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A158649
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Number of integral solutions to the equation (x_1)^3 + ... + (x_n)^3 = (x_1 + ... + x_n)^2 with 1 <= x_1 <= ... <= x_n.
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9
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1, 1, 2, 2, 4, 5, 18, 30, 94, 226, 715, 2024, 6546, 20622, 69459, 232406, 810943, 2828246
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OFFSET
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0,3
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COMMENTS
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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
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REFERENCES
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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.
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LINKS
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Max A. Alekseyev, Problem 3766, Crux Mathematicorum 38(7) (2012), 284-287.
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.
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FORMULA
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EXAMPLE
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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).
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MATHEMATICA
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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];
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CROSSREFS
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KEYWORD
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more,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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