%I #28 Sep 08 2022 08:46:13

%S 6426,7900,9614,11592,13858,16436,19350,22624,26282,30348,34846,39800,

%T 45234,51172,57638,64656,72250,80444,89262,98728,108866,119700,131254,

%U 143552,156618,170476,185150,200664,217042,234308,252486,271600,291674

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

%C 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.

%C 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.

%C There may be other numbers with this properties.

%C Because the summands have no common factor > 1 each of these two representations is called primitive. - _Wolfdieter Lang_, Aug 20 2015

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

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).

%F 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.

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

%F a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4). - _Vincenzo Librandi_, Aug 13 2015

%e a(0) = 6426 = 1^3 + 8^3 + 10^3 + 17^3 = 2^3 + 5^3 + 13^3 + 16^3.

%e a(1) = 7900 = 2^3 + 9^3 + 11^3 + 18^3 = 3^3 + 6^3 + 14^3 + 17^3.

%t CoefficientList[Series[2 (3213 - 8902 x + 8285 x^2 - 2584 x^3)/(1 - x)^4, {x, 0, 50}], x] (* _Vincenzo Librandi_, Aug 13 2015 *)

%t LinearRecurrence[{4,-6,4,-1},{6426,7900,9614,11592},40] (* _Harvey P. Dale_, Sep 30 2016 *)

%o (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

%Y Cf. A261241, A259058 (squares).

%K nonn,easy

%O 0,1

%A _Wolfdieter Lang_, Aug 12 2015