OFFSET
0,1
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
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).
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
EXAMPLE
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.
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 *)
PROG
(Magma) [(2*(n+9))*(2*n^2+36*n+357): n in [0..50]] /* or */ 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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Aug 12 2015
STATUS
approved