A267826 Numbers not of the form w^3 + 2*x^3 + 3*y^3 + 4*z^3, where w, x, y and z are nonnegative integers.

18, 22, 39, 60, 63, 74, 76, 77, 100, 103, 106, 107, 117, 126, 178, 180, 201, 215, 228, 230, 245, 271, 289, 291, 295, 315, 341, 356, 357, 393, 413, 419, 420, 480, 481, 523, 559, 606, 616, 671, 673, 705, 854, 855, 963, 980, 981, 998, 1103, 1121, 1130, 1298, 1484, 1510, 1643, 1729, 1849, 1916, 1934, 1946

Offset: 1

Author

Zhi-Wei Sun, Apr 07 2016

Keywords

• nonn

Comments

Conjecture: The sequence has exactly 122 terms the last of which is a(122) = 41405.

We have verified that there are no terms between 41406 and 2*10^5.

The conjecture implies that {P(v)+w^3+2*x^3+3*y^3+4*z^3: w,x,y,z = 0,1,2,...} = {0,1,2,...} whenever P(v) is among the polynomials a*v^3 (a = 1,5,6,7,9,10,12,15,18), b*v^4 (b = 1,2,3,5,6,12,18), c*v^5 (c = 1,2,5,12) and d*v^k (d = 5,12; k = 6,7). Moreover, it also implies that {8*t+w^3+2*x^3+3*y^3+4*z^3: t = 0,1; w,x,y,z = 0,1,2,...} = {0,1,2,...}. If a,b,c,d and m are positive integers with {m*t+a*w^3+b*x^3+c*y^3+d*z^3: t = 0,1; w,x,y,z = 0,1,2,...} = {0,1,2,...}, then we must have m = 8 and {a,b,c,d} = {1,2,3,4}.

Examples

			a(1) = 18 since it is the first nonnegative integer not in the set {w^3 + 2*x^3 + 3*y^3 + 4*z^3: w,x,y,z = 0,1,2,...}.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..122
• Zhi-Wei Sun, Universal sums u^3+a*v^3+b*x^3+c*y^3+d*z^3 with u, v, x, y, z nonnegative integers, a message to Number Theory Mailing List, April 3, 2016.

Crossrefs

Cf. A000578, A002804, A022566, A267861, A271099, A271169, A271237.

Programs

Mathematica

CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)]
n=0;Do[Do[If[CQ[m-4*z^3-3y^3-2x^3],Goto[aa]],{z,0,(m/4)^(1/3)},{y,0,((m-4z^3)/3)^(1/3)},{x,0,((m-4z^3-3y^3)/2)^(1/3)}];n=n+1;Print[n," ",m];Label[aa];Continue,{m,0,1946}]