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A234319
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Smallest sum of n-th powers of k+1 consecutive positive integers that equals the sum of n-th powers of the next k consecutive integers, or -n if none.
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2
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0, 3, 25, -3, -4, -5, -6, -7, -8, -9, -10, -11, -12, -13, -14, -15, -16, -17, -18, -19, -20, -21, -22, -23, -24, -25, -26, -27, -28, -29, -30, -31, -32, -33, -34, -35, -36, -37, -38, -39, -40, -41, -42, -43, -44, -45, -46, -47, -48, -49, -50, -51, -52, -53, -54
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OFFSET
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0,2
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COMMENTS
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a(n) is the smallest solution to m^n + (m+1)^n + ... + (m+k)^n = (m+k+1)^n + (m+k+2)^n + ... + (m+2*k)^n, or -n if no solution.
In 1879 Dostor gave all solutions for n = 2. In particular, a(2) = 25.
In 1906 Collignon proved that no solution exists for n = 3 and 4, so a(3) = -3 and a(4) = -4.
In 2013 Felten and Müller-Stach claimed to prove that no solution exists when n > 2, so if their proof is correct, a(n) = -n for n >= 3.
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REFERENCES
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Edouard Collignon, Note sur la résolution en entiers de m^2 + (m-r)^2 + ... + (m-kr)^2 = (m+r)^2 + ... + (m+kr)^2, Sphinx-Oedipe, 1 (1906-1907), 129-133.
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LINKS
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FORMULA
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a(n) = -n for n > 2.
G.f.: x*(27*x^3-50*x^2+19*x+3) / (x-1)^2. - Colin Barker, Apr 23 2014
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EXAMPLE
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m^0 + (m+1)^0 + ... + (m+k)^0 = k+1 > k = (m+k+1)^0 + (m+k+2)^0 + ... + (m+2*k)^0 for m > 0, so a(0) = -0 = 0.
1^1 + 2^1 = 3 = 3^1 is minimal for n = 1, so a(1) = 3.
3^2 + 4^2 = 25 = 5^2 is minimal for n = 2, so a(2) = 25.
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MATHEMATICA
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CoefficientList[Series[x*(27*x^3 - 50*x^2 + 19*x + 3)/(x - 1)^2, {x, 0, 50}], x] (* Wesley Ivan Hurt, Jun 21 2014 *)
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PROG
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(PARI) Vec(x*(27*x^3-50*x^2+19*x+3)/(x-1)^2 + O(x^100)) \\ Colin Barker, Apr 23 2014
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CROSSREFS
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KEYWORD
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sign,easy
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AUTHOR
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STATUS
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approved
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