OFFSET
1,1
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}.
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.
EXAMPLE
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,...}.
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}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Apr 07 2016
STATUS
approved