A259060 Numbers that are representable in at least two ways as sums of four distinct nonvanishing cubes.

6426, 7900, 9614, 11592, 13858, 16436, 19350, 22624, 26282, 30348, 34846, 39800, 45234, 51172, 57638, 64656, 72250, 80444, 89262, 98728, 108866, 119700, 131254, 143552, 156618, 170476, 185150, 200664, 217042, 234308, 252486, 271600, 291674

Offset: 0

Author

Wolfdieter Lang, Aug 12 2015

Keywords

• nonn
• easy

Comments

This is the second part of Exercise 229 in Sierpiński's problem book. See p. 20, and p. 110 for the solution. He uses the identity (n-8)^3 + (n-1)^3 + (n+1)^3 + (n+8)^3 = 4*n^3 + 390 = (n-7)^3 + (n-4)^3 + (n+4)^3 + (n+7)^3, for n >= 9.

Here n is replaced by n + 9: (n+1)^3 + (n+8)^3 + (n+10)^3 + (n+17)^3 = 4*n^3 + 108*n^2 + 1362*n + 6426 = (n+2)^3 + (n+5)^3 + (n+13)^3 + (n+16)^3, for n >= 0.

There may be other numbers with this properties.

Because the summands have no common factor > 1 each of these two representations is called primitive. - Wolfdieter Lang, Aug 20 2015

Examples

			a(0) = 6426 = 1^3 + 8^3 + 10^3 + 17^3 = 2^3 + 5^3 + 13^3 + 16^3.
a(1) = 7900 = 2^3 + 9^3 + 11^3 + 18^3 = 3^3 + 6^3 + 14^3 + 17^3.
		

References

• W. Sierpiński, 250 Problems in Elementary Number Theory, American Elsevier Publ. Comp., New York, PWN-Polish Scientific Publishers, Warszawa, 1970.

Links

• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Crossrefs

Cf. A261241, A259058 (squares).

Programs

Magma

[(2*(n+9))*(2*n^2+36*n+357): n in [0..50]]; // Vincenzo Librandi, Aug 13 2015

Magma

I:=[6426,7900,9614,11592]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..50]]; // Vincenzo Librandi, Aug 13 2015

Mathematica

CoefficientList[Series[2 (3213 - 8902 x + 8285 x^2 - 2584 x^3)/(1 - x)^4, {x, 0, 50}], x] (* Vincenzo Librandi, Aug 13 2015 *)
LinearRecurrence[{4,-6,4,-1},{6426,7900,9614,11592},40] (* Harvey P. Dale, Sep 30 2016 *)

Formula

a(n) = (2*(n+9))*(2*n^2+36*n+357) = 2*A261241(n), n >= 0. See the comment for the sum of four distinct cubes in two different ways.

O.g.f.: 2*(3213 - 8902*x + 8285*x^2 - 2584*x^3) / (1-x)^4.

a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4). - Vincenzo Librandi, Aug 13 2015