; an, {n, 0, 10}] (* _Jean-François Alcover_, Nov 06 2018 *)
%Y Cf. A227847.
%K more,nonn
%O 0,3
%A _Jens Voß_, Mar 23 2009
%E Edited by _Max Alekseyev_, Aug 18 2010
%E a(12)-a(13) from _Max Alekseyev_, Aug 20 2010
%E a(14) from _Max Alekseyev_, Sep 07 2010
%E a(15)-a(17) from _Charles R Greathouse IV_, Jun 05 2013
|
