A234319 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.

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

Offset: 0

Author

Jonathan Sondow, Dec 23 2013

Keywords

• sign
• easy

Comments

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.

Examples

			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.
		

References

• 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.

Links

• L. E. Dickson, History of the Theory of Numbers, II, p. 564.
• Georges Dostor, Question sur les nombres, Archiv der Mathematik und Physik, 64 (1879), 350-352.
• Simon Felten and Stefan Müller-Stach, A diophantine equation for sums of consecutive like powers, arXiv:1312.5943 [math.NT], 2013-2015; Elem. Math., 70 (2015), 117-124. doi: 10.4171/EM/284
• Greg Frederickson, Casting Light on Cube Dissections, Math. Mag., 82 (2009), 323-331.
• Index entries for linear recurrences with constant coefficients, signature (2,-1).

Crossrefs

Cf. A059255, A059270, A222716, A230718.

Programs

Mathematica

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 *)

PARI

Vec(x*(27*x^3-50*x^2+19*x+3)/(x-1)^2 + O(x^100)) \\ Colin Barker, Apr 23 2014

Formula

a(0) = A059270(0) = A059255(0).

a(1) = A059270(1) = A230718(1).

a(2) = A059255(2) = A230718(2).

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